foundations of euclidean constructive geometry - citeseerx

FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY

MICHAEL BEESON

Abstract. Euclidean geometry, as presented by Euclid, consists of straightedge-and-

compass constructions and rigorous reasoning about the results of those constructions.

A consideration of the relation of the Euclidean “constructions” to “constructive mathe-

matics” leads to the development of a first-order theory ECG of “Euclidean Constructive

Geometry”, which can serve as an axiomatization of Euclid rather close in spirit to the

Elements of Euclid. Using Gentzen’s cut-elimination theorem, we show that when ECG

proves an existential theorem, then the things proved to exist can be constructed by Eu-

clidean ruler-and-compass constructions. In the second part of the paper we take up the

formal relationships between three versions of Euclid’s parallel postulate: Euclid’s own

formulation in his Postulate 5, Playfair’s 1795 version, which is the one usually used in

modern axiomatizations, and the version used in ECG. We completely settle the questions

about which versions imply which others using only constructive logic: ECG’s version im-

plies Euclid 5, which implies Playfair, and none of the reverse implications are provable.

The proofs use Kripke models based on carefully constructed rings of real-valued functions.

“Points” in these models are real-valued functions. We also characterize these theories in

terms of different constructive versions of the axioms for Euclidean fields.1

Contents

1. Introduction 51.1. Euclid 51.2. The collapsible vs. the rigid compass 71.3. Postulates vs. axioms in Euclid 91.4. The parallel postulate 91.5. Polygons in Euclid 102. Euclid’s reasoning considered constructively 102.1. Order on a line from the constructive viewpoint 102.2. Logical form of Euclid’s propositions and proofs 142.3. Case splits and reductio inessential in Euclid 153. Constructions in geometry 163.1. The 48 Euclidean constructions 163.2. The elementary constructions 163.3. Circles and lines eliminable when contructing points 193.4. Describing constructions by terms 204. Models of the elementary constructions 214.1. The standard plane 214.2. The recursive model 224.3. The algebraic model 234.4. The Tarski model 23

1

2 MICHAEL BEESON

5. The geometrization of arithmetic without case distinctions 235.1. Uniform perpendicular and projection 245.2. Rotation 275.3. Addition 285.4. Multiplication and division 285.5. Square roots 315.6. Formalizing the geometrization of arithmetic 326. The arithmetization of geometry 326.1. Euclidean fields in constructive mathematics 326.2. Line-circle and circle-circle continuity over Euclidean fields 347. Other axiomatizations of geometry 367.1. Pasch 377.2. Pieri and Peano 377.3. Hilbert 377.4. Tarski and his students 387.5. Tarski’s theory of Euclidean geometry 407.6. Borsuk-Szmielew 407.7. Avigad, Dean, and Mumma 407.8. Heyting and other constructive approaches 418. The theory ECG of Euclidean Constructive Geometry 418.1. Logic of Partial Terms (LPT) 418.2. Replacing LPT and sorts with predicates if desired 438.3. Language 448.4. Intuitionistic logic and stability 458.5. Incidence and intersection axioms 468.6. Constructor and accessor axioms 478.7. Meaning of equality 488.8. Betweenness axioms 488.9. Sides of a line 518.10. Right and Left : handedness and sides of lines 528.11. Dimension axioms 548.12. Rays and segments 548.13. Congruence axioms 558.14. Line-circle continuity 568.15. Intersections of circles 589. Development of neutral geometry in ECG 619.1. Why we cannot import negative theorems from Tarski’s theory 629.2. Congruence and betweenness lemmas 649.3. Ordering of segments 689.4. Angles and triangles 709.5. Perpendicular lines 749.6. Existence of midpoints and perpendiculars 759.7. Right angles 769.8. Various forms of the Pasch axiom 829.9. Inner Pasch implies Outer Pasch and plane separation 849.10. The crossbar theorem 859.11. Ordering of angles 86

FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 3

9.12. Triangle construction and uniqueness 879.13. Euclid Books I, II, and III 929.14. Consequences of the upper dimension axiom 949.15. Right and left turns and sides of lines 999.16. Definability of IntersectCirclesSame and IntersectCirclesOpp 1019.17. Rotation and Uniform Reflection 1049.18. The other intersection point 1109.19. Defining the order of points on a line 11110. The parallel postulate 11310.1. Alternate Interior Angles 11310.2. Euclid’s parallel postulate 11510.3. The strong parallel postulate 11610.4. Playfair’s axiom 11810.5. Another version of the strong parallel postulate 12010.6. Coordinates 12110.7. The strong parallel postulate and stability of IntersectLines 12310.8. Stability of definedness and ∼= 12510.9. ECG proves Euclid’s parallel postulate 12910.10. Various forms of the parallel axiom 13111. Comparisons of ECG to other formalizations 13611.1. Comparison of ECG to Hilbert-style theories 13611.2. Comparison of ECG to Tarski’s theories 13612. Connections between geometry and Euclidean fields 13812.1. Signed addition 13912.2. Signed multiplication 14412.3. The distributive law 14912.4. Reciprocals 15012.5. Interpreting ECG in field theory 15112.6. Area in Euclid, Hilbert, and ECG 17412.7. Handedness, cross product, and linear transformations 18112.8. Interpreting field theory in ECG 19212.9. Faithfulness of the interpretations 20113. Classical geometry and constructive geometry compared 21913.1. The double negation interpretation 21913.2. The double-negation interpretation applied to ECG 22114. The relation of ECG to Tarski’s theories of geometry 22214.1. Interpretation of Tarksi’s theory in ECG 22515. Interpretation of ECG in Hilbert’s and Tarski’s theories 22715.1. Right and left handedness 22715.2. Apartness 22816. Metatheorems 23016.1. Things proved to exist in ECG can be constructed 23016.2. Disjunction properties 23217. Independence results for the parallel axioms 23317.1. Kripke models of ring theory 23317.2. Euclid 5 does not imply Axiom 58 23517.3. Playfair does not imply Euclid 5 239

4 MICHAEL BEESON

17.4. Independence of Markov’s principle 24018. Apartness 24018.1. Constructions and Apartness 24018.2. Euclidean fields and apartness 24018.3. Apartness and the parallel axioms 24019. Appendix A: Roads not taken 24219.1. Rigid compass undefinable from the collapsible compass 24219.2. Rigid compass not definable, even with degenerate circles 24419.3. Circles of zero radius, the rigid compass, and projection 24619.4. Constructing the center of a circle 24720. Appendix B: Possible reductions in the primitive construction

mechanisms 24820.1. The Mohr-Mascheroni theorem 24820.2. Strommer’s theorem 248


§1. Introduction.

1.1. Euclid. Euclid’s geometry, written down about 300 BCE, has been ex-traordinarily influential in the development of mathematics, and prior to thetwentieth century was regarded as a paradigmatic example of pure reasoning.1

During those 2300 years, most people thought that Euclid’s theory was aboutsomething. What was it about? Some may have answered that it was aboutpoints, lines, and planes, and their relationships. Others may have said thatit was about methods for constructing points, lines, and planes with certainspecified relationships to given points, lines, and planes, for example, construct-ing an equilateral triangle with a given side. In these two answers, we see theviewpoints of pure (classical) mathematics and of algorithmic mathematics rep-resented. Near the end of the nineteenth century, the Italian “Peano school”,and especially Mario Pieri, began to use the “hypothetical-deductive method”,today called the “axiomatic method.” At the same time, and apparently withoutcommunication, Hilbert took a similar approach, and published a famous book[15] in the last year of the century. According to the axiomatic method, Euclid’stheories were not about anything at all. Instead of “points, lines, and planes”,one should be able to read “tables, chairs, and beer mugs.” All the reasoningshould still be valid. The names of the “entities” were just place holders. Thatwas the viewpoint of twentieth-century axiomatics.

In this paper, we re-examine Euclidean geometry from the viewpoint of con-structive mathematics. The phrase “constructive geometry” suggests, on the onehand, that “constructive” refers to geometrical constructions with straightedgeand compass. On the other hand, the word “constructive” may suggest the useof intuitionistic logic. We investigate the connections between these two mean-ings of the word. Our method is to keep the focus on the body of mathematicsin Euclid’s Elements, and to examine what in Euclid is constructive, in the senseof “constructive mathematics”. Our aim in the first phase of this research wasto formulate a suitable formal theory that would be faithful to both the ideas ofEuclid and the constructive approach of Errett Bishop. We did achieve this aim,and the resulting theory ECG of “Euclidean constructive geometry” is presentedin this paper.

In constructive mathematics, if one proves something exists, one has to showhow to construct it. In Euclid’s geometry, the means of construction are notarbitrary computer programs, but ruler and compass. Therefore it is natural tolook for quantifier-free axioms, with function symbols for the basic ruler-and-compass constructions. The terms of such a theory correspond to ruler-and-compass constructions.

In number theory, if one proves an existence theorem, then for a constructiveversion, one has to show how to compute the desired number as a function ofthe parameters. In analysis, if one proves an existence theorem, one has to beable to compute approximations to the desired number from approximations to

1Readers interested in the historical context of Euclid are recommended to read [6], whereMax Dehn puts forward the hypothesis that Euclid’s rigor was a reaction to the first “founda-tional crisis”, the Pythagorean discovery of the irrationality of

√2.

6 MICHAEL BEESON

the parameters. In particular, the solution will depend continuously on parame-ters, at least locally. This feature of constructive analysis depends, in a way, onwhat we think it means “to be given” a number x. Whatever that may mean,it surely means that we have a way to get a rational approximation to x withinany specified limit of accuracy.2 Geometry is more like analysis than numbertheory, in the sense that we do not want to assume in advance that points canbe given to us all at once in a completely determined location; points are givenonly approximately, by dots on paper or a computer screen, or in Euclid’s case,by indentations in sand (the Greeks drew their diagrams in sand). It might bedoubtful whether two such points coincide; in such a case one would have toask the one who made the diagram to refine it. It follows that in constructivegeometry, we should have local continuous dependence of constructions on pa-rameters. We can see that dramatically in computer animations of Euclideanconstructions, in which one can select some of the original points and drag them,and the entire construction “follows along.”

Thus our theory should (i) be quantifier free, so as to allow terms for con-structions, and (ii) allow only continuous constructions. There is a third consid-eration: the axiomatization should be disjunction-free. There are two reasonsfor this: First, disjunction does not occur in Euclid. Euclid did not use first-order logic; the logical structure of his theorems will be discussed below, but fornow we just note that he never states or uses a disjunctive proposition. Second,disjunctive axioms would make trouble for our desired applications of Gentzen’scut-elimination theorem. We want to have a theory in which things proved toexist can be constructed by ruler and compass, and that will be straightforwardwith a quantifier-free, disjunction-free axiomatization. That observation is thekey: we need a theory that

• is faithful to Euclid’s Elements.• has terms for geometrical constructions, and those terms describe construc-

tions continuous in parameters.• is quantifier-free.• is disjunction free.

We present here a theory ECG that has these properties. A version of thistheory was presented in [3]. The version presented is slightly different, butis equivalent.3 Regarding the “faithfulness” to Euclid: we discovered only oneserious non-constructivity in Euclid’s Elements, namely, Book I Prop. 2, in whichEuclid shows that a collapsible compass can be simulated by a rigid compass.This issue is discussed in detail below; here we say only that the remedy is toadopt a rigid compass as primitive; essentially, we take Euclid I.2 as an axiom

2There are philosophical issues here: can we “be given” a number by using a randomprocess to generate approximations? Or must the approximations be generated by a computerprogram? It doesn’t matter for the conclusion that functions are continuous, because it is atheorem that even if we assume numbers are generated only by programs, computable functionsare still continuous. This theorem (of Kreisel-Lacombe-Shoenfield-Tseitin) is discussed in [2],p. 61.

3It is not completely obvious what the axioms of such a theory should be; in Appendix Aand Appendix B we consider alternate choices and present results justifying the choices wemade.


instead of as a theorem.4 Although it was a shock to find that the secondproposition in Euclid is not constructive, after that there are no more shocks. Itwas also necessary to make a slight strengthening of Euclid’s parallel postulate,but the necessary revision is relatively minor, and these are the only two pointsin which ECG varies from Euclid.

The second phase of this research is to study the metamathematical propertiesof ECG and related theories. First we have the result alluded to above, thatexistential theorems are “realized” by ruler-and-compass constructions. Nextthere is Tarski’s “representation problem”. We want to characterize the mod-els of ECG. Classically, the models of Euclidean geometry are well-known tobe planes over Euclidean fields (ordered fields in which nonnegative elementshave square roots). But constructively, what is a Euclidean field? It turns outthat there are three natural ways of formulating that, and that planes over suchfields correspond to three different versions of the parallel postulate. The proofof such a representation theorem (classically or constructively) involves boththe “arithmetization of geometry” and the “geometrization of arithmetic.” Thegeometrization of arithmetic (originally due to Descartes) shows how to add,subtract, multiply, divide, and take square roots of segments, and how to intro-duce coordinates in an arbitrary plane. The arithmetization of geometry (alsodue to Descartes) is using coordinates and algebra to verify that the axioms ofgeometry hold in F 2, where F is a Euclidean field. Both of these offer someconstructive difficulties; for example, Descartes’s constructions apply only topositive segments, but we need to define arithmetic on signed numbers, withoutmaking case distinctions whether the numbers are zero or not. These difficultiesare overcome and the representation theorem proved in this paper.

The third phase of this research is to investigate the provability or indepen-dence of other geometrical theorems, relative to some constructive base theory.Our most interesting result is to settle the relations between the three versionsof the parallel postulate. As it turns out, no two are equivalent. We also obtaina few other independence results.

The readers of this paper may include logicians (who may or may not havestudied axiomatic geometry or intuitionistic logic); geometers (who may or maynot have studied logic or proof theory or constructive mathematics); and con-structive mathematicians (who may or may not have studied axiomatic geom-etry and may or may not have studied proof theory). The paper is written tobe accessible to this entire audience, so experts may find they can skip certainsections.

1.2. The collapsible vs. the rigid compass. A “rigid compass” can beused to transfer a given distance, say BC, from one location in the plane toanother, thus constructing at any point A another point D such that AD = BC.A “collapsible compass”, on the other hand, can only be used to draw a circlewith a given center A passing through another given point B. The compass“collapses” as soon as you pick it up. There is a word in Dutch, passer, for a

4Euclid I.2 with the assumption that all the points in the diagram are distinct is still atheorem. Maybe that was what Euclid intended; but we need to take as an axiom the versionwithout that assumption.

8 MICHAEL BEESON

rigid compass, which was used in navigation in the seventeenth century. Butthere seems to be no single word in English that distinguishes either of the twotypes of compass from the other.

Euclid’s compass is collapsible; but a rigid compass is important in geome-try, so right at the outset, in Book I, Proposition 2, Euclid attempted to showthat a rigid compass could be simulated by a collapsible compass. This fa-mous construction shows how to use the collapsible compass to construct a pointD = e(A,B,C) such that whenever B 6= C and A 6= B then AD = BC. Thereader is advised to take out his or her Euclid and review the simple, but non-trivial and beautiful, construction.5

Our interest in this subject began with a computer animation of Euclid’sconstructions that permits the user to drag the starting points, and see how theconstruction depends on the changed starting points. The results for Euclid I.2were surprising and interesting. The first problem with this construction is thatit does not work when A = C. Of course, in that case we can just take D = B;but that case distinction requires classical logic. And the computer animationreveals that when we drag point C close to A, and then around A in a smallcircle, then the constructed point D moves around A in a circle of radius close toBC. Hence D does not depend continuously on C. This discontinuity, togetherwith the need for a case distinction just mentioned, makes it clear that from theintuitionistic point of view, there are two different versions of Euclid I.2: theversion in which (in addition to B 6= C) we take A 6= C as a hypothesis, or the“uniform” version in which we do not assume A 6= C, but assert that D can beconstructed, whatever the values of A, B, and C, as long as B 6= C. The uniformversion corresponds to a rigid compass (with which we can copy the segment BCso it has one end at A).

From the constructive point of view, what Euclid I.2 shows is that a rigidcompass can be simulated by a collapsing compass plus a “decision function”,that enables us to decide wither two given points are equal or not. The latteris not a constructive principle, in the sense of intuitionistic logic. It is also nota constructive principle, in the sense of ruler and compass constructions, since,given two fixed points A and B (which are and remain unequal), and two morepoints P and Q (which may or may not be equal), we do not know how toconstruct a point C which is equal to P if A = B and to Q if A 6= B. Moreover,since such a construction could not be continuous in P and Q, it is not a matterof ignorance: no such construction exists.

The uniform version of Euclid I.2 seems to be important in developing geome-try; we use it in constructing the projection of a point on a line, which is in turnneeded for defining addition and multiplication of segments (the geometrizationof arithmetic). Therefore, from the constructive point of view, it is a defect thatthe rigid compass cannot be simulated by the collapsing compass. We repair thisby requiring a rigid compass in ECG. That amounts to including Euclid I.2 asa new axiom.

5Should any reader not possess a copy of Euclid, we recommend [10] and [9]; or for thosewishing a scholarly commentary as well, [8].


1.3. Postulates vs. axioms in Euclid. Euclid presents his readers withboth “postulates” and “axioms”. Modern mathematicians often treat thesewords as synonyms. For Euclid and his contemporaries, however, they had quitedifferent meanings. Here is the difference, as explained by Pambuccian [22], p.130.

For Proclus, who relates a view held by Geminus, a postulate prescribes

that we construct or provide some simple or easily grasped object for the

exhibition of a character, while an axiom asserts some inherent attribute

that is known at once to one’s auditors. And just as a problem differs from

a theorem, so a postulate differs from an axiom, even though both of them

are undemonstrated; the one is assumed because it is easy to construct, the

other accepted because it is easy to know. That is, postulates ask for the

production, the poesis of something not yet given . . . , whereas axioms refer

to the gnosis of a given, to insight into the validity of certain relationships

that hold between given notions.

1.4. The parallel postulate. Euclid’s famous “parallel postulate” (Postu-late 5, henceforth referred to as Euclid 5) states that if two lines L and M aretraversed by another line T , forming adjacent interior angles on one side of Tadding up to less than two right angles, then L and M will intersect on that sideof T . Stated this way, the postulate can be viewed as a construction method forproducing certain triangles. In view of the remarks of Geminus and Proclus, itseems likely that Euclid viewd his postulate in this way, or he would have calledit an axiom.

In 1795, Playfair introduced the version that is usually used today, which isan axiom rather than a postulate: Given a line L, and a point P not on L, thereexists exactly one parallel to L through P . (Parallel lines are by definition linesthat do not meet.) Written this way, the parallel postulate does not directlyassert the existence of any specific point.

It is not immediately clear whether this version implies Euclid 5 using onlyintuitionistic (constructive) logic, although of course it does in classical logic.This question is resolved in this paper. The answer is in the negative: Playfairdoes not imply Euclid 5.

A more subtle distinction arose in the formulation of ECG. There the followingversion of the parallel postulate is used: if K is a line parallel to line L throughpoint P , and M is another line through P (different from K) then M meets P .This is Axiom 58 of ECG. We prove that Axiom 58 implies Euclid 5, and thatthe implication cannot be reversed (relative to the other axioms of ECG).

The difference between Axiom 58 and Euclid 5, but is crucial for the abilityof ECG to define (signed) arithmetic operations in a continuous way. We showbelow that the theory of Euclidean fields is related to geometry, just as in classicalmathematics, by coordinatization. The difference between Axiom 58 and Euclid5 corresponds, in the theory of ordered fields, to whether 1/x has to exist whenx 6= 0, or only when x > 0. If it exists for x > 0, it exists for x < 0; so thequestion can also be put, whether 1/x has to exist when x 6= 0, or only whenx > 0 ∨ x < 0, which is constructively not the same as x 6= 0. (Playfair’saxiom corresponds to requiring that elements without reciprocals are zero.) Indifferent sections below, we establish these results that characterize the models

10 MICHAEL BEESON

of ECG, and show that the implications between versions of the parallel axioms(or Euclidean field axioms) cannot be reversed.

1.5. Polygons in Euclid. Euclid gives 48 two-dimensional constructions (wewill ignore his three-dimensional constructions entirely). These mostly deal withconfigurations of a fixed number of points; for example, he constructs a regularpentagon, hexagon, and octagon, but does not address the general n-gon (whichof course is now known to not be constructible with ruler and compass anyway).Some small fixed number of sides would suffice for the Euclidean constructions.In Euclid, no figure with more sides than an octagon is constructed, and nofigure with more than four sides is an input to another construction, except forconstructions that work on any “figure.” But, some of the later constructions douse the word “figure”, which apparently means something like what a modernmathematician would call a “closed polygon” (and has no connection to “figure”in the sense of “ilustrative drawing”.) The general concept of a closed polygonof any number of sides may be logically problematic as it drags the concept ofinteger into geometry. We regard such theorems of Euclid formally as theoremschemata, which become theorems for each particular number n of sides of the“figure.” In this we join everyone who previously considered first-order geometrictheories.

§2. Euclid’s reasoning considered constructively. In the late twentiethcentury, contemporaneously with the flowering of computer science, there wasa new surge of vigor in algorithmic, or constructive, mathematics, beginningwith Bishop’s book [4]. In algorithmic mathematics, one tries to reduce every“existence theorem” to an assertion that a certain algorithm has a certain result.It was discovered by Brouwer that if one restricts the laws of logic suitably (to“intuitionistic” logic), then one only obtains algorithmic existence theorems, sothere is a fundamental connection between methods of proof, and the existenceof algorithms to construct the things that have been proved to exist. Brouwerthought it necessary to do more than just restrict logic; he also wanted to statesome additional principles. Bishop renounced additional principles and workedby choosing his definitions very carefully, but using only a restricted form oflogic. Results obtained this way are classically valid as well as constructively.

What happens if we examine Euclid’s Elements from this point of view? Some-what surprisingly, we only had to make two modifications to Euclid, both ofwhich we have already mentioned. Namely, we need to adopt the uniform versionof Euclid I.2 as an axiom, since Euclid’s proof is valid only for the non-uniformversion; and we had to strengthen Euclid 5 (the parallel postulate) as discussedabove. In this section, we justify this conclusion in more detail.

2.1. Order on a line from the constructive viewpoint. In this sectionwe explain how a constructivist views the relations x < y and x ≤ y on the realline. This section can be skipped by readers familiar with constructive order,but will be a necessary prerequisite to understanding constructive geometry.The theory of order is relevant to geometry, because it translates directly intothe betweenness relation in geometry, since for positive x and y we have thatx < y is the same as “x is between 0 and y”. The real numbers are given by


constructive sequences of rational approximations. For example, we could takeBishop’s definition of a recursive real as a (constructively given) sequence xn ofrationals such that for every n ≥ 1,

|xn − xm| < 1

n+

1

m.

This guarantees that the limit x of such a sequence satisfies |x−xn| ≤ 1/n. Twosuch sequences are considered equal if xn−yn ≤ 2/n; it is important that differentsequences can represent equal (or “the same”, if you prefer) real numbers. Nowobserve that if x < y, we will eventually become aware of that fact by computingfine enough approximations to x and y, so that xn + 1/n < yn − 1/n. Then wewill have an explicit positive lower bound on y − x, which is what is required toassert x < y.

The “recursive reals” are defined this way, but specifying that the sequences xn

are to be computed by a Turing machine (or other formal definition of “computerprogram”). In more detail: We write e(n) for the result, if any, of the e-thTuring machine at input n. Rational numbers are coded as certain integers,and modulo this coding we can speak of recursive functions from N to Q. A“recursive real number” x is the index of a Turing machine e, whose outpute(n) at intput n codes a rational number, such that |e(n) − x| ≤ 1/n foreach n ∈ N. The real number to which the approximations x(n) converge issometimes also called a “recursive real number”, but we call it the “value of x”.

It may be helpful to keep the “model” of recursive reals in mind, but in thespirit of “tables, chairs, and beer mugs”, modern constructivists often prefer notto make this commitment, but refer in the abstract “constructive sequences.”Thus the “classical model” is always a possible interpretation.

If x = y, we may compute for many years and still be uncertain whetherx = y or not. Hence, we are not allowed to constructively assert the trichotomylaw x < y ∨ x = y ∨ y < x, since we have no way to make the decision ina finite amount of time. One can prove (and we shall prove below) that therecursive reals definitely do not satisfy the trichotomy law, as that would implya computable solution to the halting problem. But without a commitment toa definite definition of “constructive sequence”, the most we can say is that“we cannot assert” trichotomy. This phrase “we cannot assert” in constructivemathematics is usually code for, “it fails in the recursive model.”

Note, however, that we may be able to assert the trichotomy law for varioussubfields of the real numbers. For example, it is valid for the rational numbers,and it is also valid (though less obviously) for the real algebraic numbers. Ineach of these cases, the elements of the field are given by finite objects, thatcan be presented to us “all at once”, unlike real numbers; but that propertyis not sufficient for trichotomy to hold, since the recursive reals fail to satisytrichotomy, but a recursive real can be given “all at once” by handing over aTuring machine to compute the sequence xn of approximations.

The relation x ≤ y is equivalent to ¬(y < x), either by definition, or by asimple theorem if one defines x ≤ y in terms of approximating sequences. It isdefinitely not equivalent to x < y ∨ x = y (see the refutation of the trichotomylaw in the recursive model given below). But now consider negating x ≤ y.

12 MICHAEL BEESON

Could we assert ¬x ≤ y implies y < x? Subtracting y we arrive at an equivalentversion of the question with only one variable: can we assert ¬x ≤ 0 implies0 < x? Since x ≤ 0 is equivalent to ¬0 < x, the question is whether we canassert

¬¬x > 0 → x > 0 (Markov’s principle)

In other words, is it legal to prove that a number is positive by contradiction?One could argue for this principle as follows: Suppose ¬¬x > 0. Now computethe approximations xn one by one for n = 1, 2, . . . . Note that trichotomy doeshold for xn and 1/n, both of which are rational. You must find an n suchthat xn > 1/n, since otherwise for all n, we have xn ≤ 1/n, which means x ≤ 0,contradicting ¬¬x > 0. Well, this is a circular argument: we have used Markov’sprinciple in the justification of Markov’s principle.

Shall we settle it by looking at the recursive model? There it can easily beshown to boil down to this: if a Turing machine cannot fail to halt, then it halts.Again one sees no way to prove this, and some may feel is intuitively true, whileothers may not.

Historically, the Russian constructivist school adopted Markov’s principle, andthe Western constructivists did not. It reminds one of the split between thebranches of the Catholic Church, which also took place along geographical lines.In any case, as we will see below, it seems appropriate to adopt this principle fora constructive treatment of Euclid.

From the constructive viewpoint, the main difference between x < y andx ≤ y is that x < y involves an existential quantifier; it contains the assertionthat we can find a rational lower bound on |y − x|, while x ≤ y is definedwith a universal quantifier, and contains no hidden assertions. If we take anyformula involving inequalities, and replace x < y with ¬y ≤ x, we obtain aclassically equivalent assertion no longer containing an existential quantifier. Ifin addition we replace A∨B by ¬(¬A∧¬B), we will have eliminated all hiddenclaims, and the result will be classically valid if and only if it is constructivelyvalid. To understand constructive mathematics, one has to learn to see the“hidden claims” that are made by disjunctions and existential quantifiers, whichcan make a formula “stronger” than its classical interpretation. Of course, ifexistential quantifiers or disjunctions occur in the hypotheses, then a strongerhypothesis can make a weaker theorem.

A given classical theorem might have more than one (even many) classicallyequivalent versions with different constructive meanings. Therefore, finding aconstructive version of a given theory is often a matter of choosing the rightdefinitions and axioms.

Consider the following proposition, which is weaker than trichotomy: x ≤0 ∨ x ≥ 0. This is also not constructively valid. Intuitively, no matter how longwe keep computing approximations xn, if they keep coming out zero we willnever know which disjunct is correct. As soon as we stop computing, the verynext term might have told us.

We now show that both trichotomy and x ≤ 0 ∨ x ≥ 0 fail in the recusivereals, if disjunction is interpreted as computable decidability. First considertrichotomy. We will show that there is no computable test-for-equality function,


that is, no computable function D that operates on two Turing machine indicesx and y, and produces 0 when x and y are equal recursive real numbers (i.e.have the same limiting value), and 1 when they are recursive real numbers withdifferent values. Proof, if we had such a D, we could solve the halting problemby applying D to the point (E(x), 0), where E(x)(n) = 1/n if Turing machinex does not halt at input x in fewer than n steps, and E(x)(n) = 1/k otherwise,where x halts in exactly k steps. Namely, x(x) halts if and only if the valueof E(x) is not zero, if and only if D(Z,E(x)) 6= 0, where Z is an index of theconstant function whose value is the (number coding the) rational number zero.

Now consider the proposition x ≥ 0 ∨ x ≤ 0. We can imitate the aboveconstruction, but replacing the halting problem by two recursively inseparabler.e. sets A and B, and making the number E(x) be equal to 1/n if at the n-thstage of computation we see that x ∈ A, and −1/n if we see x ∈ B; so if x is inneither A nor B, E(x) will be equal to zero. Hence x ≥ 0 ∨ x ≤ 0 fails to holdin the recursive reals.

Finally we consider this proposition:

x 6= 0 → x < 0 ∨ x > 0 (two-sides)

We call this principle “two-sides” since it is closely related to “a point ‘ is onone side or the other of the line”. (Here the “line” could be the y-axis.) Two-sides does hold in the recursive model, since, assuming that x 6= 0, if we computexn for large enough n, eventually we will find that xn +1/n < 0 or xn−1/n > 0,as that is what it means in the recursive model for x to be nonzero. But since xn

and 1/n are rational numbers, we can decide computably which of the disjunctsholds, and that tells us whether x < 0 or x > 0.

On the other hand, this verification of two-sides in the recursive model is nota proof that it is constructively valid; the following lemma shows that it is atleast as “questionable” as Markov’s principle:

Lemma 2.1. two-sides implies Markov’s principle (with intuitionistic logic).

Proof. Suppose ¬¬x > 0, the hypothesis of Markov’s principle. Then x 6= 0,so by two-sides x < 0 or x > 0; if x < 0 then ¬x > 0, contradicting ¬¬x > 0;hence the disjunct x < 0 is impossible. Hence x > 0, which is the conclusion ofMarkov’s principle. That completes the proof of the lemma.

If we assume that points on a line correspond to Cauchy sequences of rationalnumbers, then we can also prove the converse:

Lemma 2.2. If real numbers are determined by Cauchy sequences, then Markov’sprinciple implies two-sides.

Proof. We may suppose that real numbers are given by “Bishop sequences” asdescribed above, as it is well-known to be equivalent to the Cauchy sequencedefinition. A Bishop sequence for |x| is given by |x|n = max (xn,−xn). Supposex 6= 0 (the hypothesis of two-sides). Then |x| 6= 0. We claim |x| > 0. ByMarkov’s principle, it suffices to derive a contradiction from |x| ≤ 0. Suppose|x| ≤ 0. Then x = 0, contradicting |x| 6= 0. Hence by Markov’s principle, |x| > 0.Then by definition of >, for some n we have max (xn,−xn) > 1/n. But xn is arational number, so xn > 1/n or −xn > 1/n. In the former case we have x > 0;

14 MICHAEL BEESON

in the latter case x < 0. But this is the conclusion of two-sides. That completesthe proof.

Since Markov’s principle is known to be unprovable in the standard intuition-istic formal theories of arithmetic and arithmetic of finite types (see [31], pp.213 ff.), two-sides is also not provable in these theories. However, in the contextof geometry we do not assume that points are always given to us as Cauchysequences of rationals; we do not even assume that we can always construct aCauchy sequence of rationals corresponding to a given point. (There are non-Archimedean models of elementary geometry, for example.) The lemma thereforedoes not imply that two-sides and Markov’s principle are equivalent in geometry;and indeed we shall see that it is not the case. Two-sides is not provably in ourgeometric theory, even though we adopt Markov’s principle as an axiom.

It is an open (philosophical) question whether our geometrical intuitions com-pel us to accept Markov’s principle, or whether they compel us, having donethat, to also accept two-sides. In this paper we take a pragmatic approach:geometry without Markov’s principle will be too complicated, but it is possiblewithout undue complexity to consider two-sides as an added principle, which canbe accepted or not, and we choose to not accept it, since (as we show here) it isnot necessary to prove the theorems in Euclid or of the type found in Euclide.

2.2. Logical form of Euclid’s propositions and proofs. One should re-member that Euclid did not work in first-order logic. This is not because, likeHilbert, he used set-theoretical concepts that go beyond first order. It is insteadbecause he does not use any nested quantifiers or even arbitrary Boolean combi-nations of formulas. All Euclid’s propositions have the form, given some pointsbearing certain relations to each other, we can construct one or more additionalpoints bearing certain relations to the original points and each other. A modernlogician would describe this by saying that Euclid’s theorems have the form, aconjunction of literals implies another conjunction of literals, where a literal isan atomic formula or the negation of an atomic formula. One does not even findnegation explicitly in Euclid; it is hidden in the hypothesis that two points aredistinct. Often even this wording is not present, but is left implicit.

In particular there is no disjunction to be found in Euclid. In some of hispropositions, a complete proof would include an argument by cases, and Euclidhandles only one case. For example, in Prop. I.2, given B and C distinct, andgiven A, it is required to construct D with AD = BC. Euclid does not mentionthe case when A = B, presumably because it was obvious that in that case onecan take D = C, and similarly for the case A = C. Euclid treats only the caseA different from B and from C. Euclid was already criticized for this sloppinessabout case distinctions thousands of years ago. We will take up this issue in moredetail below, and show that Euclid’s sloppiness about case distinctions does notlead to any serious non-constructivities.

Euclid’s proofs have been analyzed in detail by Avigad et. al. in [1], and theyconclude:

Euclidean proofs do little more than introduce objects satisfying listsof atomic (or negation atomic) assertions, and then draw furtheratomic (or negation atomic) conclusions from these, in a simple linear


fashion. There are two minor departures from this pattern. Some-times a Euclidean proof involves a case split; for example, if ab andcd are unequal segments, then one is longer than the other, and onecan argue that a desired conclusion follows in either case. The otherexception is that Euclid sometimes uses a reductio; for example, if thesupposition that ab and cd are unequal yields a contradiction then onecan conclude that ab and cd are equal.

2.3. Case splits and reductio inessential in Euclid. It is our purpose inthis section to argue that Euclid’s reasoning can be supported in ECG, includ-ing the two types of apparently non-constructive reasoning just discussed. Thereason for this is that the apparent non-constructivities come down to the twobasic types mentioned just above, and these can be dealt with by assuming the“stability” of equality and betweenness. By the stability of equality, we mean

¬¬x = y → x = y.

This formula simply codifies the principle that it is legal to prove equality of twopoints by contradiction, and it is taken as an axiom in our theory. Similarly, ifB(a, b, c) means that b is between a and c on a line, we take as an axiom thestability of betweenness:

¬¬B(a, b, c) → B(a, b, c).

While Euclid never explicitly mentions betweenness (which is, as is well-known,the main flaw in Euclid), the stability of betweenness and equality togetheraccount for all apparent instances of nonconstructive arguments in Euclid.

A typical example of such an argument in Euclid is Prop. I.6, whose proofbegins

Let ABC be a triangle having the angle ABC equal to the angleACB. I say that the side AB is also equal to the side AC. For, if ABis unequal to AC, one of them is greater. Let AB be greater, . . .

To render this argument in first-order logic, we have to make sense of Euclid’s“common notion” of quantities being greater or less than other quantities. Thisis usually done formally using the betweenness relation. In the case at hand,we would lay off segment AB along ray AC, finding point D on ray AC withAD = AC. Then since AB is unequal to AC, D 6= C. Then “one of them isgreater” becomes the disjunction B(A,D,C) ∨ B(A,C,D). This disjunction isnot constructively valid. But its double negation is valid, since if the disjunctionwere false, we would have ¬B(A,D,C) and ¬B(A,C,D)), which would contra-dict D 6= C. The rest of Euclid’s argument shows that each of the disjunctsimplies the desired conclusion. We therefore conclude that the double negationof the desired conclusion is valid. The conclusion of I.6, however, is negative (hasno ∃ or ∨). Hence the double negation can be pushed inwards, and the stabilityof the atomic sentences applied to make the double negations disappear.

In Euclid, disjunctions never appear, even implicitly, in the conclusions ofpropositions.6 The conclusions are simply conjunctions of literals. Sometimes

6An apparent exception to this rule is Prop. I.13, “If a straight line set up on a straight linemake angles, it will make either two right angles or angles equal to two right angles.” Here

the disjunction is superfluous: we can just say, “it will make angles equal to two right angles.”

16 MICHAEL BEESON

there is an implicit existential quantifier, but if (as will always be the case in ourtheory) we can explicitly exhibit terms for the constructed points, the resultingexplicit form of the proposition will be quantifier-free. Then the double negationcan be pushed inwards as just illustrated. In this way, all arguments of theform beginning “For, if AB is unequal to AC, one of them is greater”, can beconstructivized.

Prop. I.26 gives an example of the use of the stability of equality: “. . .DEis not unequal to AB, and is therefore equal to it.” In the examples above,this principle is not really needed to reach Euclid’s desired conclusion. Sincethe conclusion concerns the equality of certain points, we can simply double-negate each step of the argument, and then add one application of the stabilityof equality at the end. In fact, this had to happen: we proved in [3], and reprovethem in this paper, explaining why the principle in question, and indeed, any usesof classical logic whatsoever, are in principle eliminable from proofs of theoremsof the form found in Euclid.

We note that if the conclusion of the proposition does not itself involve Euclid’s“greater than” or “less than”, then only the stability of equality is needed atthe end. Whether or not betweenness actually is needed in the conclusions (asopposed to the hypotheses) in Euclid, it certainly could occur in “the spirit ofEuclid”, so we include the stability of betweenness in our theory.

§3. Constructions in geometry.

3.1. The 48 Euclidean constructions. We are dealing only with the two-dimensional part of Euclid, although in some sense the culmination of Euclid isthe construction of the five Platonic solids. Thus we focus on Books I-VI. In thosefour books one finds a total of 172 propositions. Of those, 48 make a claim thatsomething can be constructed. We do not take the space to list the propositions,but with your Euclid at hand, it should suffice to list the proposition numbers:

Book I, propositions 1,2,3,9,10,11,12,22,23,31,42,44,45,46; Book II, proposi-tions 11, 14; Book III, propositions 1,2,17,25,30,33,34; Book IV, all 16 proposi-tions; Book VI, 9,10,11,12,18,25,28,29,30

These constructions are, so to speak, the “data” that our theory is to “ex-plain.”

3.2. The elementary constructions. The Euclidean constructions are car-ried out by constructing lines and circles and marking certain intersection pointsas newly constructed points. Our aim is to give an account of this process withmodern precision. We use a system of terms to denote the geometrical construc-tions. These terms can sometimes be “undefined”, e.g. if two lines are parallel,their intersection point is undefined. A model of such a theory can be regardedas a many-sorted algebra with partial functions representing the basic geomet-ric constructions. Specifically, the sorts are Point, Line, and Circle. We haveconstants and variables of each sort.

It is possible to define extensions of this theory with definitions of Arc andSegment. These are “conservative extensions”, which means that no additionaltheorems about lines and points are proved by reasoning about arcs and seg-ments. Angles are treated as triples of points. While this conservative extension


result applies to logical theories, a similar result applies to constructions con-sidered algebraically. If we replace rays and segments by lines, and angles bypairs of lines, then some terms may become defined that were not defined before,as lines may intersect where rays or segments did not, etc. But any point thatwas constructible with rays and angles will still be constructible when rays andangles are replaced by lines. Therefore it suffices to restrict attention to points,lines, and circles, which we do from now on.7

This collection of data types is adequate to cover the return types and argu-ment types of the 48 plane Euclidean constructions (given the convention aboutpolygons or “figures” discussed above).

Lines are constructed by drawing a line through two distinct points; theresulting line is Line (A,B). Circles are constructed “by center and radius”;Circle 3(A,B,C) is the circle with center A and radius BC. This term corre-sponds to a “rigid compass”. Its necessity for constructive geometry is alreadysuggested by the analysis of Euclid’s proof of Prop. I.2 above; and in AppendixA we give a proof that not only does Euclid’s simulation of the rigid compass bya collapsing compass fail to be everywhere defined, but any simulation will havethat problem: the rigid compass cannot be defined from the collapsing compass.Therefore the construction of circles from center and radius must be taken asprimitive.

We allow circles of radius zero, which we call “degenerate circles”, i.e. weconsider Circle 3(A,B,C) to be defined when B = C. Reasons for this are alsogiven in Appendix A. Starting with at least three noncollinear points, we candraw lines and circles using these constructors, and then construct more pointsusing the following six “elementary constructions”, each of which has return typePoint:

IntersectLines (LineK,LineL)

IntersectLineCircle1 (LineL,CircleC)

IntersectLineCircle2 (LineL,CircleC)

IntersectCircles1 (CircleC,CircleK)

IntersectCircles2 (CircleC,CircleK)

IntersectCircles1 and IntersectCircles2 construct the intersection points of twocircles; how the two points are distinguished will be explained below. Eucliddoes not use function symbols for the intersection points of two circles, and inparticular did not worry about how to distinguish one from the other, althoughsometimes (as in Euclid I.9, the angle bisector theorem8 ) his proofs need tobe repaired by selecting “the intersection point on the opposite side of the linejoining the centers as the given point x”, or “on the same side.” To directlysupport this kind of argument we might consider adding function symbols

IntersectCirclesSame (CircleC,CircleK,Point p)

IntersectCirclesOpp (CircleC,CircleK,Point p)

7Of course, diagrams will look better with rays and segments, so in a computer program todraw diagrams, we would want to (and did) use them; but this is a theoretical paper.

8See Heath’s commentary on I.9 in [8].

18 MICHAEL BEESON

These two constructions give the intersection point of C and K that is on thesame side (or the opposite side) of the line joining their centers as p, when p isnot on that line. It turns out that constructions meeting this specification canbe defined, and hence it is not necessary to include them as primitive symbols.However, the constructions that define them are somewhat elaborate, and in theinterest of having a system “faithful to Euclid”, in the sense that it is easy totranslate Euclid directly into the formal system, we include symbols for theseconstructions. They will be shown redundant in Theorem 9.96. Another way inwhich Euclid avoids the need for function symbols to identify the intersectionpoints is the phrase “the other intersection point.” We will show in Theorem9.109 that there is a construction to get “the other intersection point”, given twocircles and one intersection point.

In this picture of constructions, every line comes equipped with the two pointsthat were used to construct it. In hypothetical constructions, however, we maybe “given” some lines, and we need to have “accessor functions” to recover pointson those lines. Thus, pointOn1 (L) and pointOn2 (L) construct two points on aline; those will be a and b where L = Line (a, b).

Similarly, since circles can only be constructed from a given center, everycircle comes equipped with its center, so we should have an “accessor function”center(C) for the center of C, and pointOnCircle (C), to enable us to “construct”a point on a given circle.9 Think, for example, of always choosing the southernextremity of the circle.

Note that lines and circles are treated differently here: pointOnCircle (C) willbe the same for any two circles with the same points, but pointOn1 (L) willdepend on the points from which L was constructed. This will be discussedmore extensively below.

There are three issues to decide:

• when there are two intersection points, which one is denoted by which term?• In degenerate situations, such as Line (P, P ), or

IntersectCirclesSame (C,K, p) when p is on the line joining the centers,what do we do?

• When the indicated lines and/or circles do not intersect, what do we doabout the term(s) for their intersection point(s)?

We will answer these questions as follows: When the indicated lines and/or circlesdo not intersect, the term for their intersection points will be “undefined”. Whatthis means for the logic will be spelled out in due course. In other words, theoperations of these “algebras” do not have to be defined on all values of theirarguments. The same issue, of course, arises in many other algebraic contexts,for example, division is not defined when the denominator is zero, and

√x is not

defined (when doing real arithmetic) when x is negative.In degenerate situations such as Line (P, P ), we will be guided by continuity.

Thus Line (P, P ) will be undefined, since it cannot be defined in such a way asto make Line (P,Q) continuous as P approaches Q. Similarly,IntersectCirclesSame (C,K, p) and IntersectCirclesOpp (C,K, p) will be unde-fined when p is on the line joining the centers of C and K. On the other

9See Appendix A for a discussion of issues raised by Euclid III.1 in this connection.


hand, degenerate circles such as Circle (A,A) are convenient, and do no harmto continuity, so we allow them. (That decision is my “final answer” after muchvacillation; some reasons are spelled out in Appendix A.)

The question about intersection points is more complex. We note that linesare treated intensionally in our theory; that is, Line (A,B) is not equal toLine (B,A), even though the two lines are extensionally equal, i.e. the samepoints lie on them. That is, a line “comes equipped” with two points that wereused to construct it; to “be given” a line involves being given two points on thatline, and order matters. It thus makes sense to define the two intersection pointsof Line (A,B) and a circle to be such that their ordering on the line is the sameas that of A and B. While this definition works semantically in the standardplane, its axiomatic development is also possible, as we shall see.

Our original development in [3] handled circle-circle intersections using onlythe two operations that take only the two circles as arguments; but the othertwo operations, that takes an extra argument of type “point” to specify thatthe intersection point should be on the same side of the center-to-center line asthe specified point, seems more faithful to Euclid. We wish to consider all fouroperations, if only to clarify their relations. (It turns out that the second two canbe defined, so from the point of view of minimizing the number of symbols in thelanguage, they could be eliminated.) With the first two operations, we are facedwith the problem of distinguishing the two intersection points, when given onlythe circles. Unlike lines, circles do not “come with” a built-in direction. Considerintersecting circles with centers at A and B. Let the two intersection points beC and D. Then we note that one of ABC is a “right turn” and the other isa “left turn”, except in the degenerate case A = B (when the terms given byIntersectCircles1 and IntersectCircles2 will be undefined) and in the case whenthe circles are tangent, and in that case there is no need to distinguish the (iden-tical) intersection points. This observation enables us to define the semantics ofIntersectCircles1 and IntersectCircles2 , but leaves us with the problem of givingdefinitions of the notions Left (A,B,C) and Right (A,B,C), whose intended in-terpretations are that ABC is a left-hand turn or a right-hand turn, respectively.In [3], we gave a complicated solution of this problem, involving many axioms.In this paper, we give a simpler axiomatization.

3.3. Circles and lines eliminable when contructing points. It is intu-itively plausible that one does not really need to draw the lines and circles ina geometric diagram; one just uses the circles and lines to get their points ofintersection with other circles and lines; sometimes

Each of these has several “overloaded” variants, which can be defined fromthese using constructors and accessors. For example,

IntersectLines (PointA,PointB,PointC,PointD)

= IntersectLines (Line (A,B),Line (C,D))

IntersectLineCircle1 (PointA,PointB,PointC,PointD)

= IntersectLineCircle1 (Line (A,B),Circle (C,D))

IntersectLineCircle1 (PointA,PointB,CircleC)

= IntersectLineCircle1 (Line (A,B), C)

20 MICHAEL BEESON

As these three examples illustrate, One can regard circles and lines as mereintermediaries; points are ultimately constructed from other points. In a lemmaat the end of this section, we state and prove this principle precisely. On theother hand, the overloaded versions do not enable one to construct any morepoints, so for simplicity they are omitted from our axiomatic theories.

There is a second constructor for circles, which we can describe for shortas “circle from center and radius”, as opposed to the first constructor above,“circle from center and point.” Specifically Circle3 (A,B,C) constructs a circleof radius BC and center A, provided B 6= C. These two constructors for circlescorrespond to a “collapsible compass” and a “rigid compass” respectively. Thecompass of Euclid was a collapsible compass: you cannot use it to “hold” thelength BC while you move one point of the compass to A. You can only use it tohold the radius AB while one point of the compass is fixed at A, so in that senseit corresponds to Circle (A,B). The second constructor Circle3 corresponds toa rigid compass. The theory ECG includes Circle3 . That is because (as weshow in detail in Appendix A and in [3], without a rigid compass, one cannotproject a point P onto a line L, without making a case distinction between thecase when P lies on L and the case when it does not; and the ability to makesuch projections is crucial to defining a coordinate system and showing how toperform addition and multiplication on segments.

We note in the following lemma that, as far as constructing points goes, theother types are mere conveniences. The elementary constructions can be ex-pressed, as we have noted, in several ways using variables of different types.For example, we could have IntersectLines (K,L) where K and L have Line, orIntersectLines (A,B,C,D), where A, B, C, and D have type Point, and

IntersectLines (Line(A,B), Line(C,D)) = IntersectLines (A,B,C,D).

Lemma 3.1. Let t be a term of type Point, whose variables are all of typePoint. Then there is a term t∗ with the same variables as t, also of type Point,such that in the standard plane t and t∗ determine the same function, and t∗

contains only function symbols of type Point having Point arguments.

Proof. By induction on the complexity of t. Suppose that t has the formIntersectLines(r, s) where r and s are terms of type Line . None of the ele-mentary constructions has type Line, and r and s cannot be variables (since allvariables in t are of type Point), so r must have the form Line (p, q) for someterms p and q, and s must have the form Line (u, v) for some terms u and v.Then t∗ can be taken to be IntersectLines (p∗, q∗, u∗, v∗). The other elemen-tary constructions are treated similarly. The basis case, when t is a variable orconstant, is treated by taking t∗ = t.

3.4. Describing constructions by terms. It is customary to describe con-structions by a sequence of elementary construction steps. In this section, forthe benefit of readers not expert in logic, we show how terms in a logical systemcorrespond to traditional descriptions. For example, we might describe bisectinga segment this way:

Midpoint(A,B)

C = Circle(A,B); // center at A, passing through B


K = Circle(B,A);

P = IntersectCircles1(C,K);

Q = IntersectCircles2(C,K);

L = Line(P,Q); // the perpendicular bisector of AB

J = Line(A,B);

E = IntersectLines(L,J); // the desired midpoint

return E;

This description (which actually is in a precisely-defined language used for com-puter animation of constructions) looks quite similar to textbook descriptionsof constructions. We call such a description a “construction script” or justa “script.” For theoretical purposes, as in this paper, it is easier to describeconstructions officially by terms as above. When the midpoint construction isdescribed as a term, it looks like this:

IntersectLines (Line (IntersectCircles1 (Circle (A,B),Circle (B,A)),

IntersectCircles2 (Circle (A,B),Circle (B,A))),Line (A,B))

Conversely, given a term, one can “unpack” it, introducing new variablenames for subterms, and even, if desired, collapsing duplicate subterms (e.g.,Circle (A,B) need not be drawn twice). Since we have chosen readable namesfor our function symbols, such as IntersectLines instead of f , our terms some-times grow typographically cumbersome, and for human readability we will oftenuse scripts to describe constructions; but we do not define scripts formally; theyalways just stand for terms.

§4. Models of the elementary constructions. There are several interest-ing models, of which we now mention four. To define these models, we assumethere are three constants α, β, and γ of type Point.

4.1. The standard plane. The standard plane is R2, with the usual inter-pretation of points, lines, and planes, and three arbitrary points picked for α, β,and γ.

The standard plane is R2. Points, lines, and circles (as well as segments,arcs, triangles, squares, etc., in extended algebras in which such objects areconsidered) are interpreted as the objects that usually bear those names in theEuclidean plane. More formally, the interpretation of the type symbol “Point ” isthe set of points, the interpretation of “Line ” is the set of lines, etc. In particularwe must choose three specific non-collinear points to serve as the interpretationsof α, β, and γ. Let us choose α = (0, 1), β = (1, 0), and γ = (0, 0). Theconstructor and accessor functions listed above also have standard and obviousinterpretations. It is when we come to the five operations for intersecting linesand circles that we must be more specific. As discussed above, the interpre-tations of the five function symbols for the elementary constructions, such asIntersectLineCircle1 , are partial functions, so that if the required intersectionpoints do not exist, the term simply has no interpretation. In degenerate situ-ations, terms are defined if and only if they can be defined so as to make the

22 MICHAEL BEESON

interpreting function continuous; thus Line (P, P ) is undefined and Circle (P, P )is the zero-radius circle, and the only point on it is P .10

To distinguish the intersection points of two circles: IntersectCircles1 (C,K)is the intersection point P such that the angle from center (C) to center (K) toP makes a “left turn”. This is defined as in computer graphics, using the signof the cross product. Specifically, let A = center (C) and B = center (K). Thenthe sign of (A−B)× (P −B) determines whether angle ABP is a “left turn” ora “right turn”. Thus αβγ is a left turn and γβα is a right turn. In case the twointersection points are different, one of these cases must apply. This explanationhas used a case distinction as well as the cross product; later we will show how todefine “left turn” and “right turn” using only the axioms of Euclidean geometryand intuitionistic logic. For now we simply note that this notion is constructivelyappealing, because of continuity: there exists a unique continuous function of Cand K that satisfies the stated handedness condition for IntersectCircles1 whenC and K have two distinct intersection points, and is defined whenever C andK intersect (at all).

The principle of continuity leads us to make IntersectCircles1 (C,K) andIntersectCircles2 (C,K) undefined in the “degenerate situation” when circles Cand K coincide, i.e. have the same center and radius. Otherwise, as the centerof C passes through the center of K, there is a discontinuity. It makes sense,anyway, to have them undefined when C andK coincide, as the usual formulasfor computing them get zero denominators, and there is no natural way to selecttwo of the infinitely many intersection points.

IntersectCirclesSame (C,K, p) and IntersectCirclesOpp are undefined when pis on the line joining the centers of the circles; and we note that each of the twois continuous on its domain, but cannot be extended continuously to the casewhen p is on that line, because when p crosses the line joining the centers, theintersection point jumps to the “other” intersection point.

4.2. The recursive model. The recursive plane Rrec2 consists of points in

the plane whose coordinates are given by “recursive reals”, which were definedabove. It is a routine exercise to show that the recursive points in the plane areclosed under the Euclidean constructions. In particular, given approximationsto two circles (or to a circle and a line), we can compute approximations to their“intersection points”, even though it may turn out that when we compute stillbetter approximations to the circles, we see that they do not intersect at all.In the case of IntersectCirclesSame (C,K, p) or IntersectCirclesOpp (C,K, p), wecompute rational approximations pn to the point p, and rational approximationsto the centers of C and K, until we see that p cannot lie on the line joiningthe centers. Then we can compute the sign of the cross product required tosee which side of the line p is on. Then we compute IntersectCircles1 (C,K) orIntersectCircles2 (C,K), as appropriate.

We showed above that in the recursive plane, there is no computable test-for-equality function. Recall that the proof made use of a recursive real numberE(x) such that E(x) > 0 if x(x) halts and E(x) = 0 if x(x) does not halt.

10We shall also consider (briefly) the model R2− in which degenerate circles are not allowed,so Circle (A, A) is undefined.


The same example shows that we cannot test computably for incidence of a pointon a line; the number E(x) lies on the y-axis if and only if Turing machine xdoes not halt at x.

In the recursive model, we cannot decide in a finite number of steps whethera circle and a line intersect or whether two circles intersect; if they intersecttransversally we will find that out after computing them to a sufficient accuracy,but if they are tangential, we won’t know that at any finite approximation.Technically, the circle of center (0, 1) and radius 1−E(x) will meet the x-axis ifand only if the x-th Turing machine does not halt at x.

Readers familiar with recursion theory may realize that there are several waysto define computable functions of real numbers. The model we have just de-scribed is essentially the plane version of the “effective operations”. It is awell-known theorem of Tseitin, Kreisel, LaCombe, and Shoenfield, known tradi-tionally as KLS (and easily adapted to the plane) that effective operations arecontinuous. Of course, in the case at hand we can check the continuity of theelementary constructions directly.

4.3. The algebraic model. The algebraic plane consists of points in theplane whose coordinates are algebraic. Since intersection points of circles andlines are given by solutions of algebraic equations, the algebraic plane is alsoclosed under these constructions. Note that the elements of this plane are “given”all at once. We assume algebraic numbers are given by means of a rationalinterval (a, b) and a square-free polynomial f ∈ Q[x] such that f has only oneroot in (a, b).

In the algebraic plane, there is a computable test-for-equality function D. Todetermine if (a, b) and f determine the same or a different real number than(p, q) and g, first check if the two rational intervals overlap. If not, the two realsare different. If so, let (r, s) be the intersection. Now we have to determine if fand g have a common zero on (r, s). There is a simple recursive algorithm to dothat: Say g has degree greater than or equal to that of f . Then write f = gh+ rwith r of lower degree than g. Then f and g have a common zero on (r, s) if andonly if f and r have a common zero. Recurse until both polynomials are linear,when the decision is very easy to make. Similarly, we can compute whether twocircles or a circle and a line intersect.

Since algebraic numbers can be computed, the algebraic model is isomorphicto a submodel of the recursive plane.

4.4. The Tarski model. The Tarksi model is K ×K, where K = Q(√

) isthe least subfield of the reals containing the rationals and closed under takingthe square root of positive elements. This is a submodel of the algebraic plane.Its points are the points constructible with ruler and compass. Here α, β, andγ are interpreted as three fixed rational points. That the intersection points ofcircles can be computed using only the solution of quadratic equations is checkedin detail in section 6.2.

§5. The geometrization of arithmetic without case distinctions. To-day we usually think of analytic geometry as coordinatizing a plane and translat-ing geometrical relations between points and lines into algebraic equations and

24 MICHAEL BEESON

inequalities. But the converse is also possible: translating algebra into geometry,and this is important for lower estimates on the power of geometric construc-tions, for example for showing that the models of the geometry of constructionsare planes over Euclidean fields.

In modern books (such as [5]) arithmetic is geometrized as operations on con-gruence classes of segments. We operate instead on points on some fixed lineX = Line (0, 1), where 0 and 1 are two arbitrarily fixed points. As far as Ican tell, past work on coordinatization has always assumed some discontinuousconstructions, such as test-for-equality or at least apartness. Since coordinati-zation itself is patently computable and continuous, it is unesthetic to appealto discontinuous and non-computable “constructions” to achieve coordinatiza-tion and arithmetization. Although coordinatization is standard, old, and notcomplicated, we need to check that it can in fact be done from the specifiedprimitives, without using apartness or test-for-equality, by definitions that applywithout (for example) case distinctions as to whether numbers being multipliedare equal to 0 or 1 or not. It is not a priori clear that this can be done, and itis definitely not old and standard.

In this section, we give the definitions of constructions that serve to implementcoordinatization and the arithmetic operations in a continuous way. We presentthis mathematics informally, in the same spirit that Descartes presented hisoriginal coordinatization. In the process we will isolate the geometric theoremsthat are needed, so that one can later show that the work of this section canbe formalized in a specific geometric theory. Chief among those principles isthe existence of a “uniform perpendicular” construction, i.e. a construction thatproduces from a point p and a line L, a line K perpendicular to L that containsp, without a case distinction whether p is or is not on L. Second among thoseprinciples is the notion of “right turn” as applied to a triple of distinct pointsabc (and of course “left turn” is similar).

It follows from the informal discussion in this section, that in the particu-lar models of geometry discussed above, signed addition, multiplication, anddivision are defined and behave as desired. In a later section, we will give an ax-iomatic theory capable of formalizing these correctness proofs without referenceto models.

Recall that the concepts of “right turn” and “left turn” have been introducedand discussed in Section 8.10 For example, in Fig. 2, AOZ is a “right turn”,because the sign of the cross product OA×OB is positive. Intuitively, travelingfrom A to O to Z requires one to turn right at O. This definition of “right turn”is adequate for this section, since we are only concerned with models where crossproduct makes sense. (Later, we will define “right” and “left” in an axiomaticcontext.)

5.1. Uniform perpendicular and projection. In this section we showthat some fundamental constructions needed to define (signed) arithmetic canbe defined from the elementary constructions.

A very fundamental construction in constructive geometry is the uniform per-pendicular Perp(x, L), which constructs a line perpendicular to L passing through


x, without a case distinction as to whether x lies on L or not. In classical treat-ments of geometry, this case distinction is made and a different construction isused for each case.

Definition 5.1. The term Perp(x, L) is given by the following constructionscript (See Fig. 1.)

Perp(Point x, Line L)

a = pointOn1(L)

b = pointOn2(L)

Q = Circle3(b,x,a)

c = IntersectLineCircleTwo(L,Q)

C = Circle3(x,a,c)

p = IntersectLineCircle1(L,C)

q = IntersectLineCircle2(L,C)

K = Circle(p,q)

R = Circle(q,p)

d = IntersectCircles1(K,R)

e = IntersectCircles2(K,R)

return Line(d,e)

Later on, when we have given some axioms and a formal theory, we will provethe correctness of this script. For now we just give an informal argument. Thepoint is that we need to construct two points p and q on line L such that theperpendicular bisector of pq will be the desired line; thus we must have p 6= qeven if x is on L, and whether or not x = a or x = b. Well, we can get suchpoints p and q if we can draw a circle with center x, whose radius r is “largeenough.” It will suffice if we take r to be the length of x plus the length of ab.Since a 6= b, that will be a nonzero radius, and since it is more than ax, thecircle with that radius will meet L twice, in points p and q. Then we just bisectsegment pq by an ordinary method, drawing circles with centers p and q thatpass through each other’s centers. The formal proof of the correctness of thisscript can be found below in Lemma 9.56.

The construction Project(P,L) takes a point P and line L and produces apoint Q on L such that P lies on the perpendicular to L at Q. The well-knownEuclidean construction for the projection applies only if P is known not to beon L. To define Project using that construction, we would require a test-for-incidence that allows us to test whether point P is on line L or not. But nosuch test-for-incidence construction is computable over the computable plane,so the Euclidean projection construction does not lead in any obvious way to adefinition of Project. (That does not, however, constitute a proof that Projectis not definable in terms of the elementary constructions; below we give such aproof.)

Lemma 5.2. Strong extension and projection are definable from Circle3 (withcircles of zero radius allowed). Also one can, using those primitives, construct theperpendicular to line L passing through point P , without conditions as to whetherP is or is not on L. That perpendicular meets L in a point Project(P,L), the

26 MICHAEL BEESON

Figure 1. M = Perp(x, L)) is constructed perpendicular toL without a case distinction whether x is on L or not. Notebc = xa so the radius ac of C is long enough to meet L twice.

L

bx

b

ab

b

Q

b

c

C

b

pb

q

K R

b e

b d

projection of P on L, where Project is a certain explicitly given term for a ruler-and-compass construction, and the perpendicular itself is given by another suchterm.

The following gives an explicit construction for extending a non-null segmenton a given end by a (possibly null) segment:

Definition 5.3.

Extend (A,B,C,D) = IntersectLineCircle2 (Line (A,B),Circle3 (B,C,D))

Proof. To construct the projection of point P on line L, we just need somecircle with center P that intersects L in two distinct points Q and R. Thenthe projection of P on L is the midpoint of segment QR. If L is Line (A,B),then a suitable radius would be the sum of the lengths of AB and PA. Thusthe circle we need can be constructed as Circle3 (P,A,Extend (A,B, P,A)). Note


that circles of zero radius are crucial here, so that we can handle the case P = A.That completes the definition of projection.

Now, to construct the perpendicular to L at P , we simply erect the perpen-dicular to L at the projection of P on L, using the usual Euclidean construction.That completes the proof of the lemma.

Lemma 5.4. We can define a construction Para such that, for any line L andany point P (which may or may not be on L), Para(P,L) passes through P ,and if P is not on L then Para(P,L) is parallel to L, while if P is on L, thenPara(P,L) has the same points as L.

Proof. The definition of Para is

Para(P,L) = Perp(P,Perp(P,L)).

In words: First find the perpendicular to L passing through P . Then erect theperpendicular to that line at P . It is a theorem of neutral geometry (“neutral”means no parallel postulate is used) that two lines with a common perpendicularare parallel. Since Perp is everywhere defined, regardless of whether P is or isnot on L, the same is true of Para. That completes the proof of the lemma.

5.2. Rotation. We next define a construction Rotate , which requires as in-puts three distinct, non-collinear points P , O, and Q (think of angle POQ), aswell as a point A on Line (O,P ). The desired result of Rotate (P,O,Q,A) is apoint Z on Line (O,Q) such that OZ = OA and if A 6= O then AOZ is a rightturn. (See Fig. 2, but note that A might also be on the other side of O.) The

Figure 2. Z = Rotate (P,O,Q,A)

O A P

L

Q

B

Z

point is that Z is defined even when A = O (in which case it is just O, of course),and if A moves along Line (O,P ) through O, then Z moves along Line (O,Q),passing through O as A does. To construct Z, we first bisect the angle POQ(by the usual Euclidean construction, which is not problematic since the threepoints are not collinear). Let the angle bisector be line L. Then let line K bethe perpendicular to Line (O,P ) at A, let B be the intersection point of K and

28 MICHAEL BEESON

L, and let Z be the projection of B on Line (O,Q) (which is defined no matterwhether O = A or not).

Note that there are, if A 6= O, two points Z on OQ such that OZ = OA. Theone constructed by Rotate (P,O,Q,A) is such that, if POQ is a right turn, thenAOZ is a right turn when A 6= O, regardless of whether A is between O and Por not. Similarly, if POQ is a left turn, so is AOZ.

5.3. Addition. To perform addition geometrically we suppose given a lineL = Line (R,S) and an “origin”, a point O on L with S between R and O. Weneed to define a construction Add(A,B), which also depends, of course, on S,R, and O, such that Add(A,B) is a point C on L representing the (signed) sumof A and B, with O considered as origin.

Lemma 5.5. Given line L = Line(R,S), and a point 0 on L with S between Rand 0, we can construct a point Add(A,B) on L representing the signed sum ofA and B, with O considered as origin, using the elementary constructions andCircle3.

Remark. In order to appreciate that this lemma is not trivial, consider thefollowing obvious, but incorrect, attempted solution:

Add(A,B) := IntersectLineCircle2 (Line (0, B), Circle3(A, 0, B)).

This works fine for B 6= 0, whatever the “sign” of A and B, and it even workswhen A = 0, but when B = 0 it is undefined.

Proof. With Rotate in hand, we can give a construction for Add(A,B) (depend-ing also on R, S, and O). (The construction is illustrated in Fig. 3 and Fig. 4)First, we replace R with a new points on L = Line (R,S), farther away from O,so that O, A, and B are all on the same side of R, and the new R and S arein the same order on line L as R and S were before. (This can be done usingExtend , as we will prove rigorously later.) Now erect the perpendicular K toL at O, and the perpendicular H to L at B. In the process of erecting theseperpendiculars, we will have constructed a points C on K such that ROC is aright turn. Then let D be the projection of C on H and let

U = Rotate (R,O,C,A)

V = Project(U,H)

W = Rotate (D,B,R, V )

We set Add(A,B) = W . Then Add(A,B) is defined for all A,B. SupposeA 6= O. Then UV is perpendicular to both K and H . Then U and V are on thesame side of L, since if UV meets L at a point X , then XU and XO are bothperpendicular to K, which implies U = O, which implies A = O, contradictingA 6= O. It then follows from the property of Rotate that B and W occur on lineL in the same order that O and A occur. Refer to Fig. 4 for an illustration of thecase when A is negative. This implies that Add(A,B) represents the algebraicsum of A and B, since in magnitude BW = OA.

5.4. Multiplication and division. Having defined addition, we now turnto multiplication, division, and square root. The geometrical definitions of these


Figure 3. Signed addition without test-for-equality.A is rotated to U , then projected to V , then rotated to W .

b b b

b

b

b

O A B W = A+B

U V

C H

R

Figure 4. Signed addition when A is negative

A 0 W = A+B B

U V

operations go back (at least) to Descartes.11 On the second page of La Geometrie[7], he gives constructions for multiplication and square roots. We reproduce thedrawings found on page 2 of his book [7] in Figures 2 and 3. Here is Descartes’explanation of these figures:

Figure 5. La Multiplication according to Descartes

B A

C

D

E

1. Let AB be taken as unity.

11Although Descartes is usually credited with this, compare Euclid VI.12, which is verysimilar to Descartes’s treatment of multiplication and division, and Euclid VI.13, which is verysimilar to Descartes’s construction of square roots.

30 MICHAEL BEESON

2. Let it be required to multiply BD by BC. I have only to join thepoints A and C, and draw DE parallel to CA; then BE is theproduct of BD and BC.

3. If it be required to divide BE by BD, I join E and D, and drawAC parallel to DE; then BC is the result of the division.

4. If the square root of GH is desired, I add, along the same straightline, FG equal to unity; then, bisecting FH at K, I describe thecircle FIH about K as a center, and draw from G a perpendicularand extend it to I, and GI is the required root.

From the point of view of constructive geometry, there is a problem with theconstruction. Namely, Descartes has only told us how to multiply two segmentswith non-zero lengths, and at least one of whose lengths is not 1 (the length ofunity–he needs this when constructing AC parallel to DE), while we want to beable to multiply in general, without a test-for-equality construction.

Using the Para construction of Lemma 5.4 where Descartes calls for “drawingDE parallel to CA”, we no longer have a problem multiplying numbers near1 or 0. In [3], we gave a construction that successfully generalizes Descartes’smultiplication method to signed arguments. That method uses Rotate . However,Hilbert ([15], p. 54) gives another construction, whose result is equivalent to thatof Descartes for positive arguments, but which directly works correctly for signedarguments as well. It is illustrated in Fig. 6.

Figure 6. Multiplication according to Hilbert

bab

bI

ba

bb

The construction is as follows: Start with Line (0, 1). Erect a perpendicularto Line (0, 1) at 0, and let I be a point on it such that 0I = 01. Given points aand b on Line (0, 1), construct a circle passing through I, a, and b. This must bedone uniformly, so that when a = b, the circle passes through I and is tangent toLine (0, 1) at a. Then the answer is “the other point of intersection” of the circleand Line (0, I), except that this point lies not on Line (0, 1) but on Line (0, I),


so it needs to be rotated clockwise back to Line (0, 1). This construction is moreeasily visualized (for signed arguments) than that of Descartes. In order to carryout the construction uniformly, we first must be able to uniformly construct theperpendicular bisector of ab, in such a way that it is defined and perpendicular toLine (0, 1) even when a = b. Then we construct the perpendicular bisector of aI(which is unproblematic since I does not lie on Line (0, 1)), and then the centerof the circle will lie where those two lines meet. Here we must use the parallelaxiom; the existence of a circle passing through three given noncollinear points(i.e., circumscribing a triangle) is classically equivalent to the parallel postulate.

We will take up the details of both multiplication constructions in Section12.2.

Descartes’ division method is handled similarly, using Para where Descartesconstructs a parallel.

5.5. Square roots. Now we take up the geometrical construction of squareroots. Fig. 7 shows Descartes’ construction for finding the square root of HG.His answer is the length of segment IG. Here is a geometrical construction term(encoded as a program) to carry out Descartes’ construction uniformly, as longas G is not to the left of H in the illustration, regardless of whether G = H ornot. We assume that H in the diagram is 0 and the horizontal line is Line (0, 1).The point 1 is not shown in the figure.

Figure 7. Square roots according to Descartes

H K G F

I

SquareRoot(Point G)

// H in Descartes’ diagram is 0

F = Add(0,G,0,1) // so FG has unit length

K = Midpoint(F,0)

C = Circle(K,F)

L = Perp(G,Line(0,F))

I = IntersectLineCircle1(L,C)

// next rotate unit length to line L

U = IntersectLineCircle1(L,Circle(G,F))

R = Rotate(U,G,F,I)

// so now RG = IG but R is on Line(H,G), on the same side of G as F

// now we need N so that N0 = RG

MinusOne = IntersectLineCircle2(1,0,Circle(0,1))

N = Extend(MinusOne,0,G,R)

32 MICHAEL BEESON

return N

Descartes stops when he has constructed I. What we have to do extra is toconstruct a point N such that 0N = GI. In order to do that uniformly, we mustnot assume that I 6= G. In order to get a non-degenerate angle we cannot useIGF ; instead we lay off a unit length on the perpendicular GI, which has beencorrectly constructed even if 0 = H = G = I.

Thus we do not need to assume G 6= 0 for this construction to work; we onlyneed that 0 is not between G and 1, i.e. loosely speaking G ≥ 0. This worksbecause Perp is total, i.e. everywhere defined.

5.6. Formalizing the geometrization of arithmetic. It remains, of course,to prove in some axiomatic geometrical theory that multiplication and additionsatisfy the field laws. We do not take that up at this point since we have notyet discussed theories and axioms. In due course, after having presented anaxiomatic theory, we will formalize these results in deatil. But the reason forpresenting them informally is to emphasize that the ideas really do not dependon a particular formalization of constructive Euclidean geometry. All we need isthat the formalization should support the basic constructions Para, Perp, andRotate used above. Of these Perp is basic, since the other two are defined usingPerp.

Just as in classical geometry, one can define addition and prove its propertieswithout using the parallel axiom, but one needs the parallel axiom to define andprove the properties of multiplication; and one needs line-circle continuity toverify Descartes’s square root construction. The details of these points will betaken up in Sections 12.1 and 12.2.

§6. The arithmetization of geometry. Our aim in this section is to laythe groundwork for a constructive version of the classical representation theorem:the models of Euclidean geometry are exactly planes F 2 where F is a Euclideanfield, i.e. an ordered field in which positive elements have square roots. Thegroundwork in question consists in developing the theory of Euclidean fields con-structively, and verifying constructively that the axioms of line-circle continuityand circle-circle continuity hold in each Euclidean field.

6.1. Euclidean fields in constructive mathematics. We consider theanalogue for ECG, which requires us to define constructive Euclidean fields.We use a language with symbols + for addition and · for multiplication, and aunary predicate P (x) for “x is positive”.

We first discuss the axiomatization of Euclidean fields with intuitionistic logic.We take the usual axioms for fields, except the axiom for multiplicative in-

verse, which says that positive elements have multiplicative inverses. If positiveelements have inverses, it is an easy exercise to show that negative elements dotoo. We define a Euclidean field to be a commutative ring satisfying the following


additional axioms:

0 6= 1 EF0

x 6= 0 → ∃y (x · y = 1) EF1

P (x) ∧ P (y) → P (x+ y) ∧ P (x · y) EF2

x+ y = 0 → ¬(P (x) ∧ P (y)) EF3

x+ y = 0 ∧ ¬P (x) ∧ ¬P (y) → x = 0 EF4

x+ y = 0 ∧ ¬P (y) → ∃ z(z · z = x) EF5

¬¬P (x) → P (x) EF6, or Markov’s principle

Axiom EF5 says that non-negative elements have square roots. This is a strongeraxiom, intuitionistically, than simply specifying that positive elements have squareroots. We could conservatively extend field theory by a unary function symbolfor negation, −x, in which case EF5 could be more readably written ¬P (−x) →∃z (z2 = x).

As usual, we define x < y to mean ∃z(P (z) ∧ x+ z = y), or informally, y − xis positive; and x ≤ y means ¬(y < x. Then Markov’s principle is equivalent to¬(x ≤ 0) → 0 < x.

We also consider several variants of the Euclidean field axioms, which differin the axiom about which elements have reciprocals. The axiom EF1 aboverequires all nonzero elements to have reciprocals. We could replace this axiomby the weaker axiom

P (x) → ∃y (x · y = 1) EF7

Fields satisfying the system with EF7 instead of EF1 are called weakly Euclideanfields. EF1 implies EF7, because P (x) → x 6= 0, but to derive EF1 from EF7,we would have to know that each nonzero element is either positive or negative.In the language without a symbol for additive inverse, this can expressed as

x+ y = 0 ∧ x 6= 0 → P (x) ∨ P (y) EF8

Fields that satisfy EF7 and EF0 and EF2 through EF8 then automatically satisfyEF1 as well. We call these fields strongly Euclidean.

In commutative ring theory an element x is called invertible if it has a multi-plicative inverse, i.e. for some y we have x·y = 1; we also say “x has a reciprocal.”The weakest version of Euclidean field theory that we consider replaces AxiomEF1 with the following two axioms, which say that elements without reciprocalsare zero, and that if x is greater that a positive invertible element, then x isinvertible.

(∀y(x · y 6= 1)) → x = 0 EF9

P (y) ∧ P (z) ∧ y + z = x ∧ (y · v = 1) → ∃w (w · x = 1) EF10

Fields that satisfy EF0, EF2–EF6, and EF9-10 are called “Playfair fields”, be-cause (as we will see far below) they correspond to models of the Playfair parallelaxiom. EF9 enables us to verify the Playfair axiom, and EF10 enables us to verifyPasch’s axiom.

34 MICHAEL BEESON

To recap: In a Euclidean field, all nonzero elements have multiplicative in-verses, while in a weakly Euclidean field, it is only required that elements knownto be positive or negative have multiplicative inverses. In a strongly Euclideanfield, nonzero elements are either positive or negative, so the distinction doesn’tmatter. In a Playfair field, elements without reciprocals are zero, and elementsgreater than an invertible element are invertible.

Remark. Since the double negation of EF8 follows from EF4, every Euclideanfield is not not strongly Euclidean; in other words, one cannot give an example ofa Euclidean field that is not strongly Euclidean. The standard plane is stronglyEuclidean if and only if Markov’s principle holds in the reals. (To see this,suppose x 6= 0; apply Markov’s principle to |x| we have |x| > 0; then by apartness,which holds in the standard plane, we have x > −|x|/2 or x < |x|/2, and in thefirst case x > 0 while in the second case x < 0.)

6.2. Line-circle and circle-circle continuity over Euclidean fields. LetF be an ordered field. Then “the plane over F”, denoted by F 2, is a geometricalstructure determined by defining relations of betweenness and equidistance inF 2, using the given order of F . Namely, b is between a and c if it lies on theinterior of the segment ac. That relation can be expressed more formally invarious ways, for example (using the cross product) by (c− b)× (b− a) = 0 and(b − c) · (a − b) > 0. Similarly, non-strict betweenness means that b lies on theclosed segment ac, or formally, (c− b) × (b− a) = 0 and (b − c) · (a− b) ≥ 0.

The equidistance relation E(a, b, c, d), which means that segment ab is con-gruent to segment cd, can be defined by (b− a)2 = (d− c)2. Note that no squareroots were used, so these definitions are valid in any ordered field. Though wehave not yet given specific axioms for betweenness and congruence, it is reason-able to demand that any constructive axioms for betweenness and congruenceshould hold in F 2, at least for Euclidean fields F .

By line-circle continuity we understand this axiom: let point a be non-strictlyinside circle C, and point b be non-strictly outside circle C. Then Line (a, b)meets circle C in a point non-strictly between a and b. By circle-circle continuity,we mean that if circle C has a point non-strictly inside circle K, and anotherpoint non-strictly outside circle K, then there is a point on both circle C andcircle K.

Theorem 6.1 (proved constructively). Let F be an ordered field (satisfyingany of the axioms about the existence of reciprocals). Then the following areequivalent:

(i) F 2 satisfies circle-circle continuity.(ii) F 2 satisfies line-circle continuity.(iii) F is a Euclidean field (nonzero elements have reciprocals and nonnegative

elements have square roots).

Remark. We never need to divide in this proof, hence the constructive distinc-tions about the existence of reciprocals do not come into play.

Proof. A similar theorem is proved classically in [14], p. 144. We follow thatproof as closely as possible, but some details are missing there. The issue is thatit is not enough to observe that the equations for the intersection points are


quadratic. One has to translate the hypothesis that one circle has a point insideand a point outside the other circle into algebra and show algebraically that thisimplies the equations for the intersection are solvable.

First we show (i) implies (ii). If f = 0 is the equation of a circle C, and g = 0the equation of a line L, then f + g = 0 is the equation of another circle, whoseintersections with C are the same as the intersections of C and L. Hence (i)implies (ii).

Next we show (ii) implies (iii). Let b ≥ 0, let H be the origin and let G = (0, b).To show that b has a square root. Refer to Fig. 7, which shows the geometricalconstruction of square roots that goes back to Euclid. We wish to show that thepoint I = (b,

√b) exists, as the intersection point a line and circle. The circle has

center K = ((1+ b)/2, 0) and radius (1+ b)/2), and the line is x = b. So the linehas the point (0, b) non-strictly inside the circle, since |(b−(1+b)/2| ≤ (1+b)/2;and the line has points outside the circle, for example (b, (1+b). So by line-circlecontinuity, the intersection point I exists, and hence b has a square root. Notethat the case b = 0 is allowed.

Finally we have to show (iii) implies (i). Let C and K be non-concentriccircles. By making linear transformations, we can assume that the two circleshave their centers at (0, 0) and (1, 0) respectively; then the equations for thecircles are x2 + y2 = r2 and (x− 1)2 + y2 = s2.

We must make use of the hypothesis that C has a point (non-strictly) insideK and another (non-strictly) outside K. To do so we need the following lemma:

Lemma 6.2. With C and K as above, C has a point (non-strictly) inside Kand another (non-strictly) outside K if and only if r + s ≥ 1 and (s− r)2 ≤ 1.

Proof. First suppose r + s ≥ 1 and (r − s)2 ≤ 1. Then we have

r ≥ 1 − s since r + s ≥ 1

r − s ≤ 1 since (r − s)2 ≤ 1

r ≤ 1 + s.

That is, 1−s ≤ r ≤ 1+s. Then −s ≤ r−1 ≤ s. Squaring, we have (r−1)2 ≤ s2.Hence (r, 0) is a point on C and inside K.

Since (r−s)2 ≤ 1 we have s−r ≤ 1; hence −r−1 ≤ −s. Since s ≤ 0, when wesquare this inequality we get (r + 1)2 ≥ s2. Therefore, the point (−r, 0) (whichis on C) is outside K. That proves the easy half of the lemma.

Now suppose that the point (x, y) is on C and insideK. We will prove r+s ≥ 1.Since a ≤ b is equivalent to ¬(a > b), we can prove r + s ≥ 1 by contradiction.Suppose that r + s < 1. Then intuitively, the line x = (r + 1 − s)/2 separatescircles C and K, contradiction, since (x, y) is inside both circles. We prove thismore rigorously: Since (x, y) is on C and inside K, the point (x, 0) is inside bothcircles. That is, x2 ≤ r2 and (x− 1)2 ≤ s2. Since r + s < 1, we have s < 1, andsince (x, 0) is inside K, we have x > 0. Hence 0 < x ≤ r < 1 and 1 − x < s.Adding these inequalities we get x + 1 − x < r + s, or 1 < r + s, contradictingr + s < 1. This contradiction proves r + s ≥ 1.

36 MICHAEL BEESON

Now suppose in addition that the point (u, v) is on C and outside K. Thenwe claim the point (−r, 0) on C is also outside K. Here is the proof:

(u− 1)2 + v2 ≥ s2 since (u, v) is outside K

u2 − 2u+ 1 + v2 ≥ s2

r2 − 2u+ 1 ≥ s2 since u2 + v2 = r2

r2 + 2r + 1 ≥ s2 + 2r − 2u

(r + 1)2 ≥ s2 + 2(r − u)

But u ≤ r, since (u, v) is on C. Hence r − u ≥ 0, and we have

(r + 1)2 ≥ s2

which shows that (−r, 0) is outside K as claimed. Hence s ≤ r + 1. Hences− r ≤ 1. Squaring, we have the desired inequality (r−s)2 ≤ 1. That completesthe proof of the lemma.

Returning to the proof of the theorem, we want to show that the equations x2+y2 = r2 and (x−1)2+y2 = s2 have a common solution, under the hypothesis thatC has a point (non-strictly) inside K and another point (non-strictly) outsideK. Subtracting the two equations we obtain a linear equation:

x2 − (x− 1)2 = r2 − s2

2x− 1 = r2 − s2

x =1

2(r2 − s2 + 1)

Putting that back into x2 + y2 = r2 we have

1

4(r2 − s2 + 1)2 + y2 = r2

4y2 = 4r2 − (r2 − s2 + 1)2

= 4r2 − (r4 + s4 + 1 + 2r2 − 2r2s2 − 2s2)

= 2r2 − r4 − s4 − 1 + 2r2s2 + 2s2

= −(r2 − s2)2 + 2(r2 + s2) − 1

= ((r + s)2 − 1)(1 − (r − s)2)

By the lemma, both factors on the right are nonnegative; so their product isnonnegative, and we can take the square root in any Euclidean field (or weaklyEuclidean or even Playfair field). That completes the proof of the theorem.

§7. Other axiomatizations of geometry. In the previous sections, we havepresented constructive geometry informally, that is, without the use of the ap-paratus of formal logic, or even an informal axiomatic system for geometry. Itis, however, our intention to present a formal theory ECG of Euclidean con-structive geometry. Before doing so, we will review the existing formalizationsof classical geometry. It is not our purpose to review in detail the history ofaxiomatic geometry, but some history is helpful in understanding the varietyof geometrical axiom systems. In fact the modern subject of formal logic wasborn in the effort to understand the foundations of geometry; in particular to


understand the efforts to understand the relation of Euclid’s parallel postulateto the other postulates; in a sense non-Euclidean geometry was the mother ofmodern logic.

7.1. Pasch. Moritz Pasch’s 1882 book [23],[24] introduced the relation ofbetweenness that has been used in all axiomatizations of geometry since 1882.Betweenness is the relation B(a, b, c) that holds when points a, b, and c aredistinct points on a line, “in that order”. Pasch pointed out the defects inEuclid that require the use of betweenness and supplied the basic axioms forbetweenness. The most complicated of those axioms is known as Pasch’s axiom;it says that if line L meets side AC of triangle ABC, and none of A, B, or Clies on L, and L and does not meet side AB, then L meets side BC.

7.2. Pieri and Peano. The modern axiomatic method, or as Pieri called it,the “hypothetical-deductive method”, differed from Euclid not only in increasedprecision (the use of betweenness, for example) and the realization that somenotions must be taken as undefined, but also in the important idea that geometrymight have many models, rather than just “the intended model”. These threeideas arose in the second half of the nineteenth century in more than one place,and an accurate history of their development is far beyond the scope of thesebrief paragraphs. It is, however, clear that the “Italian school” made considerableprogress in this direction. For example, Pieri’s Point and Motion [25] was a fullymodern development, taking as primitive objects points and “motions”, and asa primitive relation the application of a point to a motion. Apparently neitherPieri nor Hilbert knew about the work of the other before the publication (see[20], pp. 150-151).

FINISH THIS–what about Peano?

7.3. Hilbert. Hilbert’s influential book [15] used the notion of betweennessand the axioms studied by Pasch. His theory was what would today be called“second-order”, in that sets were freely used in the axioms. Segments, for ex-ample, were defined as sets of two points, so by definition AB = BA since theset A,B does not depend on the order. Of course, this is a trivial departurefrom first-order language; but Hilbert’s last two axioms, Archimedes’s axiomand the continuity axiom, are not expressible in a first-order geometrical theory.On the other hand, lines and planes were regarded not as sets of points, but as(what today would be called) first-order objects, so incidence was an undefinedrelation, not set-theoretic membership. At the time (1899) the concept of first-order language had not yet been developed, and set theory was still fairly new.Congruence was treated by Hilbert as a binary relation on sets of two points,not as a 4-ary relation on points.

Early geometers thought that the purpose of axioms was to set down thetruth about space, so as to ensure accurate and correct reasoning about the onetrue (or as we now would say, “intended”) model of those axioms. Hilbert’sbook promoted the idea that axioms may have many models; the axioms anddeductions from them should make sense if we read “tables, chairs, and beermugs” instead of “points, lines, and planes.” This is evident from the very firstsentence of Hilbert’s book:

38 MICHAEL BEESON

Consider three distinct sets of objects. Let the objects of the first set be

called points . . . ; let the objects of the second set be called lines . . . ; let the

objects of the third set be callled planes.

Hilbert defines segments as pairs of points (the endpoints),although lines areprimitive objects. On the other hand, a ray is the set of all points on the ray,and angles are sets consisting of two rays. So an angle is a set of sets of points.Hence technically Hilbert’s theory, which is often described as second-order, is atleast third-order. Hilbert’s language has a congruence relation for segments, anda separate congruence relation for angles. Hilbert’s congruence axioms involvethe concept of angles: his fourth congruence axiom involves “angle transport”(constructing an angle on a given base equal to a given angle), and his fifthcongruence axiom is the SAS triangle congruence principle.

Hilbert’s Chapter VII discusses geometric constructions with a limited set oftools, a “segment transporter” and an “angle transporter”. These correspondto the betweenness and congruence axioms. Hilbert does not discuss the specialcases of line-circle continuity and circle-circle continuity axioms that correspondto ruler-and-compass constructions, despite the mention of “compass” in thesection titles of Chapter VII.

7.4. Tarski and his students. Hilbert’s geometry contained two axiomsthat go beyond first-order logic. First, the axiom of Archimedes (which requiresthe notion of natural number), and second, an axiom of continuity, which es-sentially says that Dedekind cuts are filled on any line. This axiom requiresmentioning a set of points, so Hilbert’s theory with this axiom included is nota “first-order theory” in a language with variables only over points, lines, andcircles. Later in the 20th century, when the concept of “first-order theory”was widely understood, Tarski formulated his theory of elementary geometry,in which Hilbert’s axiom of continuity was replaced with an axiom schemata.The set variable in the continuity axiom was replaced by any first-order formula.Tarski proved that this theory (unlike number theory) is complete: every state-ment in the first-order language can be proved or refuted from Tarski’s axioms.

Tarski’s axioms are elegantly stated using only variables for points. He replacesHilbert’s fourth and fifth congruence axioms (angle transport and SAS) with anelegant axiom, known as the five-segment axiom. We will eventually give theformal statement of this axiom, but for now, we refer to Fig. 8. The 5-segmentaxiom says that in Figure 8, the length of the dashed segment cd is determinedby the lengths of the other four segments. Formally, if there is another figurelike the one shown, and the four solid segments are pairwise congruent to thecorresponding segments in the second figure, then the dashed segments are alsocongruent.

Tarski’s 5-segment axiom is a thinly-disguised variant of the SAS criterion fortriangle congruence. To see this, think of another copy of Fig. 8 labeled withupper-case letters. The triangles we are to prove congruent are dbc and DBC.We are given that bc = BC and db = DB. The congruence of angles dbc andDBC is expressed in Tarski’s axiom by the congruence of triangles abd andABD, whose sides are pairwise equal. The conclusion, that cd = CD, give thecongruence of triangles dbc and DBC. Later we will give a formal proof of the


Figure 8. Tarski’s 5-segment axiom. cd is determined.

d

a b c

SAS criterion from the 5-segment axiom. We note that Borsuk-Szmielew alsotook this as an axiom (see [5], p. 81, Axiom C-5).

We note that an earlier version of Tarski’s theory included as an axiom the“triangle construction theorem”, which says that if we are given triangle abc,and segment AB congruent to ab, and a point x not on Line (A,B), then we canconstruct a point C on the same side of Line (A,B) as x such that triangle ABCis congruent to triangle abc. It was later realized12 that this axiom is provable.For example, one can drop a perpendicular from c to Line (a, b), whose foot isthe point d on Line (a, b), and then find a corresponding point D on Line (A,B),and then lay off dc on the perpendicular to Line (A,B) at D on the same side ofLine (A,B) as x, ending in the desired point C. Of course one must check thatthis construction can be done and proved correct on the basis of the other axioms.But as it stands, this construction demands a case distinction about the orderand possible identity of the points d, a, and c on Line (a, b). Hence, at least thisproof of the triangle construction theorem from the axioms of Tarski’s theoryis non-constructive. (The construction itself is obvious: C is the intersectionpoint of circles about A and B of radii ac and bc that lies on the same side ofLine (A,B) as x. The problem is to prove that the circles actually intersect,without using a constructively invalid case distinction.)

Tarski’s early axiom systems also included axioms about betweenness and con-gruence that were later shown [13] to be superfluous. The final version of thistheory appeared in [26]; for the full history see [30].13 The achievement of [26]is to develop a really minimal set of axioms for betweenness and congruence.14

Hilbert’s intuitive axioms about betweenness disappeared, leaving only the axiom¬B(a, b, a) and the Pasch axiom and axioms to guarantee that congruence is anequivalence relation. We shall adopt constructive versions of these axioms, and

12Acording to [30], Tarski included this principle as an axiom in his first two publishedaxiom sets, but then discovered in 1956-57 with the aid of Eva Kallin and Scott Taylor, thatit was derivable; so he did not include it in [29]. (See the footnote, p. 20 of [29].) But Tarskidid not publish the proof, and Borsuk-Szmielew take the principle as their Axiom C-7 [5].

13Note that the version mentioned in [1] is not the final version used in [26]; inner transitivityfor betweenness was eliminated in [26].

14We would like to emphasize the important contributions of Gupta, which are importantto the development in [26], and are credited appropriately there, but without a careful studyone might not realize how central Gupta’s results were. These results were apparently neverpublished under Gupta’s own name, and still languish in the Berkeley math library in hisdoctoral dissertation.

40 MICHAEL BEESON

we are able to show that negative (that is, disjunction-free and existence-free)statements that follow classically from those axioms also follow constructively.Thus, we will be able to draw freely on the labors of Tarski, Gupta, Szmielew,Schwabhauser, and others who contributed to this elegant axiom set. That willenable us to focus on the constructive aspects, rather than on derivations of (de-ceptively) simple-looking theorems, whose complications have nothing essentialto do with constructivity.

7.5. Tarski’s theory of Euclidean geometry. But Tarski’s theory is “el-ementary” only in the sense that it is first-order. It still goes far beyond Euclid.To capture Euclid’s geometry, Tarski considered the sub-theory in which the con-tinuity axiom is replaced by “line-circle continuity” and “circle-circle continuity”.These axioms assert the existence of the intersection points of lines and circles, ifsome point on the line lies inside the circle, or some point on one circle lies insidethe other circle. In the language of ECG, these axioms are expressed withoutneeding to say “there exists”; since we have terms to denote those intersectionpoints, all we have to do is say that, for instance, IntersectLineCircle1 (L,C) lieson both L and C. ECG is essentially a constructive version of Tarski’s theoryof the elementary constructions.15

7.6. Borsuk-Szmielew. This book follows Hilbert, with some axioms takenfrom Tarski, filling in many details. It treats both Euclidean and hyperbolicgeometry. The formal systems are, like Hilbert’s, second-order, and the meta-mathematics is restricted to the construction of (second-order) models. It doesintroduce a four-place equidistance relation between points, instead of treatingcongruence as a relation on sets; but it still regards, as Hilbert does, a segmentas a set of points. (Strangely, the segment ab is a, b, while the “open seg-ment” ab is the set of all points between a and b.) For example, the betweennessaxioms are those Hilbert used, including the Pasch axiom. The first edition ofthis book was in Polish, and the English edition is not merely a translation, butalso incorporated some changes in the system; although I have never seen thePolish edition, according to the introduction of the English edition, the originalfollowed Hilbert more closely, in having primitive sorts for lines and planes; theEnglish edition treats them as sets of points, in order to make the model the-ory easier. The book uses an axiom system based on Hilbert, but incorporatingsome of Tarski’s improvments, such as the five-segment axiom, which is to beexpected as co-author Szmielew worked closely with Tarski; but the manuscriptwas prepared too early to benefit from the later improvements made by Guptaand others. Today this work is superceded by [26].

7.7. Avigad, Dean, and Mumma. In [1], these three authors give a formaltheory with classical logic that is intended to faithfully reflect Euclid’s reasoning.Their theory is many-sorted, and has many primitives and many axioms, partlyin order to support educational applications (it has been used as the basis formechanical theorem-proving support of a geometry course). They distinguishdifferent kinds of inferences: diagrammatic, metric, diagram-metric transfer, and

15It is confusing that in axiomatic geometry, “elementary” sometimes refers to the elemen-tary constructions, and sometimes to the full first-order theory of Tarski. In this paper weshall not refer again to the full first-order theory.


superposition inferences. One has a sort for angles (that is, measures of angles)and a sort for lengths (measures of segments). One can perform addition andsubtraction on these sorts (but not multiplication), and compare them. Thisis used to interpret Euclid’s comparison of segments and angles. The pointof this is that there is an efficient decision procedure for this subtheory. Thediagrammatic inferences involve betweenness, axioms about the primitive notion“same side of”, including versions of Pasch’s axiom, axioms about incidence andintersection, etc. The idea is that the diagrammatic inferences are the ones thatEuclid does not state explicitly. Finally there are the diagram-metric transferinstances, in which diagrammatic information is used to infer metric information.For example, if b is between a and c, then ab+ bc = ac.

The fact that this theory uses classical logic is not very essential. Thereis nothing in the framework that would preclude a constructive version of thetheory. It is already “constructive” in the sense that it does not have continuityaxioms beyond line-circle and circle-circle. Such a theory would no doubt be aconservative extension of our theory ECG.

7.8. Heyting and other constructive approaches.

§8. The theory ECG of Euclidean Constructive Geometry. In thissection we give and explain the axioms of a first order axiomatic theory of geom-etry as close as possible to Euclid. We call it ECG, for “Elementary ConstructiveGeometry”. We will take care to formulate our axioms without quantifiers andwithout disjunction, which will be key to our applications of proof theory. Whatwe aim to do in this section is to formulate such a theory, which we feel is quiteclose in spirit to Euclid. In formulating this theory, we made use of the famousaxioms of Hilbert [15], which have been given a more modern and detailed for-mulation in the textbook of Greenberg [12]. Of course, we do not take the fullcontinuity axioms of Hilbert, but only the line-circle and circle-circle continu-ity axioms. Where possible, we formulate our axioms as correctness statementsabout the constructions; in that form they are automatically quantifier-free.Some axioms, which are not about constructions, are inherently quantifier-free.The only question of serious interest is whether disjunction can be completelyavoided. It can, as it turns out. The details of this axiomatization may seemsomewhat tedious, but the system must of course be specified in complete detailin order to use it in metamathematical proofs. Moreover, some of the details, asfar as I can determine, are actually new. In particular, we show how to definethe relations “ABC is a left turn” and “ABC is a right turn” in our theory; theexperts we consulted thought this was new. We need this in order to distinguishthe two intersection points of two circles.

8.1. Logic of Partial Terms (LPT). Since many of our function symbolsdenote “partial functions”, i.e. functions that are not always defined, we willuse the “logic of partial terms” LPT in our theories. This is a modificationof first-order logic, in which the formation rules for formulas are extended byadding the following rule: If t is a term then t ↓ is a formula. Then in additionthe quantifier rules are modified so instead of ∀x(A(x) → A(t)) we have ∀x(t ↓

42 MICHAEL BEESON

∧A(x) → A(t)), and instead of A(t) → ∃xA(x) we have A(t)∧t ↓ → ∃xA(x).Details of LPT can be found in [2], p. 97.

We could try to deal with partial terms, such as√x, by simply using an

ordinary function symbol for√

, but not saying anything in the axioms about√

of negative numbers. Thus√−1 would some real number, but we would not

know or care which one, and we would not be able to prove that its square is−1. This approach rapidly becomes awkward when complicated terms involvingsquare roots of different quantities are used, and you must add extra hypothesesto every theorem asserting that what is under every square root is positive, andwe choose to use LPT instead.

LPT includes axioms c ↓ for all constants c of any theory formulated in LPT;this is in accordance with the philosophy that terms denote things, and whileterms may fail to denote (as in “the King of France”), there is no such thingas a non-existent thing. Thus 1/0 can be undefined, i.e. fail to denote, but if aconstant ∞ is used in LPT, it must denote something.

The meaning of t = s is that t and s are both defined and they are equal. Wewrite t ∼= s to express that if one of t or s is defined, then so is the other, andthey are equal.

Definition 8.1. For terms in any theory using the logic of partial terms, t ∼= qmeans

t ↓ → t = q ∧ q ↓ → t = q.

This is read t and q are equal if defined.

Thus “∼=” is an abbreviation at the meta-level, rather than a symbol of thelanguage.

LPT contains the axioms of “strictness”, which are as follows (for each func-tion symbol f and relation symbol R in the language):

f(t1, . . . , tn) ↓ → t1 ↓ ∧ . . . ∧ tn ↓R(t1, . . . , tn) → t1 ↓ ∧ . . . ∧ tn ↓

Note that in LPT, under a given “valuation” (assignment of elements of astructure to variables), each formula has a definite truth value, i.e. we do not usethree-valued logic in the semantics. For example, if P is a formula of field theorywith a reciprocal operation 1/x, then P (1/0) is false, since 1/0 is undefined.For the same reason ¬P (1/0) is false. Hence P (1/0)∨ ¬P (1/0) is false too; butthat does not contradict the classical validity of ∀x(P (x) ∨ ¬P (x)) since we arerequired to prove t ↓ before deducing an instance P (t) ∨ ¬P (t).

As an example of the use of LPT, and also for further use below, we reformu-late the theory of Euclidean fields using the logic of partial terms. The existentialquantifiers associated with the reciprocal axioms, with the axiom of additive in-verse, and with the square-root axiom of Euclidean field theory are replaced bya function symbol

√, a unary minus −, and a function symbol for “reciprocal”,

which we write as 1/x instead of reciprocal(x). The changed axioms are


x+ (−x) = 0 (additive inverse)x 6= 0 → x · (1/x) = 1 (EF1′)P (x) → x · (1/x) = 1 (EF7′)x+ y = 0 ∧ ¬P (y) → √

x · √x = x (EF5′)

Definition 8.2. The theory EF is the theory of Euclidean fields, formulatedusing the logic of partial terms with function symbols for additive inverse, squareroot, and reciprocal, using the axioms EF1′ (nonzero elements have recipro-cals), EF2 (sums and products of positive elements are positive), EF3 and EF4(roughly, that x and −x are not both positive and not both non-positive unless xis zero), EF5′ (given just above), and EF6 (Markov’s principle).

8.2. Replacing LPT and sorts with predicates if desired. Since theusual theorem-provers do not understand the logic of partial terms, for purposesof experimentation with automated deduction, one may wish for a formulationwithout LPT. That can be done, and at the same time one can eliminate theuse of multi-sorted logic. Add unary predicates Point, Line, and Circle, andthen replace t ↓ by either Point (t), Circle (t), or Line (t), according to the sortof t, in all the axioms, including the logical axioms of LPT that involve ↓. Thenonlogical axioms then have all the variables of a given sort relativized to thecorresponding unary predicate, e.g. point variables are relativized to Point.

It may be easier to understand this procedure if we consider a more familiartheory, field theory. There is originally only one sort of variables, so we will addjust one predicate symbol F , and the axiom x 6= 0 → 1/x ↓ would becomex 6= 0 → F (1/x), and the axiom (x · 1/x) ↓ → x · (1/x) = 1 becomesF (x · (1/x)) → x · (1/x) = 1. In any model of this system, the interpretationF of F is (the carrier set of) a field, but there will be elements of the model,including an interpretation of 1/0, that are not in F . The reciprocal of zero issomething, but it doesn’t matter what, as long as it is not in F . The elementsof the model that are not in F serve as the denotations of terms that would beundefined in LPT.

One might try to not use the new symbol F , and just let 1/x be always defined,and in the field, but only assert that x 6= 0 → x · (1/x) = 1. In this systemwe can prove 1/0 = 0 (since if 1/0 6= 0 then (1/0) · 0 = 1, but anything times 0is 0 and 0 6= 1). This equation is not true in LPT, so the interpretation is notfaithful from LPT to this system. Moreover, in LPT we can prove 1/0 6= 0,since if 1/0 = 0 then 1/0 ↓ , so 1/0 · 0 = 1, but 1 6= 0, contradiction; hence1/0 6= 0 as claimed. Hence the interpretation is not even sound; one cannot justlet 1/x be an “unspecified number”; if it is defined, it has to be an “unspecifiedsomething” that is not a number.

Thus eliminating LPT necessarily involves introducing unary predicates (orsorts) with some “undefined elements” lurking in purgatory, so to speak, e.g.1/0, which cannot satisfy F .

To make these considerations precise, consider any theory T in many-sorted

predicate logic with the logic of partial terms. For each formula φ, let φ be theformula in ordinary one-sorted predicate logic obtained by replacing the sorts ofT by unary predicate symbols (with the same names as the sorts), relativizingquantifiers over sort P to the predicate symbol P , and for each sort P and term

44 MICHAEL BEESON

of sort P , replacing t ↓ by P (t). If T is a theory in predicate calculus, let T be

the theory axiomatized by the φ for φ and axiom of T .

Theorem 8.3. Let T be any theory in many-sorted predicate calculus with the

logic of partial terms. Let φ be defined as just above. Then T proves φ if and

only if φ is provable in ordinary first-order predicate logic.

Proof. Note that the interpretation commutes with implication. Therefore, bythe deduction theorem we may assume, without loss of generality, that T is theempty theory. (For readers with little background in logic, what we are sayinghere is that if T proves φ, then pure logic proves ψ → φ, for some conjunctionof axioms ψ of T , so we might as well consider ψ → φ with an empty theoryT .)

We first prove the left-to-right direction, that if many-sorted logic with LPT

proves φ, then predicate logic proves φ. We proceed by induction on the lengthof proofs. Consider the axiom of LPT,

∀xφ ∧ t ↓ → φ[x := t]

when x is a variable of sort P . This becomes

∀x(¶(x) → φ ∧ P (t) → φ[x := t].

which is provable in first-order predicate calculus. Similarly, consider the axiom

φ[x := t] ∧ t ↓ → ∃xφ.This translates to

φ[x := t] ∧ P (t) → ∃x (P (x) ∧ φ).

which is also provably in first-order predicate calculus. The other axioms andrules of inference do not involve the logic of partial terms and are straightforward.That completes the proof that if many-sorted LPT proves φ, then predicate logicproves φ.

For the converse, weStrong Parallel Axiom1

8.3. Language. For foundational purposes, it seems simplest to use onlyPoint, Line, and Circle, and that is what we do in ECG.16 We choose a multi-sorted theory, with “sorts” corresponding to those types. We use the words“sort” and “type” synonymously in this paper. It is, of course, not difficultto add sorts Ray, Arc, and Segment, and axioms making the extended theoriesconservative over ECG, but we do not do so here. Similarly, it is not difficultto eliminate all the sorts but Point. We take function symbols corresponding toconstructors and accessors for those types, described in detail below. The relationsymbols we use are standard in axiomatic geometry, B for (strict) betweennessand E for equidistance. We emphasize that B is used for strict betweenness; asHilbert put it, if C is between A and B, then A, B, and C are three distinctpoints.

16Using more sorts makes the theory closer to practice, and using fewer sorts makes it easierto analyze metamathematically.


We use on (P,L) for the incidence of point P on line L, and On (P,C) for theincidence of point P on circle C. There is a complete list of the axioms of ECGin the Appendix, for reference.

ECG has six basic function symbols of type Point, shown here with argu-ments:

IntersectLines (L,K)

IntersectLineCircle1 (L,C)

IntersectLineCircle2 (L,C)

IntersectCircles1 (C,K)

IntersectCircles2 (C,K)

IntersectCirclesSame (C,K, p)

IntersectCirclesOpp (C,K, p)

The intuitive meaning of these symbols has been discussed above. ECG doesnot have “overloaded” versions of these functions; in other words, we just write

IntersectLines (Line (A,B),Line (P,Q))

instead of having an overloaded version of IntersectLines that takes four points.

8.4. Intuitionistic logic and stability. Our underlying logic is intuition-istic. We first give the specifically intuitionistic parts of our theory, which arevery few in number. We do not adopt decidable equality, nor even the substituteconcept of “apartness” introduced by Heyting (and discussed above), primarilybecause we aim to develop a system in which definable terms (constructions) de-note continuous functions, but also because we wish to keep our system closelyrelated to Euclid’s geometry, which contains nothing like apartness. Our firstfour axioms express our intuition that there is nothing asserting existence in themeaning of equality or incidence; hence assertions of equality or incidence can beconstructively proved by contradiction. Technically, the word “stable” is appliedto a predicate Q if ¬¬Q → Q. The following axioms assert the stability ofequality, equidistance, and incidence:

¬¬x = y → x = y (Axiom S1)¬¬E(a, b, c, d) → E(a, b, c, d) (Axiom S2)¬¬on (p, L) → on (p, L) (Axiom S3)¬¬On (p, C) → On (p, C) (Axiom S4)

We also want stability for betweenness. Since we use non-strict betweenness,this stability axiom has a different character, and it will be discussed as a be-tweenness axiom below. Nevertheless we group the axiom here as a “stabilityaxiom” since it disappears with classical logic:

¬¬B(x, y, z) → B(x, y, z) (Axiom S5)

This axiom is also called “Markov’s principle for betweenness”, as we willdiscuss below.

One might consider the following candidate for an axiom (schema): for everyterm t,

¬¬t ↓ → t ↓ (S6, not an axiom)

46 MICHAEL BEESON

For example, if two lines cannot fail to intersect, then they do in fact intersect;or if two circles cannot fail to intersect, then they intersect. This seems somewhatsimilar to Markov’s principle: we just follow the lines until they intersect, andthey “must”, just as a Turing machine that cannot fail to halt “must halt”. Butof course, as with Markov’s principle, that argument is circular. If we could notprove S6, we would need to adopt it as an axiom. However, it turns out that wecan prove S6 in ECG, so we do not need to add it as an axiom. See Theorem10.16 for the proof; we cannot give it here since it relies on developing enoughgeometry in ECG to prove geometrical equivalents for the definedness of eachkind of term.

8.5. Incidence and intersection axioms. There are two incidence rela-tions, for on (P,L) for points lying on lines, and On (P,C) for points lying onand circles. We start with three “incidence axioms”:

a 6= b → on (a,Line (a, b)) (Axiom I1)a 6= b → on (b,Line (a, b)) (Axiom I2)on (p, L) ∧ on (q, L) ∧ p 6= q ∧ on (r,Line (p, q)) → on (r, L) (Axiom I3)

Axiom I3 can be paraphrased,“two points determine a line.”The following axioms express the meaning of the five main function symbols of

ECG. They do not, however, make any assertions of geometrical content, exceptthat two lines intersect in at most one point. In particular they do not say thatany intersection point of two circles is one of the two particular ones given bythe function symbols; that would involve a disjunction, which we wish to avoid;and they do not say that a term meant to define an intersection point of twocircles is actually defined, unless we already have a point on both circles. Theaxioms that say intersection points are defined are given in another section; inthis section we just say that if these terms are defined, they denote intersectionpoints.

For readability, the first five of these axioms introduce a name “p” for anintersection point; this extra variable, and the implication symbol in the axioms,can easily be eliminated by substituting the term for p; one would wish to dothis if using the axioms for automated deduction.

p = IntersectLines (L,K) → on(p, L) ∧ on (p,K) (Axiom I4)IntersectLines (L,K) ∼= IntersectLines (K,L) (Axiom I5)

(The symbol ∼= is defined in Definition 8.1.)p = IntersectLineCircle1 (L,C) → on (p, L) ∧ On (p, C) (Axiom I6)p = IntersectLineCircle2 (L,C) → on (p, L) ∧ On (p, C) (Axiom I7)p = IntersectCircles1 (C,K) → On (p, C) ∧ On (p,K) (Axiom I8)p = IntersectCircles2 (C,K) → On (p, C) ∧ On (p,K) (Axiom I9)p = IntersectCirclesSame (C,K, x) → On (p, C) ∧ On (p,K) (Axiom I10)

∧¬(B(p, y, x) ∧ on(y,Line (center(C), center(K))))

Axiom I10 says that the (open) segment px does not meet the line L joining thecenters. The next axiom, about IntersectCirclesOpp , specifies that the segmentpx does meet L; but it is allowed to meet it at p, which we need in case thecircles are tangent.


p = IntersectCirclesOpp (C,K, x) → On (p, C) ∧ On (p,K)∧ (Axiom I11)¬ (¬B(p, IntersectLines (

Line (p, x),Line (center(C), center(K)), x)∧p 6= IntersectLines (Line (p, x),Line (center(C), center(K)))

on (p, L) ∧ on (p,K) ∧ on (q, L) ∧ ¬on (q,K) (Axiom I12)→ IntersectLines (L,K) ↓

IntersectLines (L,K) ↓ ∧on (p, L) (Axiom I13)∧on (p,K) → p = IntersectLines (L,K)

on (p, L) ∧ On (p, C) → IntersectLineCircle1 (L,C) ↓ (Axiom I14)on (p, L) ∧ On (p, C) → IntersectLineCircle2 (L,C) ↓ (Axiom I15)On (p, C) ∧ On (p,K) → IntersectCircles1 (C,K) ↓ (Axiom I16)On (p, C) ∧ On (p,K) → IntersectCircles2 (C,K) ↓ (Axiom I17)

We want the intersection points of circles C andK to be undefined when circlesC and K coincide (have the same points). The following axioms so specify.

IntersectCircles1 (C,K) ↓ → center(C) 6= center(K) (Axiom I17)IntersectCircles2 (C,K) ↓ → center(C) 6= center(K) (Axiom I18)IntersectCirclesSame (C,K, x)) ↓ → center(C) 6= center(K) (Axiom I19)IntersectCirclesOpp (C,K, x)) ↓ → center(C) 6= center(K) (Axiom I20)

Similarly, we want the intersection point of two lines to be undefined whenthe lines coincide.

IntersectLines (L,K) ↓ → (Axiom I21)¬(on (pointOn1 (K), L) ∧ on (pointOn2 (K), L))

8.6. Constructor and accessor axioms. Since we allow Circle3 , we donot need Circle (x, y) as an official part of the language. We regard Circle (x, y)as an abbreviation:

Definition 8.4. Circle (x, y) := Circle3 (x, x, y)

There are function symbols corresponding to the constructor and accessorfunctions for each of the sorts. The argument and value types of these symbolsare obvious, and hence not specified here. Here are the axioms relating theconstructors and accessors.

Line (pointOn1 (L), pointOn2 (L)) = L (Axiom CA1)A 6= B → pointOn1 (Line (A,B)) = A (Axiom CA2)A 6= B → pointOn2 (Line (A,B)) = B (Axiom CA3)¬Line (A,A) ↓ (Axiom CA4)center (Circle3 (a, b, c)) = a (Axiom CA5)Circle (center (C), pointOnCircle (C)) = C (Axiom CA6)Circle3 (A,B,C) ↓ (Axiom CA7)

Note that Axiom CA7 allows for “degenerate” circles, i.e. circles of zero radius.We note that it is not necessary to have as an axiom that

pointOn1 (L) 6= pointOn2 (L)

since it can be derived; namely if A = pointOn1 (L) and B = pointOn2 (L) thenL = Line (A,B) by Axiom CA1, so Line (A,B) ↓, so by Axiom CA4 , A 6= B.

48 MICHAEL BEESON

Note that we do not have pointOnCircle (Circle (A,B)) = B. Even thoughwe have On(B,Circle (A,B)), not every circle is constructed as Circle (A,A,B);and in general we need pointOnCircle (C) to depend continuously on circle C,however C is given. For example, in models we could take pointOnCircle (C) tobe the “southernmost point” of circle C, where “south” is given by a ray fromthe center of C parallel to a certain fixed ray. Since we want all our constructionsto be continuous, we will need some axiom to guarantee that pointOnCircle (C)is continuous in the circle C; that axiom will involve the equidistance (or “con-gruence” relation) and will be given below.

8.7. Meaning of equality. Lines in ECG have direction. Line (A,B) isnot in general equal to Line (B,A). We say they “have the same points”. Lo-gicians call this “extensional equality”. We use the equality symbol betweenpoints to mean “identically equal”. Between lines, equality means “intensionalequality”. In the spirit of constructive mathematics, lines “come equipped” withtwo associated distinct points. Thus, Line (A,B) = Line (P,Q) if and only ifA = B and P = Q. This does not need to be assumed, as it follows from theaxioms given above for the accessor and constructor functions. It may, however,be confusing to those not accustomed to constructive mathematics. The notionof “extensional equality” refers to the defined relation between two lines, thatthe same points are on both lines. In practice, to avoid confusion, we rarelyif ever mention equality between lines. It should be noted, however, that ourtheory does depend on this view of lines, since the order of the two intersec-tion points of line Line (A,B) with circle C is opposite to the order of the twointersection points of Line (B,A) with C. In other words, when consideringIntersectLineCircle1 (L,C), it is essential that L is given by two points.

Unlike lines, circles do not come equipped with a direction. Two circles hav-ing the same points are considered equal. We will state this below, when theequidistance (or “congruence”) relation is introduced. This is not essential; wecould introduce a new symbol for extensional equality of circles, and keep theequality symbol to mean intensional equality; but we have no need of intensionalequality of circles for geometry; once a circle is constructed, we don’t care howit was constructed. By contrast, we do care about the direction of a line. Wedon’t care in general about the specific two points used to construct a line, so wecould identify two lines with the same direction, but there is no pressing reasonto do so.

8.8. Betweenness axioms. The basic relations of our theories are equidis-tance and betweenness, which have been recognized as fundamental at least sinceHilbert’s famous 1899 book [15]. All the arguments of these two relations havesort Point. The (strict) betweenness relation is written B(a, b, c).17 We read this“b is between a and c”. The intended meaning is that that the three points arecollinear and distinct, and b is the middle one of the three.

17Hilbert [15] and Greenberg [12] use strict betweenness, as we do. Tarski [30] used non-strict betweenness.


B(a, b, c) → on(b,Line (a, c)) (Axiom B1-i)B(a, b, c) → a 6= c (Axiom B1-ii)B(a, b, c) → a 6= b (Axiom B1-iii)B(a, b, c) → B(c, b, a) (Axiom B1-iv)

Before giving a constructive version of the remaining betweenness axioms,we discuss a related principle, “Markov’s principle for betweenness”, which hasalready been listed as Axiom S5:

¬¬B(A,B,C) → B(A,B,C)

Markov’s principle expresses the idea that by computing two points P and Q etc.to greater and greater accuracy, if they are not identical we will eventually findthat out. We want it to be provable in ECG for several reasons: it is used inEuclid (e.g. I.6 and I.26, as is discussed below); it is needed to permit argumentby cases for a conclusion that is a conjunction of equalities and betweennessstatements; and without such arguments we could not make significant headwayin geometry. Besides, it makes for a smooth metatheory. While some may con-sider Markov’s principle in number theory to be of questionable constructivity,we consider that geometry without Markov’s principle is awkward. Specifically,it would not support Euclid. As we shall see in other sections, Euclid does useit, and including it does not harm our ability to construct things that are provedto exist. In terms of order, it expresses the principle that ¬x ≤ 0 → x > 0.

Hilbert’s second betweenness axiom says that Line (P,Q) contains a pointbetween P and Q. We call this the “density” axiom. Since we want a quantifier-free axiomatization, we would like to specify the point asserted to exist. Thenatural candidate for a point between P and Q is the result of Euclid’s segment-bisection construction. We therefore take the following axiom:

B(P, IntersectLines (Line(P,Q), Line( (Axiom B2)IntersectCircles1 (Circle(P,Q), Circle(Q,P )),IntersectCircles2 (Circle(P,Q), Circle(Q,P )))), Q)

Note that, in view of the strictness axioms of LPT, this axiom implies thatthe circles and intersection points involved are defined.

Hilbert’s second axiom for betweenness says that between two distinct pointson a line there is another point. His third axiom for betweenness is, “Of anythree points on a line there exists no more than one that lies between the othertwo.”18 Here is a constructive version:

a 6= b ∧ a 6= c ∧ b 6= c ∧ on(a, L) ∧ on(b, L) ∧ on(c, L) → (Axiom B3)(¬B(a, b, c) ∧ ¬B(b, c, a) → ¬¬B(c, a, b))∧(¬B(b, c, a) ∧ ¬B(c, a, b) → ¬¬B(a, b, c))∧(¬B(c, a, b) ∧ ¬B(a, b, c) → ¬¬B(b, c, a))∧¬(B(a, b, c) ∧ B(b, c, a)) ∧ ¬(B(a, b, c) ∧ B(b, a, c))∧¬(B(b, c, a) ∧ B(b, a, c))

18In Greenberg’s system this becomes the slightly different “given three distinct points on aline, one and only one of the points is between the other two.” That formulation is too strong,constructively. (For example its translation into the recursive plane is not provable in HA plusMarkov’s principle.)

50 MICHAEL BEESON

In view of Markov’s principle for betweenness (Axiom S5), we could removethe double negations on the right. If we did so, then Axiom S5 would becomesuperfluous, as is shown in the following lemma. But we prefer to keep AxiomS5 and the Axiom B3 just given, because it facilitates the discussion of differ-ent versions of Euclid’s parallel postulate, whose relations depend on Markov’sprinciple.

Lemma 8.5. Let B3* denote Axiom B3 with the double negations removed.Markov’s principle for betweenness (Axiom S5) is provable from Axiom B3*.

Proof. Suppose ¬¬B(a, x, b). We want to prove B(a, x, b). From ¬¬B(a, x, b)we immediately have a 6= b and a 6= x and x 6= b. By Axiom B3* it suffices toprove ¬B(x, a, b) and ¬B(a, b, x). To prove ¬B(x, a, b), suppose B(x, a, b). Then¬B(a, x, b), by Axiom B3*. But that contradicts ¬¬B(a, x, b). That proves¬B(x, a, b). Similarly we have ¬B(a, b, x). That completes the proof of thelemma.

19

Hilbert’s fourth betweenness axiom is often known as Pasch’s axiom. It saysthat if line L meets (the interior of) side AB of triangle ABC then it meets (theinterior of) side AC or side BC as well. Constructively we had better say thatif it meets AB and does not meet AC, then it meets BC. Note that the mostnatural way of writing this does not further specify (as Hilbert did) that thethree points A, B, and C should not be collinear. We do not want to do thatconstructively since generally we can’t decide whether three points are or or notcollinear, and the Pasch axiom is uniformly valid, so there is no pressing reasonto require the three points to not be collinear.

Tarski refined Pasch’s axiom, considering three forms, “inner Pasch”, “outerPasch”, and “weak Pasch”, of which only the first is an axiom of ECG. Thepoint of Tarski’s refinements seems to be twofold: (i) these restricted versionscan be expressed in an AE form, which become quantifier-free when we havethe function symbol IntersectLines (this is not true of the unrestricted Paschaxiom), and (ii) the restricted forms are valid even in three-dimensional space,so they do not make an implicit dimensional assertion, as the unrestricted Paschaxiom does (it fails in three-space). The version we take is similar to the “innerPasch” Axiom 7 of [26]; it differs only in replacing an existential by a term usingIntersectLines for the point asserted to exist. See Fig. 9.

B(a, p, c) ∧ B(b, q, c) ∧ p 6= b ∧ q 6= a ∧ ¬ on(c,Line (a, q) →IntersectLines (Line (p, b),Line (a, q)) ↓ ∧B(p, IntersectLines (Line (p, b),Line (a, q)), b) ∧B(q, IntersectLines (Line (p, b),Line (a, q)), a) (Pasch’s Axiom B4)

This version differs from [26] in that the point x such that B(p, x, b)∧B(q, x, a)is explicitly given, instead of having the axiom say that some x exists with the

19Hilbert did not need this axiom because he had his “plane separation axiom.” It is derivedfrom the plane separation axioms as Greenberg’s Prop. 3.3 [12], p. 112. Tarski used non-strictbetweenness, here called T , and his version of this axiom was T(a, b, a) → a = b; it is Axiom6 in [30] and [26].


Figure 9. Inner Pasch (left) and Outer Pasch (right). Line pbmeets triangle acq in one side. The open circles show the pointsasserted to exist on the other side.

b

a xb q

bc

b b

bp

b

ab

qx

bb

b c

b

p

desired betweenness properties. Also, we use strict betweenness instead of non-strict betweenness, and the hypothesis means that the lines whose intersectionpoint is x must be defined and must not coincide, since then x would not bedefined. We will see that Tarski’s version, with non-strict betweenness, is notconstructively valid. For now let it suffice to observe that as b goes to q and pgoes to a, the position of x could be anywhere between a and q, depending howclose b is to q and p is to a, so x cannot be constructed continuously as wouldbe required for a constructive version of this axiom with non-strict betweenness.

The inner Pasch axiom can be used to derive the outer and weak Paschaxioms, as we shall see; and as Tarski, Szmielew, Gupta, and Schwabhausershowed, Tarski’s version, with non-strict betweenness, can be used with the ax-iom ¬B(a, b, a) to derive our axioms B3 and B1-iii and B1-iv; but that is nottrue with strict betweenness.

8.9. Sides of a line. Two points a and b not on line L are on opposite sidesof L if a 6= b and there is a point of L between a and b, i.e., the segment ab meetsL.

Definition 8.6.

OppositeSide (a, b, L) := B(a, IntersectLines (Line (a, b), L), b)

The definition of being on the same side is less straightforward. One obviousdefinition of SameSide (a, b, L) would be that segment ab does not meet L. Wehave to ensure that a point not on L is on the same side of L as itself, though,without implying that we can test whether two points are equal, so we wouldhave to say

∀x¬(B(a, x, b) ∧ on(x, L)).

If we try to make this quantifier-free by replacing x by IntersectLines (Line (a, b), L)we have to watch out for the case a = b. It does not work to say

¬(a 6= b ∧B(a, IntersectLines (Line (a, b), L), b))

52 MICHAEL BEESON

because (intuitively) the truth value of B(a, IntersectLines (Line (a, b), L), b) isindeterminate when a = b. More precisely, when we put a = b in this formula,the result will be equivalent to

¬B(a, IntersectLines (Line (a, a), L), a),

which contains an undefined term and is not provable, even though ∀x¬B(a, x, b)is derivable.

One can get an existential quantifier instead of a univeral quantifier by usingTarski’s definition, as given in [26]. This says that a and b are on the same sideof L if there is some c such that both a and b are on the opposite side of Lfrom c. Another advantage of this definition is that it works in more than twodimensions.

Definition 8.7.

SameSide (a, b, L) := ∃c (B(a, IntersectLines (Line (a, c), L, c) ∧B(b, IntersectLines (Line (b, c), L), c))

The question of the equivalence of this definition with the first candidate willbe addressed in Lemma 9.83 below. Evidently the upper dimension axiom willbe needed.

8.10. Right and Left : handedness and sides of lines. Consider two cir-cles C and K with different centers a and b respectively. They intersect (if atall) in two points p (which may coincide if the circles are tangent). If they do notcoincide then we define abp to be a “right turn” if p = IntersectCircles1 (C,K):

Definition 8.8.

Right (a, b, p) := p = IntersectCircles1 (Circle (a, p),Circle (b, p))

Left (a, b, p) := p = IntersectCircles2 (Circle (a, p),Circle (b, p))

Note that this counts abb as both a right turn and a left turn, but aab and aaaare neither right turns nor left turns, since the terms on the right are undefinedin those cases by Axioms I17 and I18.

We then take as axioms the statements that if abp is a right turn and abq isa right turn, and p and q are not on Line (a, b), then no point of that line isbetween p and q, and similarly for left turns:

Right (a, b, p) ∧ Right (a, b, q) ∧ ¬on(p,Line (a, b)) ∧ ¬on(q,Line (a, b)) →¬(on(x,Line (a, b)) ∧ B(p, x, q)) (Axiom H1)

Left (a, b, p) ∧ Left (a, b, q) ∧ ¬on(p,Line (a, b)) ∧ ¬on(q,Line (a, b)) →¬(on(x,Line (a, b)) ∧ B(p, x, q)) (Axiom H2)

Similarly, if abp is a right turn and abq is a left turn, and p and q are not onLine (a, b), then p and q are on opposite sides of Line (a, b):

Right (a, b, p) ∧ Left (a, b, q) ∧ ¬on(p,Line (a, b)) ∧ ¬on(q,Line (a, b)) →B(p, IntersectLines (Line (p, q),Line (a, b)), q) (Axiom H3)

Conversely, if p and q are on the same side of Line (a, b), then the turns abpand abq have the same sense. Here we use the definition of SameSide given


above. That will put an existential quantifier in the antecedent of an impli-cation; writing it out more explicitly the axiom will be quantifier-free. ThusSameSide (p, q,Line (a, b)) → φ becomes

B(p, x, y) ∧B(q, z, y) ∧ on(x,Line (a, b)) ∧ on (z,Line (a, b) → φ.

Applying this procedure to

SameSide (p, q,Line (a, b)) ∧ Right (a, b, p) → Right (a, b, q)

and to the similar formula with Left instead of Right, we have the next twoaxioms (see Fig. 10):

Figure 10. (Axiom H4) abq is a right turn if and only if abp isa right turn. Points x, y, and z witness that p and q are on thesame side of Line (a, b).

ba

bb

b

qb

p

b y

b

xbz

B(p, x, y) ∧ B(q, z, y) ∧ on(x,Line (a, b)) ∧ on (z,Line (a, b)∧Right (a, b, p) → Right (a, b, q) (Axiom H4)

B(p, x, y) ∧ B(q, z, y) ∧ on(x,Line (a, b)) ∧ on (z,Line (a, b)∧Left (a, b, p) → Left (a, b, q) (Axiom H5)

The next two axioms say that if p and q are on opposite sides of Line (a, b), thenthe turns have the opposite sense:

B(p, q, x) ∧ on (x,Line (a, b)) ∧ Right (a, b, p) → Left (a, b, q) (Axiom H6)B(p, q, x) ∧ on (x,Line (a, b)) ∧ Left (a, b, p) → Right (a, b, q) (Axiom H7)

The following axioms connects the turns in a triangle, or technically, in a list ofthree distinct points. Note that Right (a, b, c) implies a 6= b, as does Left (a, b, c),since it is defined using the intersection points of circles with centers a and b,and Axiom I17 and A18 guarantee that these intersection points are not definedfor concentric circles. That is why we do not need the hypothesis a 6= b in thenext four axioms.

Right (a, b, c) ∧ b 6= c → Right (b, c, a) ∧ Right (c, a, b) (Axiom H8)Left (a, b, c) ∧ b 6= c → Left (b, c, a) ∧ Left (c, a, b) (Axiom H9)Right (a, b, c) → Left (a, c, b) (Axiom H10)Left (a, b, c) → Right (a, c, b) (Axiom H11)

54 MICHAEL BEESON

Note that it is not necessary to specify that a, b, and c are not collinear in theabove axioms, since in that case all the turns are both left and right.

There are three constants α, β, and γ in the theory. The following axiomresolves an ambiguity that otherwise would exist (which one could think of asan ambiguity about from which side we are viewing “the plane”).

Left (α, β, γ) (Axiom H12)

Nothing in these axioms prevents the three points from being collinear, butthat will be ruled out by the dimension axioms below.

8.11. Dimension axioms. There are three constants α, β, and γ with ax-ioms saying that these three points are non-collinear. Specifically

¬on (α,Line (β, γ)) (Axiom D1)¬on (β,Line (α, γ)) (Axiom D2)¬on (γ,Line (α, β)) (Axiom D3)α 6= β ∧ α 6= γ ∧ β 6= γ (Axiom D4)

We do not have to say explicitly that α ↓ because it is part of the logic of partialterms that every constant is defined–it is nothing special to any particular theory.Axiom D4 is needed since without it there is nothing to rule out the possibilitythat α = β = γ; in that case Line (β, γ) would be undefined, so on(α,Line (β, γ))is false, and axiom D1 would still hold.

One also needs an “upper dimension axiom” to ensure that models of thetheory are planes.20 We take the one chosen by Tarski, which says that if threecircles with different centers have a common chord, their centers are collinear:

On(a, C1)) ∧On(b, C1) ∧On(a, C2) ∧On(b, C2) (Axiom D5)∧ On(a, C3) ∧On(b, C3) ∧ a 6= b ∧ center (C1) 6= center (C2)∧ center (C1) 6= center (C3) ∧ center (C2) 6= center (C3)

→ On(center (C3),Line (center (C1), center (C2)))

8.12. Rays and segments. Although we have not included rays and seg-ments in ECG, we do want to support our claim that a conservative extensionincluding them can easily be introduced; and also, we sometimes make informalarguments using rays with the implication that they can be formalized in ECG.We now show that the use of intuitionistic logic does not cause a problem aboutincidence on rays or segments. Using betweenness, we can define incidence forrays. However, there is a technicality: the origin O of Ray (O,B) is consideredto lie on the ray, i.e. “rays are closed”, while betweenness means “strictly be-tween.” It is thus easier to define the “opposite ray”: P is on the opposite rayto Ray (O,B) if P is on Line (O,B) and O is between P and B. Then Q is onRay (O,B) if it is on Line (O,B) but not on the opposite ray:

on (Q,Ray(O,B)) := on (Q,Line(O,B)) ∧ ¬B(P,O,B) (Definition 4)

This definition can be used to express informal arguments about rays in ECGwithout needing to introduce an explicit sort and axioms for rays.

20With classical logic and full first-order continuity in place of line-circle continuity, one cantake for such an axiom essentially any formula that holds in R

2 but not in Rn for n > 2, as

shown by Scott [27]. Whether this is true also with only line-circle continuity and classicallogic, let alone with intuitionistic logic, I do not know.


In a similar way we could define incidence for segments, so that “segments areclosed”, but we do not do so.

8.13. Congruence axioms. The equidistance relation is officially writtenE(A,B,C,D). Axioms for equidistance are sometimes called “congruence ax-ioms” since equidistance can be thought of as congruence of segments. Fol-lowing a tradition that goes back to Euclid, we write AB = CD instead ofE(A,B,C,D). Hilbert’s first congruence axiom, paraphrased from [12], is closelyrelated to the uniform version of Euclid’s Book I, proposition 2:

A 6= B ∧ C 6= D → (not an axiom of ECG)∃R(on (R,Ray(A,B)) ∧AR = CD

This axiom permits us to “lay off” segment CD along Ray (A,B). Since we areseeking a quantifier-free axiomatization, we want to specify the point R. Thiswe do by taking

R(A,B,C,D) = IntersectLineCircle2 (Line (A,B),Circle3 (A,C,D)).

It should be IntersectLineCircle2 rather than IntersectLineCircle1 because weintend that those two intersection section points should occur on Line(A,B) inthe same order as A and B. Since we do not take Ray as a primitive of the lan-guage, we need to eliminate it, replacing on (R,Ray(A,B)) by on(R,Line (A,B)∧¬B(R,A,B). Since we allow degenerate circles, we drop the hypothesis C 6= Das well. Our version of the axiom is thus

a 6= b → on (R(a, b, c, d), Line(a, b))

∧¬B(R(a, b, c, d), a, b) ∧ E(a,R(a, b, c, d), c, d)

Note that the strictness axioms then will imply that R(a, b, c, d) is defined whena 6= b. It is allowed that c = d since we are allowing degenerate circles. Theofficial version, with R replaced by is definition, is Axiom C1:a 6= b → (Axiom C1)

(on (IntersectLineCircle2 (Line (a, b),Circle3 (a, c, d)), Line(a, b))∧ ¬B(IntersectLineCircle2 (Line (a, b),Circle3 (a, c, d)), a, b)∧ E(a, IntersectLineCircle2 (Line (a, b),Circle3 (a, c, d)), c, d)

We do not need to require the uniqueness of R in the axiom, since we can provethe uniqueness from other axioms; this will be demonstrated in Lemma 9.4 below.

Just below are four more congruence axioms, adapted from Tarski’s system.The first two turn out be enough to guarantee that congruence of segments isan equivalence relation (which we prove in a subsequent section); Hilbert statedthat in his axioms. Note that Axiom C2 is not exactly transitivity of congruence,although it has a similar flavor. Hilbert did not need C3 (which was introducedby Tarski), since he did not allow degenerate segments; and he did not need C4since he treated segments as (unordered) pairs of points. In our theory, ab = cdis a 4-ary relation on points, so we need this axiom. Axiom C3 can be thoughtof as saying that addition is well-defined on congruence classes of segments. Wetherefore refer to it as the “additive congruence axiom.”

ab = cd ∧ ab = pq → cd = pq (Axiom C2)ab = cc → a = b (Axiom C3)ab = ba (Axiom C4)

56 MICHAEL BEESON

One might take the additivity of congruence

B(a, b, c) ∧ B(p, q, r) ∧ ab = pq ∧ bc = qr → ac = pr

as an axiom, except that it would be redundant to do so, as will be shown inLemma 9.3 below.

Using equidistance, we define incidence for circles:

On (x,Circle3 (a, b, c)) ↔ ax = bc (Axiom C5)

Since we are allowing circles of zero radius, we do not need to specify A 6= Q.We need Axiom C3 to guarantee that the only point on a circle of zero radius isits center.

Now to complete the list of congruence axioms, we have to choose whether tofollow Hilbert or Tarski. Hilbert’s axioms are perhaps closer in spirit to Euclid,though Euclid did not perceive the necessity of any congruence axioms beyondwhat has already been stated. (He tried to prove SAS, using an argument by“superposition”.) In [3] we followed Hilbert, but here we choose to follow Tarski,as that enables us to complete the list of axioms immediately, and separate thediscussion of angles and triangles into another section; and it also makes it easierfor us to make use of the results of [26], which is based on Tarski’s axioms. Thecrucial axiom is Tarski’s 5-segment axiom, which is illustrated in Fig. 8 anddiscussed near that figure.

We take the following variant of Tarski’s 5-segment axiom as our last congru-ence axiom:

a 6= b ∧ B(a, b, c) ∧ B(A,B,C) ∧ ab = AB ∧ bc = BC (Axiom C6)∧ad = AD ∧ bd = BD → cd = CD

Note that Axiom C6 does not allow a = b or b = c, but it does allow d to lie onLine (a, b). Constructively this axiom may seem inadequate, because we wouldlike the length of cd to be determined even when c is between a and b, and withouthaving to make a case distinction about that; but we will show in Lemma 9.10that an extended version of the axiom is already provable. In essence, one canargue by cases to prove a congruence relation, because congruence is stable.

8.14. Line-circle continuity. We now formulate the axioms of line-circlecontinuity. The first two of these just tell us when a line and a circle intersect–namely, when there is a point on the line closer (or equally close) to the centerthan the radius of the circle.21 But we have not defined inequalities for segmentsyet, so the formal statement is a bit more complex. Moreover, we have to includethe case of a degenerate circle or a line tangent to a circle, without making a casedistinction.22 Therefore we must find a way to express “P is inside the closedCircle (A, Y )”. For that it suffices that there should be some X non-strictlybetween A and Y such that AX = AP . Since this will appear in the antecedentof the axiom, the “some X” will not involve an existential quantifier. Nonstrictbetweenness T can be defined, since we have Markov’s principle for betweenness:

21Note that in spite of the use of the word “circle” the axiom is valid in n-dimensionalEuclidean space, where it refers to the intersections of lines and spheres.

22Avigad et. al. count only transverse intersection, not tangential intersection, as“intersection.”


Definition 8.9. Non-strict betweenness is defined by

T(a, b, c) := ¬(a 6= b ∧ b 6= c ∧ ¬B(a, b, c))

This disjunction-free definition is equivalent to

¬¬(a = b ∨ b = c ∨ B(a, b, c))

Definition 8.10. ab < cd (or cd > ab) means ∃x(B(c, x, d) ∧ ax = ab).ab ≤ cd (or cd ≥ ab) means ∃x(T(c, x, d) ∧ ax = ab), where T is non-strict

betweenness.

Definition 8.11. Let C be a circle with center a. Then point p is strictlyinside C means there exists a point b on C such that ap < ab, and p is insideC, or non-strictly inside C, means ap ≤ ab.

We could equivalently say “for every b on C” or “for b = pointOnCircle (C)”.The first axiom for line-circle continuity will then say that if line L has a point

p inside circle C, then both intersection points of the line L and Circle (A, Y )are defined.

We first express this in terms of T. See Fig. 11

Figure 11. Line-circle continuity. p is inside the circle, so Lmeets the circle.

ba

L

bb b

x

b

p

on(p, L) ∧ ax = ap ∧T(a, x, b) →IntersectLineCircle1 (L,Circle (a, b)) ↓IntersectLineCircle2 (L,Circle (a, b)) ↓

Now we express this in terms of the primitives of the language, eliminating theabbreviation T:

on(p, L) ∧ ax = ap ∧ ¬(a 6= x ∧ x 6= b ∧ ¬B(a, x, b)) →IntersectLineCircle1 (L,Circle (a, b)) ↓ (Axiom Cont 1)IntersectLineCircle2 (L,Circle (a, b)) ↓

58 MICHAEL BEESON

Our next axiom says that if L contains a point strictly inside C then the twointersection points are distinct.

on(p, L) ∧ ax = ap ∧ ¬(a 6= x ∧ ¬B(a, x, b)) ∧ ax = ap →IntersectLineCircle1 (L,Circle (a, b)) 6= (Axiom Cont 2)IntersectLineCircle2 (L,Circle (a, b))

This axiom deserves some discussion. Consider the propositions (i) if L is aline and e is a point not on L then there is a perpendicular from e to L, and (ii)if L is a line containing a point x strictly inside circle C, then L meets C in atleast two points. It is not hard to prove (i) from (ii), but if we follow Euclid, weneed (ii) for (i). Also (i) can be proved from the “triangle existence theorem”(Lemma 9.67 below): let a and b be two points on L, and construct triangle abEcongruent to abe with E on the other side of L from E; then eE ⊥ L. The triangleexistence theorem is an axiom in Hilbert’s system, so with Hilbert’s axioms wecan dispense with Axiom Cont 2. Actually, Axiom Cont 2 is superfluous, since(i) can also be proved in Tarski’s system, which we are following here, withoutappealing to line-circle continuity at all. This development is presented in [26]and is completely constructive. However, it is long and complicated, and farfrom Euclid. We have therefore chosen to include this “harmless little axiom”,and follow Euclid in the construction of perpendiculars.

Our next axiom says that the intersection points depend extensionally on thecircle:

A = center (C) ∧A = center (K)∧ (Axiom Cont 3)On (P,C) ∧ On (Q,K) ∧AP = AQ →

IntersectLineCircle1 (L,C) ∼= IntersectLineCircle1 (L,K)∧IntersectLineCircle2 (L,C) ∼= IntersectLineCircle2 (L,K)

Note that these axioms do not guarantee the continuous dependence (on thepoint and circle) of the intersection points defined by each of the two functionsymbols. In other words, we have not yet specified which of the two intersectionpoints is which. We want to say that the two intersection points occur in thesame order on Line (a, b) as a and b occur. That is done by the following axioms:

¬on(z, L) ∧Right (z, a, b) → Right (z, IntersectLineCircle1 (Line (a, b), C),IntersectLineCircle2 (Line (a, b), C)) (Axiom Cont 4)

¬on(z, L) ∧ Left (z, a, b) → Left (z, IntersectLineCircle1 (Line (a, b), C),IntersectLineCircle2 (Line (a, b), C)) (Axiom Cont 5)

The order of the arguments is important here, because if the line is tangent tothe circle, the two intersection points will coincide, so we may have Right (p, a, a),which is true, but Right (a, a, p) is always false. Note that the axiom is still correctif the two intersection points coincide, since pxx is both a right and a left turn.

8.15. Intersections of circles. We next give the axiom(s) known as line-circle continuity. The axioms in question should say that if point p on circleK lies(non-strictly) inside circle C, and point q lies (non-strictly) outside C, then bothintersection points of the circles are defined. (Incidence axioms already giventhen guarantee they are on both circles.) To express that p is non-strictly inside(or outside) Circle (A, Y ) we use the same technique as just above. Namely, p


is inside C if ap = ax for some X non-strictly between a and a point y on C.The situation is illustrated in Fig. 12. We need this axiom to apply even todegenerate circles, and to points that are on C rather than strictly inside, so wecannot say B(x, y, z) as we must allow x = y or y = z, and we must even allowa = x = y = z; hence the point w in the axiom is needed so that we definitelyhave a line connecting a and w on which x, y, and z must lie, non-strictly inthat order. The axiom expresses this using the primitive relation B.

Figure 12. Circle-circle continuity. p is inside C and q is out-side C, as witnessed by x, y, and z, so the intersection points 1and 2 exist. ab1 is a right turn and ab2 is a left turn.

b

bb

a

b

z

b

w

by

b

x

b

q

b

p

2

1

KC

ap = ax ∧ aq = az ∧ on (q,K) ∧ on (p,K) ∧ on (y, C) (Axiom Cont 6)∧ ¬B(y, x, z) ∧ ¬B(x, z, y) ∧ B(a, x, w) ∧ B(a, y, w) ∧ B(a, z, w) →

IntersectCircles1 (C,K) ↓ ∧ IntersectCircles2 (C,K) ↓

In Section 8.10, we have already provided axioms to distinguish the two pointsof intersection, according as the triple formed by the two centers and the pointof intersection is a right turn or a left turn.

The axiom for IntersectCirclesSame is similar (but does not involve Right andLeft):

On(P,C) ∧On(Q,C) ∧AP = AX ∧AQ = AZ∧ (Axiom Cont 7)on(X,Line (A, Y )) ∧ on(Z,Line (A, Y ))∧¬B(X,A, Y ) ∧ ¬B(A, Y,X) ∧ ¬B(A,Z, Y ) ∧ ¬B(Z,A, Y ) ∧ Z 6= A∧ ¬B(A, Y,X) ∧AX = AP ∧On(P,C)∧ ¬on(x,Line (center(C), center(K))→ IntersectCirclesSame (C,K, x) ↓ ∧IntersectCirclesOpp (C,K, x) ↓

60 MICHAEL BEESON

We have already provided in Axiom I10 that once IntersectCirclesSame (C,K, x)is defined, it lies on the same (closed) side of the line joining the centers of C andK as x does, and IntersectCirclesOpp (C,K, x) lies on the opposite side. In otherwords, if the circles are tangent, and x is not on the line L joining the centers,then IntersectCirclesSame (C,K, x) is still defined. This axiom and I10 togetherdo not specify anything about what happens when x lies on L (since nothing isspecified, we cannot use IntersectCirclesSame unless we know x is not on L).

Our next axiom specifies that the intersection points of two circles dependextensionally on the circles:

A = center (C1) ∧A = center (C2)∧ (Axiom Cont 8)On (P,C1) ∧ On (Q,C1) ∧AP = AQ →

IntersectCircles1 (C1,K) ∼= IntersectCircles1 (C2,K)∧IntersectCircles2 (C1,K) ∼= IntersectCircles2 (C2,K)∧IntersectCircles1 (K,C1) ∼= IntersectCircles1 (K,C2)∧IntersectCircles2 (K,C1) ∼= IntersectCircles2 (K,C2)IntersectCirclesSame (C,K, x) ∼= IntersectCirclesSame (C,K, x)IntersectCirclesOpp (C,K, x) ∼= IntersectCirclesOpp (C,K, x)

This contrasts with the line-circle intersection operation, where if we changeLine (A,B) to Line (B,A), the two intersection points with a given circle switchlabels. That is absolutely necessary, as the following example shows: let line Lpass through the center of circle C. Then continuously rotate line L through 180degrees. That switches the two points of intersection, and changes Line (A,B) toLine (B,A). We definitely want the points of intersection to vary continuously, sowe cannot have them depend extensionally on the line. Consider what happenswhen we rotate circles instead. Start with two intersecting circles of the samesize, and rotate both circles 180 degrees, around the midpoint of the segmentconnecting their centers center of one around the center of the other. Then theintersection point that was originally at the top of the diagram is now at thebottom. But the circles have also switched places. So there is also a label-switching property here, but it involves switching the labels of the centers of thecircles as well as the labels of the intersection points. That is,

IntersectCircles1 (Circle3 (a, x, y),Circle3 (b, x, y)) =

IntersectCircles2 (Circle 3(b, x, y),Circle3 (a, x, y))

This does not contradict Axiom Cont 7.We need an axiom to guarantee the continuity of pointOnCircle . The intu-

itive idea is that pointOnCircle (C) should always be a point on C in a canonicaldirection (say “north”) from the center of C. But there are no “canonical direc-tions”, so we have to express continuity indirectly. Consider two circles C andK, and let p = pointOnCircle(C) and q = pointOnCircle (K), and let a and bbe the centers of C and K, respectively. If the two circles are the same size wecould require that pq = ab. If they are not the same size, then if t is a point onRay (b, q) with bt = ap, we require pt = ab. See Fig. 13 for an illustration. Wewant the axiom to guarantee that the figure is accurate, but to do that we alsoneed to specify that p and t are on the same side of Line (a, b) (if a 6= b). Ofcourse a = b is also allowed, and also all four points could be collinear.


Figure 13. Axiom Cont 9: if bt = ap then pt = ab and segmentpt does not meet ab.

b

a

b

p

b b

b

q

b t K

C

a = center (C) ∧ b = center (K) (Axiom Cont 9)∧ p = pointOnCircle (C) ∧ q = pointOnCircle (K)∧ ¬ (¬B(b, t, q) ∧ t 6= q ∧ b 6= t) ∧ ap = bt∧B(u, b, q) ∧ On (u,K) ∧ B(p, x, t)→ pt = ab ∧ ¬B(a, x, b)

In the figure, point v is IntersectLines (Line (p, u),Line (a, b)). The hypothesisB(u, b, q) means the axiom will not apply when K is a degenerate circle, but inthat case there is only one possibility for q anyway. The formula ¬ (¬B(b, t, q)∧t 6= q∧b 6= t) is just the translation into primitive symbols of T(b, t, q) (non-strictbetweenness).

§9. Development of neutral geometry in ECG. It will be necessaryto prove some basic theorems in ECG, in spite of the existence of the doublenegation interpretation and the work in [26]; we will explain this point in moredetail below. The word “neutral” here means that part of geometry not usingthe parallel axiom. In fact, for much of this development we also do not use theupper dimension axiom; it is used for the first time in section 9.14.

In this version of ECG, we have followed Tarski’s lead in choosing a small set ofshort, simple axioms for betweenness and congruence, while keeping the largerlanguage of points, lines, and incidence introduced by Hilbert. The detaileddevelopment of elementary geometry from Tarski’s axioms has been carried outin [26] in almost formal detail, so it would be nice if could use some of this workrather than going over similar territory but paying attention to the constructivityof the proofs. As mentioned above, we have used strict betweenness B insteadof non-strict betweenness T as Tarski did, and consequently we have a few morebetweenness axioms, and our other axioms, similar in form to Tarski’s, maybe weaker or different because they use B instead of T. However, in carryingout proofs, one notices repeatedly that one argues by cases, for example onB(a, b, c) ∨ B(b, c, a) ∨ B(c, a, b) for three distinct points on the same line, andthen one notes that since the double negation of this disjunction is valid, the

62 MICHAEL BEESON

double negation of whatever one has proved by cases is valid. If that conclusionis, for example, an atomic formula or conjunction of atomic formulas, then itsdouble negation implies itself, by the axioms of ECG.

We would haved liked to formulate a metatheorem about this phenomenonthat would have enabled the wholesale importation of a useful class of theoremsfrom [26]. The first subsection below explains why this cannot be made towork. The rest of the section gives proofs of some basic theorems of constructivegeometry in ECG(without using any parallel axiom). We avoid the parallelaxiom so that these results can be used in our study of the provability relationsbetween different versions of the parallel axiom, and so that they can be used(in subsequent work) to study constructive non-Euclidean geometry.

Since [26] works with non-strict betweenness, we must be careful not to confuseour betweenness B with non-strict betweenness, which we here denote by T.Classically, T(a, b, c) means that a = b ∨ b = c ∨ B(a, b, c); constructively, wedefine it by taking the double negation and pushing one negation in:

T(a, b, c) := ¬ (a 6= b ∧ b 6= c ∧ ¬B(a, b, c).

9.1. Why we cannot import negative theorems from Tarski’s theory.Godel introduced [11] his double-negation interpretation for this purpose. Thisinterpretation assigns a formula A− to every formula A, by replacing ∃ by ¬∀¬and replacing A ∨B by ¬(¬A ∧ ¬B). For atomic formulae, A− is defined to be¬¬A. The rules of intuitionistic logic are such that if A is classically provable(in predicate logic) then A− is intuitionistically provable. Hence, if we havea theory T with classical logic, and another theory S with intuitionistic logic,whose language includes that of T , and for every axiom A of T , S proves A−,then S also proves A− for every theorem A of T . In case the atomic formulas inthe language of T are stable in S, i.e. equivalent to their double negations, thenof course we can drop the double negations on atomic formulas in A−.

One might like to try this with T taken to be Tarski’s theory as used in [26],and S taken to be ECG. Indeed that is straightforward, and we shall present thedetails in a later section, after developing some geometry in ECG. The questionaddressed here is whether we could possibly avoid some of that development.The short answer is “no”. We will explain the difficulties, in order to justify thenecessity of going over the ground plowed by [26] again from a constructivistviewpoint.

Having the double negation interpretation from Tarski’s theory into ECGonly shows us that with classical logic the two theories are equivalent; that willmake only negative theorems automatically constructively provable. But manyof the theorems in which we are interested are not negative. To give a specificexample, the “outer Pasch” theorem, Satz 9.6 of [26] asserts the existence of apoint x, which (in the non-degenerate case) is given as the intersection of twolines, but the double negation interpretation is only going to tell us that sucha point cannot fail to exist. If it does exist, it will be the intersection point oftwo specific lines, but the interpretation is only going to tell us that those twolines cannot fail to intersect, not that they actually do intersect. To concludethat they do intersect we need the stability principle ¬¬t ↓ → t ↓, which weconsidered taking as Axiom S6, but did not, because it can be proved using


other axioms of ECG, in particular in this case the strong parallel axiom. Butwe want to prove that inner Pasch implies outer Pasch without using the strongparallel axiom; we need to work without the strong parallel axiom in order tostudy the independence relations of the various parallel axioms.

We might have considered taking S6 as an axiom if it would enable us toimport the negative theorems of [26] in one blow. But there is a second difficulty:Tarski’s axioms have many “degenerate” cases, in which the points that appearin the diagram are collinear or some of them coincide. These cases are veryuseful in developing Tarski’s theory and contributed to the small number ofaxioms required: for example, the symmetry property of betweenness is provedusing the five-segment axiom in a degenerate case. (That is why ECG has tohave the symmetry property as an axiom.) One cannot define explict Skolemfunctions in ECG to cover the degenerate cases. Therefore, one cannot avoidthe need for S6 by first extending Tarski’s theory by Skolem functions. Onecan extend Tarski’s theory by Skolem functions, of course, but then one cannotinterpret them by definable terms of ECG, unless one is also willing to changethe axioms of Tarski’s theory to avoid degenerate cases. But if one makes suchchanges, then one is no longer able to import theorems wholesale from [26].

A second example of a theorem derived using degenerate cases of the axioms isthe “inner transitivity” of betweenness, B(a, b, d)∧B(a, b, c) → B(b, c, d). Thiswas an axiom of early versions of Tarski’s theories, but was proved in Gupta’sthesis, and occurs only halfway through [26], and the proof uses degenerate casesof the axioms. As it happens those particular degenerate cases are also allowedin ECG, so that proof can be constructivized. But there is no way to distinguishthese two examples without looking at the details.

The conclusion is that there is no cheap way to import the results of [26].One must actually go through the mathematics and establish the constructivecontent. Moreover, there are other reasons why that is worth doing: one alsohas to constructivize the definitions, i.e. to make sure that the defined conceptscan be constructivized, and to choose the correct way of doing so. For example,what does it mean to construct a triangle congruent to abc “on a given side of”segment AB (assuming AB is congruent to ab?

One may ask for a specific counterexample to the conjecture that negativetheorems of Tarski’s theory are provable in ECG. For example, the inner Paschaxiom of Tarski’s theory is negative, and its translation into ECG is also negativewhen non-strict betweenness is given a negative interpretation, and the result isnot provable in ECG(because as the given points tend to a degenerate case, thedependence of the constructed point is not continuous). Details of this result aregiven in Theorem ??.

In addition to these issues, there are other reasons for developing some geom-etry in ECG. There are a number of definitions to be made (ordering of seg-ments, angles, congruence of angles, ordering of angles, etc.) that are necessaryto support Euclid, and there are some interesting constructive distinctions to bemade. In other words, the development of geometry in ECG has independentinterest, and is not just a matter of taking the double negation interpretationof the classical development. On the other hand, much of the development in[26] is essentially constructive, and generally speaking, using strict betweenness

64 MICHAEL BEESON

makes the proofs easier, not more difficult. Therefore we do not repeat lengthyarguments from [26] when there is no significant new twist require for the con-structivization.

9.2. Congruence and betweenness lemmas. In this section we provesome basic lemmas about congruence and betweenness. In some cases we havefollowed proofs from [26]. Note how the five-segment axiom is used, even whenall the segments lie on the same line. All the lemmas in this section are provedin ECG. In Tarski’s theories, the betweenness axioms have almost disappeared,compared to Hilbert; only the axiom that T (x, y, x) → x = y. One recoversthe missing betweenness axioms by clever applications of Pasch’s axiom and thefive-segment axiom, often applied in ‘degenerate cases” where three vertices ofa “triangle” lie on a line, etc. Some of the degenerate cases introduced by usingnon-strict betweenness are not constructively valid, and we have chosen to usestrict betweenness. Since Euclid did not mention betweenness at all, one cannotuse “faithfulness to Euclid” to justify the choice, and no doubt the theory couldbe developed either way. But, for example, in ECG one can no longer (at leastnot by a similar proof) derive the symmetry axiom T(a, b, c) → T(c, b, a) fromPasch’s axiom, is done in Tarski’s theories (see [26], Satz 3.2, p. 30).

Lemma 9.1. Congruence is an equivalence relation on segments.

Remark. For Hilbert and for Borsuk-Szmielew, this was not an issue, sincesegments were defined as pairs of points.

Proof. Reflexivity. Suppose given points p and q. By Axiom C4 we have qp = pq.Now Axiom C2 yields pq = pq, taking both hypotheses of Axiom C2 to be thesame congruence qp = pq.

Symmetry. Suppose ab = cd. By reflexivity, which we have just proved,ab = ab. Then by Axiom C2, since ab is congruent to cd and to ab, we havecd = ab.

Transitivity. Suppose ab = cd and cd = pq. By symmetry (just proved) wehave cd = ab. Then cd is congruent to both ab and pq, so by Axiom C2 we haveab = pq. That completes the proof of the lemma.

Lemma 9.2. All null segments are congruent, i.e. aa = bb.

Proof. Extend segment ba to a point x on Ray (b, a) such that ax = bb, by AxiomC1. Then by Axiom C3, we have a = x. Hence aa = bb. That completes theproof.

Lemma 9.3 (Additivity of congruence). Suppose B(a, b, c) and B(A,B,C) andab = AB and bc = BC. Then ac = AC.

Proof. Since B(a, b, c) we have a 6= b. We wish to apply the five-segment axiom,Axiom C6, with d = a and D = A. Then ad = AD means aa = AA, which wasproved in Lemma 9.2. The condition db = DB becomes ab = AB, which is trueby hypothesis. Hence Axiom C6 is applicable, and it yields cd = CD, which isca = CA. By Axiom C4 and Lemma 9.1, we have ac = AC. That completes theproof of the lemma.


Lemma 9.4 (Uniqueness of extension).

B(q, a, x) ∧B(q, a, y) ∧ ax = bc ∧ ay = bc → x = y.

Proof. Since ax = bc and ay = bc, by Lemma 9.1 we have ax = ay. ByLemma 9.3 we have qx = qy. Now we apply the five-segment axiom, AxiomC6, with (a, b, c, d) = (q, a, x, x) and (A,B,C,D) = (q, a, x, y). We concludexx = xy. By Lemma 9.1, xy = xx. By Axiom C3, x = y. That completes theproof.

Sometimes it is more convenient to phrase the uniqueness of extension in termsof rays:

Lemma 9.5. Suppose x and y are two points on Ray (a, b), which is to say thatthey are on Line (a, b) and ¬B(x, a, b) and ¬B(y, a, b), and suppose ax = ay.Then x = y.

Proof. Let q be a point with B(E,A,B), and apply Lemma 9.4.

Lemma 9.6 (Subtractivity of congruence). Suppose B(a, b, c) and B(A,B,C)and ac = AC and ab = AB. Then bc = BC. Similarly if instead of ab = AB weassume bc = BC, then ab = AB.

Proof. It suffices to prove the first assertion, since the second follows from thefirst using the axiom, interchanging the roles of a and C and the roles of A andC using Axiom B1-iv, which says B(a, b, c) implies B(c, b, a).

Assume B(a, b, c) and B(A,B,C), and ac = AC and ab = AB. Then there isa point x on Ray (A,B) such that Bx = bc, and by Lemma 9.4, x is the uniquesuch point. By Lemma 9.3, we have Ax = ac. Since ac = AC, by Lemma 9.1,we have Ax = AC. Let R be a point on Ray (B,A) such that RA = AB. ThenB(R,A,B), and x and C are two points on Ray (R,A) such that Ax = AC.Then by Lemma 9.4, x = C. That completes the proof.

Lemma 9.7 (Inner transitivity). B(a, b, d) ∧ B(b, c, d) → B(a, b, c)

Remark. This (with non-strict betweenness) was an axiom of Tarski’s theoriesuntil and including the thesis of Gupta [13], where many other earlier axiomswere proved. It is listed as Ax. 15 in [30], who says that Szmielew used results of[13] to show that it could be replaced with the identity axiom for betweenness,which with non-strict betweenness is ¬B(a, b, a). It appears as Satz 3.5 in [26],and we give the proof from there, presumably due to Szmielew, to check that itstill works with non-strict betweenness and constructive logic.

Proof. We prove only the first claim. Suppose B(a, b, d) ∧ B(b, c, d). ApplyingAxiom B4 (Inner Pasch), we obtain a point x such that B(b, x, b) ∧ B(c, x, a).(Note that it was important that Axiom B4 allows all the points to be collinear.)But B(b, x, b) contradicts Axiom B1-ii. That completes the proof of the firstclaim. The other three are proved similarly.

Lemma 9.8 (Outer transitivity).

B(a, b, c) ∧B(b, c, d) → B(a, c, d) ∧ B(a, b, d).

66 MICHAEL BEESON

Proof. Assume B(a, b, c) and B(b, c, d). It suffices to prove B(a, b, d), for then wecan apply that lemma with (a, b, c, d) replaced by (d, c, b, a) to conclude B(d, c, a),and then by Axiom B1-iv we have B(a, c, d).

Extend segment bc to a point x such that B(a, c, x) and cx = cd. Using thesymmetry of betweenness (Axiom B1-iv) we have B(c, b, a) and B(x, c, a). ByLemma 9.7 we have B(x, c, b), and by Axiom B1-iv we have B(b, c, x). Then byLemma 9.4 we have x = d. But since B(a, c, x) and x = d we have B(a, c, d) asrequired. That completes the proof.

Lemma 9.9 (Transitivity extended).

B(a, b, d) ∧ B(b, c, d) → B(a, c, d)

B(a, b, c) ∧ B(a, c, d) → B(b, c, d) ∧ B(a, b, d)

Proof. These are easy consequences of the previous two lemmas and the symme-try of betweenness.

From a constructive viewpoint, the formulation of Tarski’s 5-segment axiom isinadequate, at least on the face of it, since it depends on the ordering of pointsb and c in Fig. 8. We want segment dc to be determined even if c is between aand b. A remedy for this is to require that cd be determined by ad, ab, bd, andthen ac instead of bc. To say that a, b, and c are collinear without specifying theorder of b and c, and to stay in Tarski’s language (i.e. without mentioning linesand incidence), we must mention another point e, which would be on the line tothe right of b and c, as shown in Fig. 14:

Figure 14. cd is determined, but the order of b and c is notspecified.

d

a b c e

This leads us to the following formulation, which turns out to be provable:

Lemma 9.10 (6-segment lemma, see Fig. 14).

a 6= b ∧B(a, b, e) ∧ B(a, c, e) ∧ B(A,B,C) ∧B(A,B,E)

∧ ab = AB ∧ ac = AC ∧ ad = AD ∧ bd = BD ∧ ae = AE → cd = CD

Proof. We first show that in each of the three cases B(a, b, c) or b = c orB(a, c, b), the theorem holds. If b = c then dc = db, and db = DB by hypothesis,so dc = DB by Lemma 9.1. Then we claim C = B. We have AC = ac byhypothesis, and ab = ac since b = c, and ab = AB by hypothesis, so AB = ACby Lemma 9.1. Then B = C by Lemma 9.4. That disposes of the case b = c.


Now assume B(a, b, c). We have ac = AC and ab = AB, so by Lemma 9.6, wehave bc = BC. Then by Axiom C6 we have cd = CD as required, disposing ofthat case. Now we assume B(a, c, b). Then bc = BC by Lemma 9.6, since wehave ab = AB and ac = AC. We have ae = ae = AE = EA by hypothesis andAxiom C4. Since ab = AB by hypothesis, we have be = BE by Lemma 9.6; soeb = EB by Axiom C4. Now we apply Axiom C6 with (a, b, c, d) = (a, b, e, d)and (A,B,C,D) = (A,B,E,D). The hypotheses have been verified, so we canconclude ED = ed. Next we apply Axiom C6 again, this time with (a, b, c, d) =(e, b, c, d) and (A,B,C,D) = (E,B,C,D). The hypotheses are B(e, b, c) andec = EC and eb = EB and ed = ED. We have already verified ed = EDand eb = EB and bd = BD is given by hypothesis. We have ea = EA andB(a, c, e) by hypothesis, so ec = EC by Lemma 9.6. that leaves just B(e, b, c)to check. We have B(a, c, b) by the hypothesis of this case, and B(a, b, e) byhypothesis. By Axiom B1-iv (symmetry of betweenness) we have B(b, c, a) andB(e, a, b). Hence by the inner transitivity of betweenness, Lemma 9.7, with(a, b, c, d) in the lemma replaced by (e, b, c, a) here, we have B(e, b, c) as desired.Hence Axiom C6 is applicable as claimed. The conclusion is cd = CD, disposingof the third case.

We have thus shown that if B(a, b, c) ∨ b = c ∨ B(a, c, b), then the theoremholds. But because of the stability of equality and betweenness, the doublenegation of the theorem implies the theorem. Hence it suffices to prove thedouble negation of B(a, b, c) ∨ b = c ∨ B(a, c, b), i.e. to prove that not all threecases fail. For proof by contradiction, then, assume that all three cases fail.Then by Axiom B3 and Axiom B1-iv, we have B(b, a, c). Since B(a, c, e), byLemma 9.8 we have B(b, a, e). Now we have B(a, b, e) by hypothesis. These twobetweenness relations contradict Axiom B3. That completes the proof of thelemma.

The following lemma shows that a segment cannot be congruent to a subseg-ment:

Lemma 9.11. If B(a, b, c) then not ab = ac.

Proof. Let d be a point with B(d, a, b). Suppose ab = ac; then b and c are onRay (d, a) and by Lemma 9.4 we have b = c. That completes the proof.23

The following lemma shows that we did not need to provide for uniqueness inthe congruence axiom about laying off a segment along a ray.

Here is a corollary that surprises one by not coming earlier in the development.

Lemma 9.12. The center of a nondegenerate circle is between the endpoints ofany diameter; that is, if line L contains the center p of circle C and two distinctadditional points a and b, then B(a, p, b).

Proof. We have ap = bp by Axiom C5. Since C is nondegenerate, a 6= p andb 6= p. By hypothesis a 6= b. By Axiom B3 and Markov’s principle for between-ness, it suffices to rule out the possibilities B(b, a, p) and B(p, b, a). SupposeB(b, a, p). By axiom B1-iv we have B(p, a, b). Since ap = bp, Lemma 9.11

23Narboux [21] says that this lemma is needed to fill gaps in [26], Ch. 5. Since our Lemma 9.4is Satz 2.12 in [26], the addition required to [26] is minimal.

68 MICHAEL BEESON

yields a contradiction. On the other hand, if B(p, b, a), then by Axiom C1 andLemma 9.11 we have b = a, contradiction. That completes the proof.

9.3. Ordering of segments. Euclid uses the notion of one segment being“greater than” another, without defining it. In [26], ab > cd means that thereexists an x between a and b (or equal to a, because null segments are allowed)such that ax = cd. We adopt a similar definition. 24

Lemma 9.13. The points of an equilateral triangle do not all lie on a line.That is, we cannot have three distinct points on a line with ab = bc = ac.

Proof. Suppose three such points exist. Suppose B(a, b, c). Then segment acis congruent to its proper subsegment ab, contradicting Lemma 9.11. Hence¬B(a, b, c). Similarly, ¬B(b, c, a) and ¬B(c, a, b). Then by Axiom B3, the threepoints do not all lie on a line. That completes the proof.

Lemma 9.14. Suppose a and d lie on Ray (b, a) and are not equal to b. Thenthere is a point e such that B(b, a, e) and B(b, d, e).

Proof. Extend segment ba by bd to find point x with B(b, a, x) and ax = bd.Extend segment bd by ba to find point y with B(b, a, y) and dy = ba. Then byLemma 9.3 we have xb = by, since xa = bd and ab = dy. Hence bx = by. Sincea and d both lie on Ray (b, a), x and y both lie on Ray (b, a) as well. Hence byLemma 9.5, x = y. Take e = x. Then B(b, a, e) since B(b, a, x) and e = x; andB(b, d, e) since B(b, d, y) and e = y. That completes the proof of the lemma.

Recall that ab < cd was defined in Definition 8.10.

Lemma 9.15 (Stability of ab < cd). ¬¬ab < cd → ab < cd.

Proof. By definition of ab < cd, ¬¬ab < cd means

¬¬∃x(B(c, x, d) ∧ cx = ab).

But by Lemma 9.5, there is one and only one possible point p on Ray (c, d)such that cx = ab. Hence ab < cd is equivalent to B(c, p, d), and ¬¬ab < cdis equivalent to ¬¬B(c, p, d). By Markov’s principle for betweenness, we have¬¬ab < cd → ab < cd. That completes the proof.

Lemma 9.16 (Ordering of segments is transitive). If ab < cd and cd < efthen ab < ef .

Proof. Suppose ab < cd and cd < ef . Let K be the circle with center e andradius cd; let it meet Ray (e, f) in point y; then since cd < ef we have B(e, y, f).Let C be the circle with center c and radius ab; let it meet Ray (c, d) in pointz; then since ab < cd we have B(c, z, d). Let J be the circle with center e andradius ab; let it meet Ray (e, f) in point x; we must show that B(e, x, f). Wehave cz = ex since both are equal to ab, and cd = ey by construction of y, and

24This approach to the ordering of segments seems to be faithful to Euclid, in the sense thatit accounts for Euclid’s use of the concept, and does not go much beyond it. In particular,we do not associate any number to a segment to “measure its length”, and we do not performarithmetic, even addition, on the lengths of segments. The formalization of Euclid in [1] takesa different approach, using linear arithmetic on lengths and on the measures of angles. Thatis convenient for computerizing proofs, since it permits the use of decision procedures for thelimited arithmetic involved.


B(c, z, d), and we wish to prove B(e, x, y). Since x lies on Ray (e, f), by AxiomB3, not not x = 0∨ x = y ∨ ∃gB(y, x, g) If we show that each of the three casesimplies B(e, x, y), or implies the final goal ab < ef (or is contradictory), then wecan push the double negation in and apply Markov’s principle for betweenness(which implies the stability of ab < ef), to finish the proof. Therefore we canproceed to argue by cases, as we would classically.

Case 1, e = x. Then because cz = ex we have cz = ee and hence by axiomC3, c = z, and hence by Axiom C3 again, a = b. But then ab is a null segment,so ab < cd is automatic.

Case 2, x = y. Then cz = ex = ey = cd, so segment cd is congruent to itssubsegment cz, contradicting Lemma 9.11.

Case 3, B(y, x, g) for some g. Then let g′ lie on Ray (z, d) with dg′ = yx. Theng′ 6= z since if g′ = z then y = x, contradicting B(y, x, g). Then since cd = eywe have cg′ = eg by Lemma 9.3. Then cz = ex = cg′. We have B(z, d, g′) byAxiom B3, since g′ lies on Ray (z, d) and is not equal to z. Then segment cg′ iscongruent to its subsegment cz, contradicting Lemma 9.11.

Thus each case has been shown to lead to the desired conclusion (remember acontradiction implies anything). That completes the proof of the lemma.

Lemma 9.17.

ab = AB ∧ ac = AC ∧ B(a, b, c) ∧ on(B,Line (A,C)) → B(A,B,C).

Proof. Suppose ab = AB ∧ ac = AC ∧ B(a, b, c) ∧ on(B,Line (A,C)). Wemust show B(A,B,C). Suppose, in order to refute it, that B(A,C,B). Extendsegment ac by bc to find point x such that B(a, c, x) and cx = bc. Now applythe five-segment axiom with (a, b, c, d) replaced by (a, c, x, b) and (A,B,C,D)replaced by (A,C,B,B). The hypotheses hold since ac = AC, cx = CB = bc,ab = AB, and cb = CB. The conclusion is xb = BB. By Axiom C3, x = b.Now since B(a, c, x) and x = b we have B(a, c, b). But also we have B(a, b, c),contradicting Axiom B3. Hence ¬B(A,C,B). If C = B then ab = AB =AC = ac, so segment ac is congruent to its proper subsegment ab, contradictingLemma 9.11. Hence C 6= B. If B = A then AA = AB = ab, so by Axiom C3,a = b, contradicting B(a, b, c). Now assume B(B,A,C). Then extend segmentca to find point x with B(x, a, c) and xa = ba = BA. Now apply the five-segment axiom with (a, b, c, d) replaced by (c, a, x, b) and (A,B,C,D) replacedby (C,A,B,B). Checking the hypotheses, we have ac = AC, ax = AB, cb = CB,and ab = AB. The conclusion is xb = BB. Then by Axiom C3, x = b. SinceB(x, a, c) we have B(b, a, c), which together with B(a, b, c) contradicts AxiomB3. This contradiction refutes B(B,A,C). We have now shown that B cannotbe equal to A or C and we cannot have B(B,A,C) or B(A,C,B). By Axiom B3we therefore have ¬¬B(A,B,C). Then by Markov’s principle for betweenness,we have B(A,B,C) as desired. That completes the proof of the lemma.

Lemma 9.18 (Ordering respects congruence).

ab < cd ∧ ab = AB ∧ cd = CD → AB < CD.

70 MICHAEL BEESON

Proof. Define

X = IntersectLineCircle2 (Line (C,D),Circle3 (C,A,B))

x = IntersectLineCircle2 (Line (c, d),Circle3 (c, a, b)

In order to prove AB < CD, according to the definition of AB < CD, wemust show B(C,X,D). Since ab < cd we have B(c, x, d). We have cx = ab =AB = CX , so by Lemma 9.1 we have cx = CX . Since cd = CD, we can applyLemma 9.17 with (a, b, c) = (c, x, d) and (A,B,C) = (C,X,D). The conclusionis B(C,X,D). That completes the proof of the lemma.

Lemma 9.19 (Extended transitivity).

ab ≤ cd ∧ cd ≤ ef → ab ≤ ef

ab ≤ cd ∧ cd < ef → ab < ef

ab < cd ∧ cd ≤ ef → ab < ef

Proof. Similar to the proof of Lemma 9.16.

Lemma 9.20. ab ≤ cd↔¬(cd < ab).

Proof. Suppose ab ≤ bc and cd < ab. By Lemma 9.19, ab < ab. Then bydefinition of <, there is a point x between a and b with ax = ab. But thatcontradicts Lemma 9.5. That completes the proof.

For those new to constructive mathematics, it is not true that ¬(y < x) →x < y; intuitively, x < y requires us to produce an explicit lower bound on y−x,which ¬(y ≤ x) does not require. It is also not true that x ≤ y is equivalent tox < y ∨ x = y.

9.4. Angles and triangles. Angles and triangles feature prominently in Eu-clid, yet for logical simplicity we have not included in ECG any variables andpredicates that speak directly about angles and triangles. In this section wejustify this choice, by showing how Euclid’s use of angles and triangles can be(almost) directly expressed in ECG.

When Euclid mentions a triangle, it is always by mentioning the vertices. Wecan therefore just regard a triangle as three (distinct) points. Order matters,because in Euclid triangle ABC is not (usually) congruent to triangle CBA. Therelation “triangle ABC is congruent to triangle DEF” is expressed by

AB = DE ∧BC = EF ∧ CA = FD.

That is, the SSS criterion is taken as the definition of congruence.Euclid uses the terms “line” and “straight line”, but not the terms “ray” and

“segment”. He defines an “angle” as “the inclination to one another of two linesin a plane which meet one another and do not lie in a straight line”. He takesfor granted that an angle can be “less than” or “greater than” another angle;what this means is indicated by his “common notion” that “the whole is greaterthan the part.”

Even though Euclid doesn’t mention rays, one really needs them in ordermake sense of “angle” (even to make sense of Euclid’s own definition). We havealready explained, in connection with Axiom C1, that on(p,Ray (A,B)) shouldmean on(p,Line (A,B)) ∧ ¬B(p,A,B).


We take an angle, like a triangle, to be given by three distinct points, as in“∠ABC”, in which B is the vertex of the angle. However, if A′ is another pointon Ray (B,A) (different from B) and C′ is another point on Ray (B,C) (differentfrom B), then angle A′BC′ is “the same angle” as angle ABC. Classically wewould define an angle to be an equivalence class under this equivalence relation.Constructively we continue to regard an angle as (given by) three points, butwe have the notion of two triples of points “determining the same angle.” SinceEuclid uses the phrase “equal angles” to mean “congruent angles”, it would betoo confusing to use the word “equal” for this equivalence relation, or to usethe symbol ‘=’, so we will just use the English “same angle”, as in “ABC andA′BC′ are the same angle.”

Whenever we define a relation involving angles, we will check that it respectsthis equivalence relation, i.e. if angles are replaced by other triples of pointsdefining the same angles, then the new relation is preserved. That way, whenwe are “given an angle ABC”, then we can “without loss of generality” replaceA and B by any other more convenient points not equal to B on Ray (B,A) andRay (B,C) respectively.

Note, however, that abc is not the same angle as cba. We will prove below thatthey are congruent, but they are not the same angle. This decision is somewhatarbitrary; we could have chosen to define the relation “same angle” more broadly.

Definition 9.21. Angle abc is congruent to angle ABC if triangle a′bc′ iscongruent to triangle A′BC′, where the primed points are constructed so thata′ lies on Ray (b, a), c′ lies on Ray (b, c), A′ lies on Ray (B,A), C′ lies onRay (B,C), and all the segments ba′, bc′, BA′, and BC′ are congruent to afixed segment αβ. (Here α and β are two of the points mentioned in the lowerdimension axiom.)

This definition has the advantage that it is quantifier-free. There are otherpossible definitions we could take25 :

(i) We could require that there exist points a′, c′, A′, and C′ on Ray (b, a),Ray (b, c), Ray (B,A), and Ray (B,C) respectively, such that triangle a′bc′ iscongruent to triangle A′BC′.

(ii) We could require that if a′ and c′ are on Ray (b, a) and Ray (b, c) re-spectively, and A′ and C′ are on Ray (B,A) and Ray (B,C) respectively, andba′ = BA′ and bc′ = BC′, then a′c′ = A′C′.

Note that choices (i) and (ii) use a quantifier (universal for (ii) and existentialfor (i)), while (iii) is quantifier-free. All the choices make it apparent that con-gruence is symmetric and respects the “same-angle” equivalence relation. Thefollowing lemma shows that the possible definitions are all equivalent.

Lemma 9.22. The three possible definitions of angle congruence listed above,i.e. (i) and (ii) and the actual definition, are equivalent.

Proof. Evidently the actual definition implies (i) and (ii) implies the actualdefinition, so it suffices to show that (i) implies (ii). Suppose we are given

25In [26], p. 95, a different way of constructing canonical points is used, similar to theconstruction in the proof of Lemma 9.14. This definition implies (i) and is implied by (iii).

72 MICHAEL BEESON

triangle abc congruent to triangleABC, and points a′, c′, A′, andC′ on Ray (b, a),Ray (b, c), Ray (B,A), and Ray (B,C) respectively, and a′ 6= b, c′ 6= b, A′ 6= B,and C′ 6= C, and ba′ = BA′ and bc′ = BC′. We must prove a′c′ = A′C′.

By Lemma 9.14, there is a point e on Ray (b, a) such that B(b, a, e) andB(b, a′, e), and a point f be a point on Ray (b, c) such that B(b, c, f) and B(b, c′, f).We intend to use these points in applying Lemma 9.10, so we need correspondingpoints E and F : let E and F be points on Ray (B,A) and Ray (B,C) respectivelysuch that BE = be and BF = bf . Then we claim B(B,A,E) and B(B,A′, E)and B(B,C, F ) and B(B,C′, F ). We prove B(B,A,E) (the other three claimsare similar). We claim A 6= E. Assume that A = E. Then ba = BA = BE = be,so ba = be, contradicting Lemma 9.11. Hence A 6= E. Next assume B(B,E,A).In that case we have BE < BA by definition of <, and BA = ba by hypothesis,so BE < ba by Lemma 9.18, and ba < be by definition of <, so BE < be byLemma 9.16, and since be = BE we have BE < BE by Lemma 9.18. But thenby definition of <, some proper subsegment of BE is congruent to B, contradict-ing Lemma 9.11. Hence the assumption B(B,E,A) is false, i.e. ¬B(B,E,A).But E lies on Ray (B,A), and is not equal to B or to A, and is not between Band A. Hence by Axiom B3 and Markov’s principle for betweenness, we haveB(B,A,E) as desired. Similarly, we have the other three betweenness relationsinvolving E and F claimed above.

We do not know the order of a and a′ on Ray (b, a), so the Pasch axiom doesnot directly apply, but the modified version in Lemma 9.10 comes to the rescue.Applying that lemma with (a, b, c, d, e) set to (a′, a, b, c, e) (and similarly for(A,B,C,D,E), we obtain a′c = A′C. Applying it again with (a, b, c, d, e) set to(b, c, c′, a′, f) we find a′c′ = A′C′ as desired. That completes the proof that (i)implies (ii), and with it, the proof of the lemma.

Lemma 9.23 (SAS). Suppose angles abc and ABC are congruent. Supposeab = AB and bc = BC. Then triangles abc and ABC are congruent; that is,ac = AC.

Proof. By Lemma 9.22. Specifically, this follows immediately using definition(ii) of angle congruence.

Lemma 9.24. Congruence of angles is an equivalence relation. Moreover angleabc is congruent to angle cba.

Proof. By the SAS lemma, the problem of proving that angle congruence is anequivalence relation reduces to showing congruence of triangles is an equivalencerelation; and this in turn reduces to the fact that congruence of segments is anequivalence relation, which was proved in Lemma 9.1. As for the second claim ofthe lemma, given abc, let c′ be constructed on Ray (b, c) have bc′ = ba, by AxiomC1. Then triangle abc is congruent to angle cba, since ac = ca by Axiom C4.

Lemma 9.25. Congruence of angles is stable, in the sense that if angle abc isnot not congruent to angle ABC, then angle abc is congruent to ABC.

Proof. Because we chose definition (iii) for congruence, the congruence of anglesis expressed by a quantifier-free conjunction of the three atomic statements ex-pressing the congruence of the sides. Pushing the double negation inwards, wecan then use the stability of E (Axiom S2). That completes the proof.


Definition 9.26. Suppose triangles ABC and abc are congruent. Then thecorresponding angles of the triangles are the pairs ABC and abc, BCA andbca, and CAB and cab.

Lemma 9.27. Corresponding angles of congruent triangles are congruent.

Proof. Suppose triangle ABC is congruent to triangle abc. We show that angleABC is congruent to angle abc. We have AB = ab and AC = ac and BC = bcby Lemma 9.22.

Lemma 9.28. The base angles of an isosceles triangle are congruent.

Proof. Let triangle ABC be isosceles; assume AB = BC. Then angle ABCis congruent to angle CBA, by Lemma 9.24. Then triangle ABC is congruentto triangle CBA by SAS. The base angles of triangle ABC are correspondingangles in this congruence, and hence are equal. That completes the proof.

Definition 9.29. Angles ABC and CBE are supplements if B(A,B,E).Angles ABC and DBE are supplements if D is on Ray (B,C) and B(A,B,E).

Lemma 9.30. Supplements of congruent angles are congruent.

Proof. Let ABC and DBE be supplements. Suppose that angle ABC is con-gruent to angle abc, with B(A,B,E), and e is a point with B(a, b, e) and d apoint on Ray (b, c), so angle abc and angle dbe are supplements. We must proveangle DBE is congruent to angle dbe. Without loss of generality we can assumeD = C and d = c and ab = AB and bc = BC and be = BE. We can there-fore apply the five-segment axiom, Axiom C6, to conclude that CE = ce. Thatimplies that angle DBE is congruent to angle dbe. That completes the proof.

Lemma 9.31. An angle congruent to a right angle is a right angle.

Proof. Let abc be a right angle and let d be a point with B(a, b, d) and ab = bd.Since abc is a right angle we have ac = dc. Let angle ABC be congruent to angleabc. Without loss of generality we may assume ab = AB and bc = BC. Let Dbe a point with AB = BD and B(A,B,D). Then BD = bd since congruenceof segments is transitive. Then by the five-segment axiom also dc = DC. ThenAC = ac = dc = DC. Hence Line (B,C) is perpendicular to Line (A,B), andABC is a right angle. That completes the proof.

Definition 9.32. Angles amb and cmd are vertical angles if B(a,m, c) andB(b,m, d), or if B(a,m, d) and B(b,m, c).

Lemma 9.33. Vertical angles are congruent.

Proof. An immediate consequence of Lemma , together with the fact (proved inLemma 9.24) that angle abc is congruent to angle bca.

Lemma 9.34. If angle abc is not not congruent to angle ABC, then angle abcis congruent to angle ABC.

Proof. Suppose angle abc is not not congruent to triangle ABC. Let a′, b′, A′

and B′ be as in Definition 9.21. Then ¬¬ ba′ = αβ, ¬¬BA′ = αβ, ¬¬ bc′ = αβ,and ¬¬BC′ = αβ. By Axiom S1 (stability of segment congruence), the doublenegations can be removed. Then by Definition 9.21, we have angle abc congruentto angle ABC. That completes the proof.

74 MICHAEL BEESON

9.5. Perpendicular lines.

Definition 9.35. Lines L and K, meeting at point m, are perpendicularat m if there are points a and b on L with B(a,m, b) and c on K, with c not on Land a and b not on K, such that triangle amc is congruent to triangle bmc. LinesL and K are perpendicular if they are perpendicular at IntersectLines (L,K).We write ab ⊥ bc to mean Line (a, b) is perpendicular to Line (b, c), and L ⊥ Kto mean L is perpendicular to K.

Figure 15. L and K are perpendicular at m if a, b, and c existmaking triangles amc and bmc congruent.

L b

m

K

b b

b

a b

c

Remark. Note that if L and K are perpendicular, they do not coincide, by theclause in the definition that c is not on L.

Lemma 9.36. If lines L and K meet at m, and a and b are any points on Lwith B(a,m, b), and c is any point on K different from m, then angles amc andbmc are congruent if and only if L and K are perpendicular.

Proof. If L and K are perpendicular, then the congruence of the angles fol-lows immediately from the definition of “congruent angles” and the definition of“perpendicular.” Conversely, suppose the angles are congruent. Then let b′ bea point on L, on the opposite side of m from a, with mb′ = ma. Then trianglecmb′ is congruent to triangle cma by SAS, so K and L are perpendicular.

Lemma 9.37. If lines L and K are perpendicular, and lines ℓ and k have thesame points as L and K, respectively, then ℓ and k are also perpendicular.

Remark. Remember that equality between lines is intensional, so this is notautomatic.

Proof. The points a, b, and c that witness the perpendicularity of L and Kaccording to the definition, also witness the perpendicularity of ℓ and k.

Lemma 9.38. If L and K are perpendicular, then K and L are perpendicular.

Proof. Suppose L and K are perpendicular. Let m be their intersection point.By definition of “perpendicular”, there are points a and b on L with B(a,m, b)and am = mb, and c be on K with c 6= m, such that triangle amc is congruentto triangle bmc. Extend segment cm to a point d such that B(c,m, d) andcm = md. Then angle bmd and angle amc are vertical angles, and hence by


Lemma 9.33 they are congruent. Angle amc is congruent to angle bmc, sincethey are corresponding angles of congruent triangles. hence by Lemma 9.24,angle bmc is congruent to angle bmd. Then triangle bmc is congruent to trianglebmd, by SAS. Then by definition of perpendicular, K and L are perpendicular.That completes the proof.

Lemma 9.39 (Stability of perpendicular). If L and K are not not perpendic-ular, then they are perpendicular

Proof. To prove this we must define perpendicular in a quantifier-free way,getting rid of the existential quantifier in the definition. That we can do so is aconsequence of Lemma 9.36. Suppose L and K meet m. Then let C be the circlewith center m and radius αβ. Then let p and q be the two intersection points of Land C given by IntersectLineCircle1 (L,C) and IntersectLineCircle2 (L,C), andsimilarly let a and b be two intersection points of K and C. Then the trianglespma and pmb are congruent if and only if L and K are perpendicular at m. Thatis, L and K are perpendicular at m if and only if pa = pb. Since p, a, and bare explicitly defined by terms in L and K (putting m = IntersectLines (L,K)),we have an equation between terms of type Point. This is stable by Axiom S1.That completes the proof.

9.6. Existence of midpoints and perpendiculars. Most mathematiciansare familiar with the construction in Euclid I.1, spelled out in section 3.4. It is,however, possible to construct midpoints without assuming line-circle continuity,cf. Prop. 4.3, page 167 of [12] or Satz 8.22 of [26]. It is also possible to constructa perpendicular to line L from a point c not on L without using a line-circlecontinuity or circle-circle continuity. However, at present we are allowing thefull use of ruler and compass, so we do not stop to analyze these constructions.Instead, we prove the correctness of the Euclidean constructions. However, theconstruction Euclid gives for bisecting a segment contains an error (besides notproving the circles intersect), as we will point out below.

Lemma 9.40. Every segment ab with a 6= b has a midpoint, i.e. a point m be-tween a and b with ma = mb, and a perpendicular bisector, i.e. a line perpendic-ular to Line (a, b) at m. Both m and a perpendicular bisector can be constructed(defined) by terms of ECG.

Remark. The following proof follows Euclid I.1 and I.9 and I.10, but it fills somegaps and corrects an error. Specifically, the circles constructed in I.1 must beshown to actually intersect, the intersection points must be shown not to lie onLine (a, b), and indeed must be shown to lie on opposite sides of ab, all of whichis omitted in Euclid; then in I.9, the equilateral triangle that Euclid constructsmight have its apex at the vertex of the angle being bisected. One can show thatthe construction made there will not be continuous in parameters; in a computeranimation, one can drag the sides of the angles to make the two points allegedto determine the angle bisector coincide. Of course, one should then take the“other equilateral triangle”, but that cannot be done continuously. In essencethe following proof repairs this defect by constructing both equilateral trianglesand drawing the angle bisector between their apexes, instead of using the vertex

76 MICHAEL BEESON

of the angle being bisected. In other words, there is a non-constructivity in theproof of Euclid I.9, which can be easily removed by a slightly different proof.

Proof. Given segment ab, construct circle C with center a and radius ab, and cir-cle K with center b and radius ab. Extend segment ab to point d with B(a, b, d)and bd = ab. Then d lies on K and is outside circle C, because B(a, b, d) and abis the radius of C. On the other hand point a lies inside C and lies on circle K.Hence by Axiom Cont 6 (circle-circle continuity), both IntersectCircles1 (C,K)and IntersectCircles2 (C,K) are defined. Let these points be p and q respec-tively. Then pab is an equilateral triangle. Then p does not lie on Line (a, b), byLemma 9.13. Similarly, abq is equilateral, and hence q does not lie on Line (a, b).Now, by Definition 5, abp is a right turn and abq is a left turn. By AxiomH3 then, p and q are on opposite sides of Line (a, b); that is, Line (p, q) meetsLine (a, b) in a point m between p and q. Thus the defects of Euclid’s proof areboth succesfully dealt with in ECG.

Since ap = ab and bp = ab, and congruence is an equivalence relation, wehave ap = bp. Similarly aq = bq. Since pq = pq (every segment is congruent toitself) we have triangle apq congruent to triangle bpq. Then the correspondingangles apq and bpq are congruent. But these are the same angles as apm andbpm (here “same angles” has been precisely defined above). Hence triangle apmis congruent to triangle bpm by SAS. Hence the corresponding sides am and bmare congruent. That is, m is the midpoint of ab.

Moreover, the supplementary angles pma and pmb are congruent as corre-sponding angles of congruent triangles. Hence, by definition of perpendicular,pm is perpendicular to Line (a, b). That completes the proof.

Here is a two-dimensional version of Markov’s principle:

Lemma 9.41. ECG proves that, if point P does not lie on line L, then somecircle with nonzero radius and center P lies entirely on the same side of L as P .

Proof. Let point P not lie on line L; we will construct a circle with center P lyingon the same side of L as P . Let point K on L be the foot of the perpendicularfrom P to L. Then K 6= P . Hence the two circles C1 = Circle (K,P ) and C2 =Circle (P,K) that are used to bisect PK have different centers and each containsa point inside the other. Hence the points X = IntersectCircles1 (C1, C2) andY = IntersectCircles2 (C1, C2) are defined. If X = Y then X is between Pand K, contradicting PX = PK. Hence the midpoint of PK is given by M =IntersectLines (Line (X,Y ),Line (P,K)). Hence Circle (P,M) lies on the sameside of L as P , because K is nearer to P than any other point of L. Thatcompletes the proof of the lemma.

9.7. Right angles.

Definition 9.42. abc is a right angle if Line (a, b) is perpendicular toLine (b, c) at b.

Lemma 9.43. A right angle is congruent to its supplement.

Proof. Let abc be a right angle, and let d be a point with B(a, b, d), so thatangle cbd is the supplement of angle abc. Since abc is a right angle, Line (b, c)is perpendicular to Line (a, b), and by definition of “perpendicular”, there are


points A on Ray (b, a) and C on Ray (b, c) and D on Ray (b, d) such that Ab = eband AC = eC. That is, triangle AbC is congruent to triangle ebC. Hence bydefinition of angle congruence, angle abc is congruent to angle dbc.

Lemma 9.44. All right angles are congruent. In other words, if abc and ABCare right angles with ab = AB and bc = BC then ac = AC.

Remark. Euclid took this as his Postulate 4. Hilbert ([15], p. 20) remarks thatthis was “unjustified”, and says that the proof of it goes back to Proclus. Wenote that Hilbert’s proof makes use of a constructively unacceptable argumentby cases, but since the conclusion is disjunction-free, we could just double-negatehis whole argument. However, since his axioms differ from ours, that still willnot produce a proof in our system. We have not yet proved the uniqueness of theperpendicular to a line at a given point, and this lemma will be used to proveit; so the lemma has to be proved without it.

Proof. This is Satz 10.12 in [26]. We sketch the proof, with attention to checkingthat the argument is constructive. We note that the conclusion is stable, i.e.implied by its double negation; hence we can argue classically if we like; inparticular a case distinction whether B = b or not is allowed. One begins byproving that reflection in a point, and reflection in a line, both preserve rightangles; this is easy since both kinds of reflection preserve congruence of segments(reflections are isometries). Now reflect triangle ABC about the midpoint ofsegment bB (if b 6= B, or about b if B = b). Replacing ABC by its reflection,we can assume b = B. Again, since the conclusion is stable, we can argue bycases on whether c = C or not. If c 6= C, then N let y be the midpoint ofcC. Reflection in line by takes c into C, since bc = bC. similarly reflect on theline joining b to the midpoint of cC. This enables us to assume, without loss ofgenerality, that c = C. Then we have two right angles Abc and abc with ab = Ab.If A and a are on opposite sides of Line (b, c), then reflect A in Line (b, c); sincethe desired conclusion is stable, we may argue by cases on whether A and aare on opposite sides of Line (b, c) or not; hence, without loss of generality wecan assume they are not on opposite sides. Then aA does not meet Line (b, c).Again, since the desired conclusion is stable, we can argue by cases; if a = A weare finished, so we can assume a 6= A. Let z be the midpoint of aA. Then sinceba = bA, bza is a right angle and A is the reflection in Line (b, z) of a. Let e bethe reflection of c in point b. Then since ab ⊥ bc we have ae = ac. Since Ab ⊥ bcwe have Ac = Ae. Hence Ae = Ac = ac = ae, so Ae = ae. Hence triangle Abe iscongruent to triangle abe. Hence angle Abe is congruent to angle abe. But angleabe is congruent to angle abc and angle Abe is congruent to angle Abc. Henceangle abc is congruent to angle Abc. That completes the proof.

Lemma 9.45. a 6= b ∧ on(c,Line (a, b)) ∧ ac = ac′ ∧ bc = bc′ → c = c′.

Remark. The possible constructive difficulty here is that we do not know theorder relations of c with a and b. But Lemma 9.10 takes care of this. This is akey lemma needed to prove the uniqueness of perpendiculars.

Proof. We apply the six-segment lemma (Lemma 9.10) with (a, b, c, d) replacedby (a, b, c, c), and (A,B,C,D) replaced by a, b, c, c′. The conclusion is cc = cc′.Hence by Axiom B1-ii, we have c = c′.

78 MICHAEL BEESON

Lemma 9.46 (Uniqueness of perpendicular from a point not on a line).Suppose abc and acb are right angles. Then b = c.

Proof. (following [26] Satz 8.7) By the stability of equality, we may use proofby contradiction. Suppose b 6= c. Then extend line segment cb to point c′ withbc = bc′ and B(c, b, c′), and extend ac to a point a′ with B(a, c, a′ and ac = a′c.Then because ab ⊥ bc, we have ac = ac′, and similarly a′c = a′c′. But sinceB(a, c, a′), we have from Lemma 9.45 that c = c′. Then from B(c, b, c′ we obtainB(c, b, c), and hence by Axiom B1-ii we have b = c. That completes the proof.

Lemma 9.47. A supplement of a right angle is a right angle.

Proof. Let abc be a right angle and B(a, b, d); we must show abd is a rightangle. Let e be a point with B(a, b, e) and ab = be. By the definition of “rightangle”, it suffices to show abe is a right angle.

We distinguish between “dropped perpendiculars”, dropped from a point noton a line to the line, and “erected perpendiculars”, erected from a given pointon the line. We have just proved the uniqueness of dropped perpendiculars; theuniqueness of erected perpendiculars is harder, and will be proved in anothersection.

We now draw some corollaries from the uniqueness of dropped perpendiculars.

Lemma 9.48. Suppose lines K and M are both perpendicular to line L and donot coincide. Then K and M do not meet (i.e., they are parallel).

Proof. If they do meet in point p, then they meet L in different points, since twopoints determine a line. But then K and M are two different perpendiculars toL from point p, contradicting Lemma 9.46. That completes the proof.

Lemma 9.49. The line from the vertex of an isosceles triangle to the midpointof the base is perpendicular to the base.

Proof. Let a be the vertex and m the midpoint of bc, and ab = ac. Then sinceam = bm, triangle abm is congruent to triangle acm. Hence am is perpendicularto bc. That completes the proof.

Lemma 9.50. Two right triangles with hypotenuse and one leg congruent, arecongruent. In symbols: if ac = AC and ab = AB and angles abc and ABC areright triangles, then triangle abc is congruent to triangle ABC.

Proof. Let a′ be the reflection of a in Line (b, c), so ab = a′b. Let A′ be thereflection of A in Line (B,C), so AB = A′B. Then AA′ = aa′ by Lemma 9.3,and ab = a′b since abc is a right angle, and AB = A′B since ABC is a rightangle. Hence triangle aba′ is congruent to triangle ABA′, since all three sidesare pairwise congruent. Hence angle baa′ is congruent to angle BAA′. But anglebac is the same angle as angle baa′, and angle BAC is the same angles as angleBAA′, so angle bac is congruent to angle BAC since angle congruence respectsthe “same angle” relation. Then triangle abc is congruent to triangle ABC bySAS. That completes the proof.

Lemma 9.51. A line meets a circle in at most two points.


Remark. This is closely related to Euclid III.2, which is discussed in Section9.13; but this lemma is easier to prove.Proof. Let a and b lie on the circle with center e. Suppose B(a, c, b) and c lieson the circle. Then ea = ec, so triangle eac is isosceles. Let m be the midpointof ac; then by Lemma 9.49, em ⊥ ac. Let q be the midpoint of ab; then sinceea = eb, eq ⊥ ab by the same Lemma. But a, b, and c are collinear by hypothesis;hence m = q by the uniqueness of dropped perpendiculars (Lemma 9.46). Butthen am = aq, and b and c are both extensions of segment am by am, and bothb and c lie on Ray (a, b), by the assumption B(a, c, b). Hence by Lemma 9.5,b = c, contrary to the assumption B(a, c, b). That completes the proof.

Lemma 9.52. If line L contains a point strictly inside circle C, then L meetsC in exactly two points.

Remark. See the discussion following Axiom Cont 5.

Proof. By Lemma 9.51 there are at most two points of intersection, and byAxiom Cont 5, there are at least two. That completes the proof.

Definition 9.53. Line L is tangent to circle C at x if L meets C at x andonly at x.

Remark. Some books take the property in the following lemma as the definitionof tangent instead.

Lemma 9.54. (i) If circle C with center e is tangent to L at x then ex ⊥ L.(ii) Given line L, point x on L, and two points a and e not on L, with ex ⊥ Land e lying on a perpendicular bisector of xa, then Circle (e, a) is tangent to Lat x.

Remark. We have to say “a perpendicular bisector” rather than “the perpendic-ular bisector” since the perpendicular bisector cannot be proved unique until weallow the use of the upper dimension axiom. This lemma does not require theupper dimension axiom.

Proof. Ad (i). See Fig. 16. Drop a perpendicular K from e to L, and let zbe intersection point of K and L. Suppose z 6= x. Then extend segment xzto a point w such that B(x, z, w) and xz = wz. Then since ez ⊥ L, we havetriangle ezx congruent to triangle ezw. Then the corresponding sides ex andw are equal, so w is a point of intersection of C and L. Then w = x by thedefinition of “tangent”, Definition 9.53. Since B(x, z, w) and w = x we haveB(x, z, x). Hence by Axiom B1-ii, we have a contradiction. Hence ¬z 6= x. Bythe stability of equality, we have z = x. But by construction ez ⊥ L; henceex ⊥ L. That completes the proof of (i).

Ad (ii). See Fig. 17. Let C = Circle (e, a). Let m be the midpoint of ax. Thenem ⊥ ax, so triangles emx and ema are congruent. Hence their correspondingsides ex and ea are equal. Hence L meets C at x. Now suppose L meets C ata point y 6= x. Let p be the midpoint of xy. Then triangles epx and epy arecongruent, since ex = ey = ea and px = py because p is the midpoint of xy, andthe third side ep is shared. Then ep is perpendicular to L at p. Then p = x bythe uniqueness of the perpendicular from e to L (Lemma 9.46). But then x = y,

80 MICHAEL BEESON

Figure 16. if ez ⊥ L and xz = zw then ex = ew. So if C istangent to L, z 6= w is impossible.

be

L

bx

z

w

contradiction. Hence L does not meet C at a point y 6= x. Hence L is tangentto C at x, by Definition 9.53. That completes the proof of the lemma.

Figure 17. Construction of a circle tangent to L passingthrough a. If y lies on L and C and px = py then ep ⊥ L.

be

L

bx

b

a

bm

by

b

p

Lemma 9.55. There is a point not on a given line.

Remark. Classically we could show that one of the three constants α, β, γ isnot on L. But constructively we cannot do that, as the choice cannot be madecontinuously in L. The point that we choose is the one from Euclid Book I Prop.


1, the point E that makes an equilateral triangle with two points A and B on L.Famously Euclid provided no proof that the two relevant circles actually intersectat E; but here we have to prove something equally obvious, which Euclid alsooverlooked: that E does not lie on L. Constructively or classically, this is equallydifficult: why can’t all the points of an equilateral triangle lie on a line? Bothbetweenness and congruence axioms are needed.

Proof. This was proved in the proof of Lemma 9.40; the vertex p of the equilateraltriangle constructed on ab does not lie on Line (a, b), by Lemma 9.13.

Remark. We note that CircleThree was not used in the above construction; thelemma is thus somewhat remarkable in view of the theorem that any constructiondefined using only circles constructed from center and point with only one vari-able must be somewhere undefined. The above construction E(a, b) constructsa point not on Line (A,B), and indeed it is undefined when A = B, so it doesnot contradict the theorem. It is surprising that we cannot find a construction(without Circle3 ) that always constructs a point different from a given point,but we can find a construction that always constructs a point different from agiven line.

Lemma 9.56 (Uniform construction of perpendiculars). Let Perp(x, L) be de-fined as in Definition 5.1. For every point x and line L, Perp(x, L) is defined,and Perp(x, L) is perpendicular to L, and x lies on Perp(x, L).

Remark. The important point here is that no case distinction is required as towhether x is or is not on L; that is, there is no distinction in this constructionbetween dropped perpendiculars and erected perpendiculars. That is good, sinceconstructively we cannot decide if a given point x is or is not on a given line, yetwe still need to be able to project x onto the line.

Proof. Here is the construction script (whose parsing yields the term Perp of thelemma) given in Definition 5.1.

Perp(Point x, Line L)

a = pointOn1(L)

b = pointOn2(L)

Q = Circle3(b,x,a)

c = IntersectLineCircleTwo(L,Q)

C = Circle3(x,a,c)

p = IntersectLineCircle1(L,C)

q = IntersectLineCircle2(L,C)

K = Circle(p,q)

R = Circle(q,p)

d = IntersectCircles1(K,R)

e = IntersectCircles2(K,R)

return Line(d,e)

Proof. We go through the script defining Perp(x, L) and show that the term onthe right of each line is defined. Lines 1 and 2 are defined since pointOn1 (L)and pointOn2 are always defined, by Axiom CA1. In line 3, Circle3 (b, x, a) is

82 MICHAEL BEESON

defined because Circle3 is everywhere defined, by Axiom CA7. Since the centerof that circle is b and b lies on L, b is inside that circle, so by the line-circlecontinuity axiom, c in line 4 is defined. Circle C in line 5 is defined by AxiomCA7. To show that p and q in lines 5 and 6 are defined and distinct, we mustshow that L has a point strictly inside C. That point is a, as we now prove. Lete be any point on C. Then ec = ac since C = Circle 3(x, a, c). Since a 6= b, Cis not a degenerate circle; in fact ab ≤ xe (equality might hold if x = a) andax < xe. Hence a is strictly inside C. Hence points p and q in lines 6 and 7 ofthe script are defined, and by Lemma 9.52 they are distinct. Line (d, e) is theperpendicular bisector of segment pq, which exists by Lemma 9.40; the proof ofthat lemma provides the explict term mentioned in the script for constructingLine (d, e). That completes the proof of the lemma.

9.8. Various forms of the Pasch axiom. The intuition behind the Paschaxiom is this: if a line enters a triangle, it must come out again. In making thisprecise we must watch out for the case when the line just touches the triangle,either at a vertex or along one side. Perhaps the most natural version is this:“if abc is a triangle (that is the three points are distinct) and line L containsa point x on the closed segment ab (that is T(a, x, b)) then line L contains apoint y with T(a, y, c) ∨ T(b, y, c).” That version is not constructively valid, aswe cannot continuously decide which alternative holds. More precisely, given afixed triangle abc in the standard model, we can construct a line Lx dependingon a real number x such that Lx meets the midpoint of ab, and if Lx meets theclosed segment ac then x ≤ 0 and if it meets bc then x ≥ 0. Hence this versionof Pasch would imply the constructively invalid principle x ≤ 0 ∨ x ≥ 0.

To get a constructively valid principle we could simply take the double negationof the conclusion. That is equivalent to saying that it is impossible that L shouldmeet the closed segment ab and not meet the closed segment ac and not meetthe closed segment bc. Here is that principle:

a 6= b ∧ a 6= c ∧ b 6= c ∧ on (x, L) ∧ B(a, x, b) → (Negative Pasch)¬∀y(on(y, L) → (y 6= a ∧ y 6= b ∧ y 6= c ∧ ¬B(a, y, c) ∧ ¬B(b, y, c)))

Negative Pasch has two strikes against it as a candidate axiom: it is notquantifier-free, and it does not actually produce the point of exit of L from thetriangle when given the entrance point. Asserting that existence gives us thefollowing principle:

a 6= b ∧ a 6= c ∧ b 6= c ∧ on (x, L) ∧ B(a, x, b) → (Strong Pasch)∃y(on(y, L) ∧ ¬(y 6= a ∧ y 6= b ∧ y 6= c ∧ ¬B(a, y, c) ∧ ¬B(b, y, c))

It turns out that Strong Pasch is provable, but as an axiom it is unappealing,as the construction of the required point y by ruler and compass is not quiteevident. Indeed one may suspect at first that it might not be possible to constructy uniformly, since y depends continuously but not differentiably on the line L.But it is not the case that all the functions definable in ECG are differentiableon their domains; for example,

IntersectLineCircle2 (Line (a, b),Circle (b, c))

does not depend differentiably on c as c passes through b. (If we did not allowcircles zero radius, constructions would be differentiable on their domains.)


Negative Pasch is too weak; Strong Pasch is too strong; we now give an axiomwith a Goldilocks quality: it makes an existence assertion, but one that can berealized with ruler and compass. It says if line L does not meet side ab of triangleabc but does meet side bc then it must meet side ac as well:

∀x(on(x, L) → ¬B(a, x, b) ∧ x 6= a ∧ x 6= b)∧ B(a, IntersectLines (Line (a, c), L), c) →

B(b, IntersectLines (Line (b, c), L), c) (Explicit Pasch)

Note that that “triangle” abc is allowed to be degenerate, in that b might lieon Line (b, c). Explicit Pasch would be a constructively acceptable axiom, butit is not quantifier-free, and we are aiming for a quantifier-free axiomatization,so we are not willing to take Explicit Pasch as an axiom. We will prove it inLemma ??.

Some of the versions of Pasch considered so far all mention incidence as wellas betweenness. That is not essential, as on(y,Line (a, b)) is equivalent to

¬¬(B(y, a, b) ∨ y = a ∨B(a, y, b) ∨ y = b ∨B(a, b, y).

To use this way of expressing incidence, we must have a way to get a and b fromL. In ECG, that is built-in, but Tarski needed versions of Pasch that specify theline L by using an auxiliary point specified to be on L. Direct translations of theabove forms using this idea are too long to be elegant. Several special cases canbe succinctly formulated. Three in particular, “Inner Pasch”, “Outer Pasch”,and “Weak Pasch” have been studied by Tarski and his students. We exhibitthese formulations in order to examine them constructively. The first one, innerPasch, is the one taken as an axiom in [26]. Here T is non-strict betweenness. Itconsiders a line Line (p, b) entering a “triangle” caq. We put “triangle” in quotesbecause it is allowed that some or all of the points be equal, and they can alsobe collinear. The line is specified by a point b on cq extended.

T(a, p, c) ∧ T(b, q, c) → ∃x(T(p, x, b) ∧ T(q, x, a)) (Inner Pasch)

A closely related version is axiom B4 of ECG, which we repeat here for compar-ison. We will also refer to Axiom B4 as “Explicit Inner Pasch.” The existentialquantifier has been replaced by a suitable construction term, so the result isquantifier-free. In addition non-strict betweenness has been replaced by strictbetweenness, and the three points are required to form a triangle, i.e. to bedistinct and non-collinear. Without the non-collinearity requirement, the termgiving the interection point in B4 could possibly be undefined.

B(a, p, c) ∧ B(b, q, c) ∧ p 6= b ∧ q 6= a ∧ ¬on(c,Line (a, q) →IntersectLines (Line (p, b),Line (a, q)) ↓ ∧B(p, IntersectLines (Line (p, b),Line (a, q)), b) ∧B(q, IntersectLines (Line (p, b),Line (a, q)), a) (Pasch’s Axiom B4)

If the line Line (p, b) is specified by b on the other side of segment cq we getOuter Pasch26 :

T(a, p, c) ∧ T(q, c, b) → ∃x(T(b, p, x) ∧ T(a, x, q)) (Outer Pasch)

26Note that it is T(a, x, q) rather than T(q, x, a) as in Inner Pasch. That does not matter,after the symmetry axiom has been included (as in ECG) or derived from inner Pasch (as in[26]), but we follow [30] to the letter here.

84 MICHAEL BEESON

As with Inner Pasch, one can replace ∃x by a term of ECG (the same oneas in Inner Pasch) to obtain a quantifier-free version in ECG. That version isgiven here:

B(a, p, c) ∧ B(q, c, b) ∧ ∧p 6= b ∧ a 6= q ∧ ¬on(c,Line (a, q) →IntersectLines (Line (p, b),Line (a, q)) ↓ ∧B(p, b, IntersectLines (Line (p, b),Line (a, q))) ∧B(a, IntersectLines (Line (p, b),Line (a, q)), q) (Explicit Outer Pasch)

9.9. Inner Pasch implies Outer Pasch and plane separation. OuterPasch was an axiom (instead of, not in addition to, Inner Pasch) in versions ofTarski’s theories until 1965, when it was proved from Inner Pasch in Gupta’sthesis [13].27 Outer Pasch appears as Satz 9.6 in [26], and the plane separationtheorem follows it as Satz 9.8. The proofs as given in [26] are constructive, andare valid also for strict betweenness.

Lemma 9.57 (Gupta [13]). Explicit Outer Pasch is provable in neutral ECG.

Proof. This is Theorem 3.70 in [13], or Satz 9.6 in [26].

Lemma 9.58 (Plane separation, Gupta [13]).

SameSide (a, b, L) ∧ OppositeSide (a, c, L) → OppositeSide (b, c, L).

Proof. See [26], Satz 9.8.

Remark. We note that both theorems are proved using Satz 9.5, which is thespecial case of the plane separation theorem when Line (a, b) meets L. Neverthe-less there is no argument by cases according as that line does or does not meetL; the proof is completely constructive.

Lemma 9.59. SameSide (a, b, L) ∧ SameSide (b, c, L) → SameSide (a, c, L).

Proof. Suppose SameSide (a, b, L) ∧ SameSide (b, c, L). Then there exists apoints x and y such that OppositeSide (a, x, L) and OppositeSide (b, x, L). SinceSameSide (b, c, L), by Lemma 9.58 we have OppositeSide (c, x, L). Hence by def-inition of SameSide , we have SameSide (a, c, L). That completes the proof.

Note that OppositeSide (a, b, L) ∧ OppositeSide (b, c, L) → SameSide (a, c, L)is a trivial consequence of the definition of SameSide .

Lemma 9.60. If SameSide (a, b, L) then no point of L lies between a and b.Equivalently, no line meets all three (open) sides of a triangle.

Proof. The two statements are equivalent since SameSide (a, b, L) means that Lmeets two sides of some triangle abx. Suppose that B(a, u, x) and B(b, v, x) andB(a,w, b). Then OppositeSide (x, b, L) and OppositeSide (b, a, L), so SameSide (v, a, L).But since OppositeSide (a, v, L), by Lemma 9.58 we have OppositeSide (a, a, L).Hence for some y we have B(a, y, a), contradiction. That completes the proof.

The converse will be taken up in section 9.14

27But apparently, judging from footnote 4 on p. 191 of [30], Tarski knew as early as 1956-57that Outer Pasch implies Inner Pasch; in that footnote Tarski argues against replacing OuterPasch with Inner Pasch as an axiom, as Szmielew and Schwabhauser chose to do. Also onp. 196 of [30], Tarski attributes the idea for the proof of Inner Pasch from Outer Pasch tospecific other people; the history is too long to review here, but he credits only Gupta withthe derivation of Outer Pasch from Inner Pasch.


9.10. The crossbar theorem. The following lemma is called the “crossbartheorem” on p. 116 of [12]. It is needed to prove the basic theorems about theordering of angles. One special case of it occurs as Satz 3.17 in [26], but I do notfind the complete theorem there. We prove it here from Outer Pasch and InnerPasch; but since by Lemma 9.57, Outer Pasch is derivable from Inner Pasch, thecrossbar theorem is provable in ECG.

Lemma 9.61 (Crossbar Theorem). Suppose given triangle abc, and points Aon Ray (b, a) and C on Ray (b, c), with A 6= B and C 6= B, and point e withB(a, e, b), and point E the intersection point of Line (b, e) with Line (A,C). ThenB(A,E,C).

Proof. If A = E then e is on Line (b, a) and Line (a, c) and hence it is equal toa, since those two lines do not coincide because c is not on Line (b, a). HenceA 6= E. Similarly C 6= E.

We argue by cases on the disjunction B(b, c, C) ∨ C = c ∨ B(b, C, c). We willshow that in each case B(A,E,C). That will suffice since the double negationof this disjunction holds, so we will then have ¬¬B(A,E,C) and by Markov’sprinciple for betweenness we conclude B(A,E,C). Within each case, we willalso argue by cases according as B(b, a, A) ∨ A = a ∨ B(b, A, a), with a similarjustification. That gives a total of nine cases; but the problem is symmetrical ifvariables a,A are changed with variables c, C, so some of the nine cases can beomitted.

We first take up the case c = C. If a = A then E = e and we are done. IfB(b, a, A) then we apply outer Pasch (Lemma 9.57), with (a, p, c, q, b) replacedby (c, e, a, A, b). The hypotheses of outer Pasch under this substitution becomeB(c, e, a) and B(A, a, b), both of which we have by the assumptions and the sym-metry of betweenness. The conclusion is that the intersection point of Line (e, b)and Line (A,C), which is E, satisfies B(c, E,A). Since c = C we have B(A,E,C)by the symmetry of betweenness, as desired. Now suppose B(b, A, a). We nowapply inner Pasch (Axiom B4) with (a, p, c, q, b) replaced by (b, A, a, e, c). Thehypotheses become B(b, A, a) and B(c, e, a), which we have. The conclusion isB(A,E, c); since c = C we have B(A,E,C) as desired. That completes the casec = C. The case a = A is symmetrical, so we may drop subcases below whena = A.

We next take up the case, B(b, c, C). First assume B(b, a, A). We apply outerPasch (Lemma 9.57), with (a, p, c, q, b) replaced by (a, e, c, C, b). The hypothesesof outer Pasch under this substitution become B(a, e, c) and B(C, c, b), bothof which we have. The conclusion tells us that there is a point x such thatB(a, x, C) and B(b, e, x). Now, we apply outer Pasch again, this time with(a, p, c, q, b) replaced by (C, x, a,A,B). Then the hypotheses become B(C, x, a)and B(A, a,B) both of which we have (by the assumptions and the symmetryof betweenness). The conclusion is that there exists a point y with B(b, e, y)and B(A, y, C). But that point y is E, since only one point lies on Line (b, e)and Line (A,C), and that point is E. That completes the subcase B(b, a, A). Asnoted, we can drop the subcase a = A; we take up the subcase B(b, A, a). Weapply inner Pasch with (a, p, c, q, b) replaced by (C, c, b, A, a). The hypothesesare B(C, c, b) and B(a,A, b), which we have by the symmetry of betweenness.

86 MICHAEL BEESON

The conclusion is B(c, E, a) ∧ B(A,E,C), the latter of which is our goal. Thatdisposes of the case B(b, c, C).

The last case is B(b, C, c). The subcase B(b, a, A) is symmetric to the one wejust finished, so it is done. The subcase a = A is also done. That leaves as thelast subcase, B(b, A, a). This is treated similarly to the first case, except thatnow we use inner Pasch instead of outer Pasch. Let x be the intersection poitof Line (b, c) and Line (A, c). Then we claim B(b, x, e) as can be proved by oneapplication of inner Pasch. Then we prove B(C,E,A) in one more applicationof inner Pasch. That completes the proof.

9.11. Ordering of angles. Euclid often mentions a relation < between an-gles, so in order to claim that ECG is faithful to Euclid, we must define thisnotion in ECG.

Definition 9.62. Point p lies in the interior of angle abc if B(a, e, c)and ¬B(p, b, e), where e = IntersectLines (Line (b, p),Line (a, c). Point p lies inangle abc if T(a, e, c) and ¬B(p, b, e) , where T is non-strict betweenness.

Lemma 9.63. If abc and ABC are the same angle and p lies in the interior ofabc then p lies in the interior of ABC. Similarly, if p lies in abc then p lies inABC.

Proof. Since ABC and abc are the same angle, B = b, and A lies on Ray (b, a)and C lies on Ray (b, c). Let L = Line (b, p) and let e be the intersection pointof L with Line (a, b), and let E be the intersection point of L with Line (A,B).Suppose p lies in the interior of abc; then by definition B(a, e, c) and ¬B(p, b, e).We must show B(A,E,C) and ¬B(p, b, E). Now p is irrelevant and we can applyLemma 9.61. That completes the proof of the first part.

The second part is similar, but we only assume T(a, e, c) instead of B(a, e, c)and must prove T(A,E,C). If e = A then Line (b, p) and Line (b, a) coincide, soE = A and T (A,E,C). Similarly if e = b then E = B and T (A,E,C). But sinceT(a, e, c), by definition of T we have ¬¬(e = c∨ e = c∨B(a, e, c). Since in eachof the three cases we have T(a, e, c), we have ¬¬T(a, e, c), and hence T(a, e, c).That completes the proof of the lemma.

Definition 9.64 (Ordering of angles). abc < ABD if there is a point C in theinterior of angle ABD such that angle abc is congruent to angle ABC. Similarly,abc ≤ ABD if there is a point C in angle ABD such that angle abc is congruentto angle ABC.

Remark. There is no need to use the notation ∠ABC < ∠DEF since ABC <DEF could not have any other meaning.

Lemma 9.65 (Angle ordering respects congruence). If abc < def and anglesabc and def are congruent to angles ABC and DEF respectively, then ABC <DEF . Similarly with ≤ in place of <.

Proof. Since abc < def , there is a point x in the interior of angle def suchthat angle abc is congruent to angle dex. By Lemma 9.63, we may assume thatB(d, x, f). Without loss of generality we can assume ED = ed and EF = ef ,since changing the points E and F on Ray (B,E) and Ray (B,F ) makes thesame angle. Then by SAS, DF = df . Let X be a point on Ray (D,F ) such that


DX = dx. Then triangle dex is congruent to triangle DEX . Hence angle dex iscongruent to angle DEX . It only remains to show B(D,X,F ). By SAS, triangledef is congruent to triangle DEF , so df = DF . Now we have B(d, x, f) anddx = DX and df = DF , and X is on Ray (D,F ). If X = F then segment df iscongruent to its proper subsegment dx, contradiction; hence X 6= F . If X = Dthen non-null segment dx is congruent to null segment DX , contradiction. Nowassume B(D,F,X). Then DF < DX , so by Lemma 9.18, df < dx. Hence thereis a point p between d and x such that dp = df . Since p and f are both onRay (d, f), we have p = f by Lemma 9.5. But B(d, x, f) and B(d, p, x) impliesB(p, x, f) by Lemma 9.7, and so we have B(f, x, f) since p = f . This contradictsAxiom B1-ii. Hence ¬B(D,F,X). But X is on Ray (D,F ). Hence by Axiom B3,¬¬B(D,X,F ), and by Markov’s principle for betweenness, B(D,X,F ). Thatcompletes the proof of the lemma.

9.12. Triangle construction and uniqueness. The triangle constructionand uniqueness principles proved in this section were axioms of Tarksi’s earlytheories, and they are closely related to Hilbert’s axiom about laying off a givenangle at a given point on a given line. That something so fundamental is aconsequence of Pasch and SAS (in the form of the five-segment axiom) is ratheramazing. We check here that it also holds constructively. According to [30],these beautiful theorems were first proved in the year 1956-57, along with otherprinciples that had been axioms until then, “as a result of joint efforts by EvaKallin, Scott Taylor, and Tarski”.

Lemma 9.66. Let a, b, and x be points on a line L with a 6= b. Let AB = ab.Then we can find a point X on Line (A,B) with AX = ax and BX = bx.

Remark. We are not told anything about the order relations of x with a and b.

Proof. By Lemma 9.14, construct a point e on Line (b, a) such that B(e, a, b)and B(e, x, b). Then extend segment BA by ea to construct a point E withB(E,A,B) and EA = ea. Then find a point X on Ray (E,A) with EX = ex.We claim that X is the desired point, i.e. AX = ax and BX = bx. Sinceequality is stable, we can prove this by cases on the possible order relations ofa, b, and x. (The point is that the construction of X did not demand suchcase distinctions; a single term with parameters (a, b, x, A,B) gives X .) Case1, B(a, x, b). Then since ex = EX and ea = EA we have by Lemma 9.6 thatax = AX . Since ab = AB, applying Lemma 9.6 again we have xb = XB (andhence BX = bx). Case 2, x = a. Then since ex = ea we have X = A byLemma 9.4. Hence ax = aa = AA since all null segments are congruent; andbx = ba = BA = BX , so bx = BX . Case 3, B(x, a, b). Then since ea = EA andex = EX we have xa = XA by Lemma 9.6; so ax = AX . Then since ab = ABwe have xb = XB by Lemma 9.6, so bx = BX . The other cases reduce to thesecases by interchanging a with b and A with B. That completes the proof of thelemma.

Lemma 9.67 (Triangle construction theorem). Let abc be a triangle (so thevertices are distinct and not collinear). Let AB = ab and let z be a point noton Line (A,B). Then there is a point C such that triangle ABC is congruent to

88 MICHAEL BEESON

triangle abc and zC does not meet L. (In fact point C exists if we only assumea, b, and c are distinct, without assuming the are non-collinear.)

Remark. The conclusion says zC does not meet L, rather than SameSide (z, C, L).We will prove the equivalance of these two statements, but only later, using theupper dimension axiom.

Proof. Let x be a point on Line (a, b) such that cx ⊥ ab. The existence of suchan x was proved in Lemma 9.56. By Lemma 9.66, we can find a point X onL = Line (A,B) such that ax = AX and bx = bX . By Lemma 9.56, we can finda line K perpendicular to Line (A,B) at X . Now we want to take C to be apoint on K with CX = cx, and C on the same side of L as z. To construct C,drop a perpendicular from z to K (which we can do without knowing whetherz is on K or not, by Lemma 9.56). Let w be the foot of that perpendicular,so w is on K. Then wz does not meet L, by Lemma 9.48, since wz ⊥ K andK ⊥ L. In particular w 6= X , since X lies on L, so Line (X,w) is defined, andit makes sense to speak of a point lying on Ray (X,w). Then we can constructC as a point on Ray (X,w) with XC = xc. Note that we have not yet usedthat a b, and c are non-collinear; so the parenthetical remark in the lemma isestablished. But now, since c does not lie on Line (a, b), axc is a triangle, andX does not lie on Line (A,B), so AXC is a triangle. Then since all right anglesare congruent, we have triangle axc congruent to triangle AXC, and triangleBXC congruent to triangle bxc. Hence AC = ac and BC = bc. Since we aregiven AB = ab, triangle ABC is congruent to triangle abc by SSS (the definitionof triangle congruence. We must still show that Cz does not meet L. Supposethat Cz meets L in point p. Then C 6= w since zw does not meet L. Since Cis on Ray (X,w), we have ¬¬(B(X,C,w) ∨ B(X,w,C)). If B(X,w,C) then weapply Inner Pasch, with (a, p, c, b, q) replaced by (z, p, C,X,w). The conclusionof Inner Pasch gives as point r with B(p, r,X) and B(w, r, z). But p and X areon L, so r is on L; but then since B(w, r, z), L meets wz, contradiction. Hence¬B(X,w,C). If B(X,C,w) we instead apply Outer Pasch, with (a, p, c, q, b)replaced by (z, p, C,w,X). The conclusion gives us a point r with B(z, r, w) andB(r, p,X). Again since r and p are on L, B(r, p,X) implies r is on L, and thenB(z, r, w) shows that zw meets L, contradiction. Hence ¬B(X,C,w). But nowwe have contradicted ¬¬(B(X,C,w) ∨ B(X,w,C); hence Cz does not meet Lafter all. That completes the proof.

Note that intuitively, C is the intersection point of two circles, circle C withcenter A and radius ac, and circle K with center B and radius bc. The problemis to prove those circles intersect. If there is a shorter way to show the circles dointersect than the above proof, I do not know it.

The uniqueness of the triangle, which is the next lemma, is more difficult thanthe existence. Even if we knew those circles intersected, we have not proved thattwo circles intersect in only one point on a given side of the line joining theircenters; indeed, this lemma will be used to prove that.

Lemma 9.68 (Triangle uniqueness theorem). In the previous lemma the pointC is unique in the sense that if ABC and ABD are both congruent to abc andC and D are on the same side of Line (A,B) as x, then C = D.


Remark. We follow the general idea of Satz 10.16 of [26], but appealing to ourLemma 9.56 for the existence of perpendiculars. Note that uniqueness in thesense that Cx does not meet Line (A,B) will not hold in three dimensions andwe have not yet used the upper dimension axiom; but uniqueness in the same-sidesense does hold in higher dimensions.

Proof. Suppose C andD are two points on the same side ofAB as x, and trianglesABC and ABD are both congruent to triangle abc. Then by Lemma 9.1, triangleABC is congruent to triangle ABD. Let s be a point on Line (A,B) such thatCs ⊥ AB; such a point exists by Lemma 9.56. Extend segment Ds (not Cs) topoint C∗ such that B(D, s, C∗) and Ds = sC∗. Now C∗ and D lie on oppositesides of Line (A,B), as witnessed by point s, and D and C lie on the same sideof Line (A,B), by hypothesis. By Lemma 9.38, Line (A,B) is perpendicular toLine (D,C∗), and hence, if s 6= B, trianglesDsB andC∗sB are congruent, and soDB = C∗B. If s = B then DB = Ds = sC∗ = BC∗. Since ¬¬(s = B ∨ s 6= B),we have ¬¬DB = C∗B, and hence DB = C∗B. Since triangles ABD and ABCare congruent, we also have DB = CB, so by Lemma 9.1 we have DB = CB.

By the plane separation principle, C andC∗ lie on opposite sides of Line (A,B).Hence there is a point t on Line (A,B) between C and C∗. Since trianglesABC and ABD are both congruent to triangle ABC, we have AC = AD andBC = BD. Angle BsC∗ is congruent to angle BsC by Lemma 9.43, so bySAS, triangle BsC∗ is congruent to triangle BsC. Hence BC∗ = BC. Similarly,exchanging A and B in the argument, we have AC∗ = AC.

We claim tC = tC∗. To prove this, we have to argue by cases on the possibleorderings of A, t, and B on Line (A,B). If we show in every case tC = tC∗ thensince not all the cases can fail, we conclude ¬¬tC = tC∗ and hence tc = cC∗ bythe stability of equality. First assume B(A, t, B) or B(A,B, t). By Lemma 9.14,let E be a point on Line (A,B) such that B(A, t, E) and B(A,B,E). ApplyLemma 9.10 with (a, b, c, d, e) replaced by (A, t, B,C,E) and (A,B,C,D,E) re-placed by (A, t, B,C∗, E). The hypotheses are satisfied since AC = AC∗ andBC = BC∗. The conclusion is tC = tC∗. The next case is t = B; thentC = BC = BC∗ = tC∗. The next case is t = A; then tC = AC = AC∗ = tC∗.The remaining cases reduce to these cases by interchanging A and B. HencetC = tC∗.

Now we have AC = AC∗ and BC = BC∗ and tC = tC∗. Assume A 6= t, soAtC and AtC∗ are triangles. Then triangle AtC is congruent to triangle AtC∗,by definition of triangle congruence (SSS). Then angle AtC is congruent to angleAtC∗, as corresponding angles of congruent triangles. Then AB ⊥ CC∗, bydefinition of perpendicular. Then Cs and C∗t are two perpendiculars from C∗

to Line (A,B). Hence by Lemma 9.46 we have s = t. Then both C and D lieon Line (C∗t), and both extend segment C∗t by tC, since Dt = Ds = sC∗ byconstruction of C∗, and Ct = tC∗ so since s = t we have Dt = sC∗ = tC∗ = Ct.But then by Lemma 9.4, C = C∗. That is the desired conclusion, but it wasreached under the assumptionA 6= t. Under the assumptionB 6= t, interchangingA and B in the argument we also reach the desired conclusion C = C∗. But sinceA 6= B, we have ¬¬(t 6= A ∨ t 6= B) (which is equivalent to ¬(t = A ∧ t = B)).Hence ¬¬C = C∗; hence C = C∗. That completes the proof.

90 MICHAEL BEESON

Lemma 9.69. [SAA] If angles abc and acb are congruent respectively to anglesABC and ACB, and ab = AB, then triangles abc and ABC are congruent.

Proof. By Lemma 9.67 there exists a point E on the same side of Line (A,B)as C such that triangle ABC is congruent to triangle abc. Then angle ABE iscongruent to angle ABC since both are congruent to angle abc. Let x be a pointon Ray (A,C) such that ac = AX . Then since angle ABC is congruent to angleABE, we have triangle ABC congruent to triangle ABE. But since E and Care on the same side of Line (A,B), by Lemma 9.68, E = C. Hence triangleABC is congruent to triangle abc, since triangle ABE is congruent to triangleabc. That completes the proof.

Lemma 9.70. Suppose angle abc is congruent to angle abd. Then d lies onRay (b, c).

Proof. Let A lie on Ray (b, a) with AB = αβ, and let C lie on Ray (b, c) withBC = αβ, and D on Ray (b, d) with bd = αβ. By the definition of angle congru-ence, triangle AbC is congruent to triangle AbD. Hence, by Lemma 9.68, C = D.Then Ray (b, c) coincides with Ray (b, d), so d lies on Ray (b, c) as claimed. Thatcompletes the proof.

Lemma 9.71. Let z be a point not on line L. Two circles with distinct centerson L intersect in at most one point C such that Cz does not meet L.

Proof. Let C and D be two intersection points of circles C and K with centersA and B respectively. Triangle ABC is congruent to ABD, so by Lemma 9.68,not both Cz and Dz can fail to meet L. That completes the proof.

Lemma 9.72 (Stability of angle ordering).

¬¬ abc < def → abc < def

Remark. The proof is nontrivial; after the definitions are unwound, one sees thekey idea, but it still requires several of the difficult theorems already proved tocomplete the proof. The proof is illustrated in Fig. 18. Note also that this is thefirst time we have used IntersectCirclesSame .

Figure 18. Stability of angle ordering. Given angle dep con-gruent to angle abc, we must have R = r.

bb b

c

ba

be br

b d

bA

b f

b pbqb

R


Proof. By the definition of angle ordering, abc < def means that there existsa point p in the interior of angle def such that angle abc is congruent to angledep. By Definition 9.62, p is in the interior of angle def if and only if B(d, q, f)∧¬B(p, e, q), where

q = IntersectLines (Line (e, p),Line (d, f)).

That is, p lies on Ray (e, q).The key observation is that there is a canonical choice of point p, namely the

one produced by copying triangle abc onto segment de with its vertex at b. (Thatobservation is key, because it means that we do not have to pass a double negationthrough an existential quantifier ∃p ; the quantifier can be replaced by a termdefining p, as we shall see in detail.) Let C = Circle (e, a, b) and let A be a pointon Ray (e, d) such that eA = ba; then A = IntersectLineCircle2 (Line (e, d), C),so A is given by a term. Let K = Circle (A, a, c), and let

r = IntersectCirclesSame (C,K, f).

By the triangle construction theorem (Lemma 9.67) we see that r is defined.We claim that abc < def if and only if r is in the interior of angle def . Thatis, if there is any p in the interior of angle def such that abc is congruent todep, then this r is such a p. To prove this, we argue as follows. Suppose thatabc is congruent to dep and p lies in the interior of angle dep. Triangle Aeris congruent to triangle abc by definition of triangle congruence, since bc = erand ac = Ar by the definition of r, and ab = Ae by construction of A. Henceangle abc is congruent to angle Aer. Hence angle Aer is congruent to angle dep,since both are congruent to angle abc. Let point R on Ray (e, p) have eR = er.Then triangle Aer is congruent to triangle AeR by SAS. We claim r = R. LetL = Line (e,A). By the triangle uniqueness theorem, it suffices to prove thatr and R are on the same side of L. By construction, r is on the same side ofLine (e,A) as f , since L joins the centers of C and K. Since p is in the interiorof angle def , the point q defined above lies between d and f , so by Lemma 9.83,q and f are on the same side of L, since segment qf does not meet L. But Rlies on Ray (e, p). Hence Rq does not meet L, since Line (e, p) meets L in e andq does not lie on L. Hence by Lemma 9.83, R and q are on the same side ofL. Since we already proved q and f are on the same side of L, by the planeseparation theorem, r and R lie on the same side of L. That completes the proofthat if any p in the interior of angle def has abc congruent to dep, then r is sucha p. In other words,

abc < def↔B(d, q, f) ∧ ¬B(r, e, q)

where

q = IntersectLines (Line (e, r),Line (d, f))

and where r is given by a specific term with variables a, b, c, d, and e, as definedabove. Hence

92 MICHAEL BEESON

¬¬ abc < def ↔ ¬¬ (B(d, q, f) ∧ ¬B(r, e, q))

↔ ¬¬B(d, q, f) ∧ ¬¬¬B(r, e, q)

↔ ¬¬B(d, q, f) ∧ ¬B(r, e, q)

↔ B(d, q, f) ∧ ¬B(r, e, q)

↔ abc < def by the stability of betweenness

That completes the proof of the lemma.

9.13. Euclid Books I, II, and III. We are now in a position to prove thetheorems in the first three books of Euclid. We mention a few of these that wewill need later. Since we are still working in neutral geometry, we will explicitlymention which parallel axiom is needed. Euclid’s first use of the parallel postulateis in Prop. I.29, and it is also needed for theorems about circles in Book III.These propositions, even the ones in Book I, could not be proved earlier, sincemany of them mention the ordering of angles, and some of them require us tolay off an angle on a given side of a given segment; our ability to do that requiresthe triangle construction lemma; indeed Euclid I.23 is the triangle constructionlemma.

For example, a corollary of the congruence of vertical angles and the definitionof angle ordering is the exterior angle theorem, Euclid 1.16. (Without the parallelaxiom, one only gets an inequality, not the stronger theorem that an exteriorangle can be divided into two angles, each congruent to one of the interior andopposite angles.)

Moreover, in order to support arguments by cases on whether one angle islarger than another or not, we need the stability of angle ordering (Lemma 9.72),which as we have seen was not trivial to prove.

Lemma 9.73 (Exterior angle theorem). Suppose B(B,C,D) and A does notlie on Line (B,C). Then angle ACD is greater than angle BAC. That is, thereis a point E such that B(A,E,C) and and point F such that B(B,E, F ) suchthat angle BAC equals angle ACF .

Proof. This is Euclid I.16. We note that according to our precise definitionof angle ordering, the meaning of the proposition is as stated here. CompareEuclid’s proof to the proof in [12], Theorem 4.2, page 165, which requires themuch deeper crossbar theorem (Lemma 9.61); Euclid’s proof does not require it.In particular it is not required to prove that F and G in Euclid’s diagram lie onopposite sides of Line (B,C).

Lemma 9.74. A leg of a right triangle is less than the hypotenuse.

Proof. Euclid does not state this explicitly, but by the exterior angle theorem(Euclid I.16), or Lemma 9.73 above, the right angle is greater than each of theother two angles, so the lemma follows from Euclid I.19, which in turn followsfrom Euclid I.18. Euclid’s proofs make use of an argument by contradiction onangle ordering, but since angle ordering is stable by Lemma 9.72, Euclid’s proofsare essentially constructive. That is as much as we will say about this proof.


Lemma 9.75 (Euclid III.2). Let line L meet circle C in two points a and b,and let point x on L be different from both a and b. Then x is inside c if (andonly if) x is between a and b, and otherwise x is outside c.

Remark. Euclid’s proof has been criticized (see e.g. Heath’s commentary on theproposition in [8]), since III.2 uses III.1 and vice-versa. The alleged flaw is thatEuclid’s proof of III.2 starts by using III.1 to find the center of the given circle,but III.1 implicitly uses III.2. That criticism does not apply in our theory, sincecircles always “come equipped” with their centers. That is, instead of needingIII.1, we can just take the center to be center(C). To put the matter anotherway: the issue is that if one views Euclid III.1 as an attempt to show thatcenter(C) is definable and hence redundant, the attempt is circular, and hencefails; but that is a different matter from the correctness of Euclid III.2.

Proof. Euclid’s proof proceeds by contradiction, but that is constructively ac-ceptable due to the stability of segment ordering, proved in Lemma 9.15. Hisproof appeals to the equality of the base angles of an isosceles triangle (Lemma 9.28)and to euclid I.16 and I.19, whose proofs have been discussed above.

Lemma 9.76. If line L meets chord pq of circle C, then the two intersectionpoints of L with C lie on opposite sides of Line (p, q).

Figure 19. Lemma 9.76. B(p, x, q) implies B(a, x, b).

b e

bp

bq

ba

b b

b x

Remark. This innocent-looking lemma requires Euclid III.2, which as we haveseen requires the exterior angle theorem and the triangle construction and unique-ness theorems. See Fig. 19.

Proof. Let L meet pq in point x, so B(p, x, q), and let a and b be the intersectionpoints of L with C. We have to show B(a, x, b). Let e be the center of C. Thenby Lemma 9.75, since x lies on the chord pq, x is strictly inside C, i.e. ex < ep.Suppose B(a, b, x). Then by Lemma 9.75, x is strictly outside C, i.e. ex > ep.

94 MICHAEL BEESON

This is a contradiction; hence ¬B(a, b, x). Similarly ¬B(x, a, b). Also x 6= a andx 6= b since x is strictly inside C. Hence by axiom B3, B(a, x, b) as required.That completes the proof.

Lemma 9.77 (Euclid I.29). The Playfair parallel axiom implies that the alter-nate interior angles formed by a transversal to parallel lines are congruent.

Proof. By Lemma 9.34, if the alternate interior angles are not not congruent,they are congruent. Hence we can prove the proposition by contradiction. ThenEuclid’s proof is valid. That completes the proof.

Lemma 9.78 (Euclid III.20). The Playfair parallel axiom implies that an ex-terior angle of a triangle is equal to the sum of the alternate interior angles.

Remark. The use of the word “sum” does not imply addition. It merely meansthat the exterior angle can be decomposed into two angles, each congruent to oneof the opposite interior angles. Addition of angles has not been defined (eitherhere or in Euclid).

Proof. Let abc be the triangle and led d be a point on Line (a, c) with B(a, c, d),and consider the exterior angle bcd. The line mentioned in the remark isPara(c,Line (a, b)). Then angle abc is congruent to the alternate interior anglebce (by Lemma 9.77), and angle bca is congruent to angle ecd, as can be provedclassically from the parallel postulate. Double negating this proof, and usingthe stability of angle congruence (Lemma 9.34), we find that these congruenceassertions are derivable constructively from Playfair. That completes the proof.

Lemma 9.79 (Euclid III.20). The Playfair parallel axiom implies: An angleinscribed in a semicircle is a right angle. That is, if ea = eb = ec and B(a, e, b),then angle acb is a right angle. More generally, if a and b are any points on thecircle, and c is any point on the circumference, angle aeb is double angle aec.

Remark. Note that, although we allow degenerate circles, for this lemma thesemicircle must be non-degenerate, as otherwise the notion of “inscribed angle”makes no sense.

Proof. The proof in Euclid is constructive; see Fig. ??. Euclid’s proof dependson Euclid I.32, that an exterior angle of a triangle equals the sum of the oppo-site interior angles; we have proved I.32 above from the Playfair axiom. Thatcompletes the proof.

9.14. Consequences of the upper dimension axiom. So far, we havenot used the upper dimension axiom. Now we will use it.

Lemma 9.80. Suppose K and M are lines perpendicular to L at m. Then Kand M coincide (have the same points).

Remark. This result depends on the upper dimension axiom, since in 3-spacethere are many perpendiculars to L at m.

Proof. Suppose K and M are perpendicular to L at m. Let a be a point onK not on L. Extend segment am to b such that am = mb and B(a,m, b).Let c be a point on M with mc = ma. Let p and q be two points on L with


pm = mq (to be specific take pm = mq = ma although the distance pm isnot essential). Then since K ⊥ M we have triangle amp congruent to amqand triangle bmp congruent to bmq, and triangle cmp congruent to cmq. LetCa = Circle (a, q), let Cb = Circle (b, q), and Cc = Circle (c, q). Then p and q lieon all three circles. By the upper dimension axiom, their centers a, b, and c arecollinear. By Lemma 9.11 we do not have B(m, c, a) or B(m, a, c) or B(m, b, c)or B(m, c, b). Since we also have c 6= m, by Axiom B3 we have ¬¬(c = b∨c = a).If c = b ∨ c = a then K and M coincide, so in either case c is on K; thereforewe have ¬¬on(c,K). By stability axiom S3, we can drop the double negation,so on(c,K). Hence K and M coincide. That completes the proof.

Lemma 9.81. Suppose K and M are lines perpendicular to L and x lies onboth K and M . Then K and M coincide.

Remark. The point is that we do not require a case distinction as to whether xis or is not on L.

Proof. Let u be on K; we must show u is on M . If x is on L, then Lemma 9.80applies, so on (u,M) in that case. If x is not on L, then Lemma 9.46 ap-plies, so on (u,M) in that case. But we have ¬¬ (on (x, L) ∨ ¬on (x, L). Hence¬¬ on(u,M). But then by Axiom S3 we have on (u,M). That completes theproof.

Lemma 9.82. Let ab be a chord of circle C. Then the center of C lies on theperpendicular bisector of ab.

Proof. Let m be the midpoint of ab and p the center of C. Then pa = pb andma = mb and pm = pm, so triangle pam is congruent to triangle pbm. Hencemp is perpendicular to ab at m. But also the perpendicular bisector of ab isperpendicular to ab at m; so by Lemma 9.80, p is on the perpendicular bisector.That completes the proof.

Figure 20. If ab does not meet L then a and b are on the sameside of L because bc meets L. (Lemma 9.83).

L

ba

b bbw

b

rb

x

bm

bc

z

Lemma 9.83. If a and b are not on L and no point of L is between a and bthen SameSide (a, b, L). Indeed a specific point c, namely the reflection of a in

96 MICHAEL BEESON

L, is constructible (i.e. given by a term t(a, L)) such that c is on the oppositeside of L from both a and b.

Remark. Recall that SameSide (a, b, L) means a and b are both on the oppositeside of L from some point c. This lemma obviously fails in three-space, since ifa and b are on the same side of L then they are coplanar with L, but it may bethat segment ab does not meet L even though they are not coplanar.

Proof. Drop a perpendicular from a to L and let x be its foot, so x is on L andax ⊥ L. By Lemma 9.56, construct a line K perpendicular to ax; let w be apoint on both K and Line (a, x). Then w 6= x since then, by Lemma 9.80, L andK would coincide, so b would lie on L, contrary to hypothesis. Let m be themidpoint of segment wx. Let rm = mb with B(r,m, b).

Assume w 6= b. Then angles rmx and bmw are vertical angles, so they arecongruent by Lemma 9.33. Then triangles rmx and bmw are congruent, by SAS.

On the other hand (still assuming w 6= b), let s be a point on L with sx = wb.Since w 6= b, we have s 6= x. Since these are both right angles, by Lemma 9.44they are congruent. Then triangles sxm and bwm are congruent, by SAS. Thenrx and sx are both perpendicular to K at x. By Lemma 9.80, Line (x, r) andLine (x, s) coincide. Therefore r lies on L. (For all we know s might be r or itmight be the reflection of r in x, so we cannot say s = r, but we are finishedwith s now that we know r is on L.) Let c be the reflection of a in x, that is,the point on K with B(c, x, a) and ax = cx.

Assume B(m,x, c). We wish to apply Outer Pasch with (a, p, c, q, b) replacedby c, x,m, b, r). We check the hypotheses of Outer Pasch under this substitution.One is B(b,m, r), which we have by construction of r. The other is B(m,x, c),which we have assumed. Hence Outer Pasch can be applied. The result is apoint z with B(c, z, b) and B(z, x, r). Since B(z, x, r), z is on L, and hence bcmeets L. Then b 6= c since B(c, z, b), so z = IntersectLines (L,Line (b, c)). Wehave thus proved

w 6= b → on(IntersectLines (L,Line (b, c))

under the assumptions w 6= b and B(m,x, c).Now keep the assumption w 6= b, but replace B(m,x, c) with the assumption

B(m,x, a). Interchanging a and c in the above argument we find by Outer Pascha point z on L with B(a, z, b), contradicting the assumption that ab does notmeet L. Hence ¬B(m,x, a). We also have m 6= x, since if m = x then wx isa null segment, so bw and L are both perpendicular to ax at x, contradictingLemma 9.80. Hence by Axiom B3,

¬¬(B(x,m, a) ∨m = a ∨ B(x, a,m).

We claim that in each of the three cases we have B(m,x, c). We have B(a, x, c) byconstruction of c, so if m = a we have B(m,x, c) immediately. If B(x,m, a) thenB(a,m, x); together with B(a, x, c) we get B(m,x, c) by Lemma 9.9. If B(x, a,m)then from B(c, x, a) we obtain B(c, x,m) by outer transitivity (Lemma 9.8) andhence B(m,x, c). Hence all three cases imply B(m,x, c). Hence ¬¬B(m,x, c).By Markov’s principle for betweenness we have B(m,x, c). But above we haveshown that this assumption leads to the conclusion that bc meets L.


Now drop all betweenness assumptions, and assume w = b. Then b 6= c, sinceba does not meet L, but bc meets L in x. Hence Line (b, c) is defined and equalsK. Hence

IntersectLines (L,Line (b, c)) = x

and hence IntersectLines (L,Line (b, c)) lies on L.We have thus proved

w = b ∨ w 6= b → on(IntersectLines (L,Line (b, c))

But we have ¬¬ (w = b ∨ w 6= b). Hence

¬¬ on(IntersectLines (L,Line (b, c)).

By stability axiom S3, we can drop the double negation. Then b and c are onopposite sides of L, by definition of OppositeSide . Since B(c, x, a), a and c areon opposite sides of L. Then a and b are on the same side of L, by definition ofSameSide . That completes the proof of the first part of the lemma.

For the second part, we use the fact that there is a term representing everyruler and compass construction, and we showed how to construct c, which wenow summarize.

p = pointOn1(L)

q = pointOn2(L)

C = Circle(p,a)

K = Circle(q,a)

u = IntersectCircles1(C,K)

v = IntersectCircles2(C,K)

M = Line(u,v);

x = IntersectLines(L,M);

c = IntersectLineCircle2(Line(a,x),Circle(x,a))

There is, of course, a term t(a, L) representing this construction. That com-pletes the proof.

Lemma 9.84. Not both SameSide (a, b, L) and OppositeSide (a, b, L).

Proof. Suppose SameSide (a, b, L) and OppositeSide (a, b, L). Then some c is onthe opposite side of L from both a and b. Since b is on the opposite side from cand a is on the opposite side from b, a is on the same side as c. Since a and care on opposite sides, and a and c are on the same side, by the plane separationlemma (Lemma 9.58), a and a are on opposite sides. Hence there is a point uwith B(a, u, a), contradicting Axiom B1-ii. That completes the proof.

Lemma 9.85 (Stability of SameSide ).

¬¬SameSide (a, b, L) → SameSide (a, b, L).

Proof. Suppose a is not on L and ¬¬SameSide (a, b, L). We claim that no pointof L is between a and b. Assume that B(a, z, b). Then a and b are on oppositesides of L. Then ¬SameSide (a, b, L), by Lemma 9.84. But that contradicts¬¬SameSide (a, b, L). Hence there does not exist any z between a and b. Thenby Lemma 9.83, a and b are on the same side of L. That completes the proof.

98 MICHAEL BEESON

Remark. It is not hard to show that the stability of OppositeSide is equivalentto the special case of the principle ¬¬t ↓ → t ↓, when t is IntersectLines (L,K).This principle, which we have called S6, is not taken as an axiom of ECG, butwe will show below that it follows from the strong parallel axiom, so in ECGOppositeSide is also stable. We do not prove that here since we wish to put theconsequences of the parallel axiom in a separate section.

Lemma 9.86. Explicit Pasch is provable in ECG.

Remark. Clearly ExplicitPasch can fail in three-space, so it should require theupper dimension axiom to prove it. By contrast, Inner Pasch and Outer Paschdo not require the upper dimension axiom.

Proof. Suppose B(a, IntersectLines (Line (a, c), L), c) and suppose L does notmeet ab and that a and b do not lie on L. Then by Lemma 9.83, a and b areon the same side of L. By definition of opposite sides, a and c are on oppositesides of L. Hence, by the plane separation theorem (Lemma 9.58), b and c are onopposite sides of L. By definition of opposite sides, there is a point y on L withB(b, y, c). Since point b is not on L, IntersectLines (Line (b, c), L) is defined,by Axiom I12. Then by Axiom I13, y = IntersectLines (Line (b, c), L). SinceB(b, y, c), the conclusion of Explicit Pasch is verified. That completes the proofof the lemma.

Lemma 9.87. Explicit Outer Pasch implies Negative Pasch.

Proof. Suppose abc is a triangle, i.e. the points are distinct and non-collinear,and suppose line L meets ab in a point p between a and b. Suppose L meetsLine (b, c); let x be the point of intersection (there is only one since a, b, and care not collinear). We have to prove not all of the following fail: a, b, or c is onL, or L meets ac, or L meets bc. Suppose then that they all do fail. SupposeL meets Line (b, c). Let x be the point of intersection. Since L does not meetbc and is not equal to b or c, then ¬¬(B(x, b, c) ∨ B(b, c, x)). If B(x, b, c), thenby Axiom B4 (explicit inner Pasch), L meets ac, contradiction. If B(b, c, x),then by Explict Outer Pasch, L meets bc, contradiction. Hence L does not meetLine (b, c). Then b and c are on the same side of L. Since L meets segmentab, a and b are on opposite sides of L. By the plane separation theorem, a andc are on opposite sides of L. Hence L meets segment ac, contradiction. Thatcompletes the proof of the lemma.

Tarski also considered the following variant of Pasch’s axiom:

B(a, t, d) ∧ B(b, d, c) → ∃x∃y((¯a, x, b) ∧ B(a, y, c) ∧ B(y, t, x)) (Weak Pasch)

Of course Tarski used T instead of B, but in case a, b, and c are collinear,or t = a, or t = d, one can take x = y = t, so weak Pasch with T is classicallyequivalent to weak Pasch with B. Weak Pasch differs from the others in thatthe points x and y are neither unique nor in any way specified by the axiom.Nevertheless they can be constructed:

Lemma 9.88. Explicit Pasch implies weak Pasch.

Proof. Construct a parallel K to L = Line (a, d) through t and apply ExplicitPasch to triangle abd. Line K meets side ad at t and does not meet side bd since


it is parallel to L. Hence by Explicit Pasch, it meets side ab in x. Similarly,using triangle acd, K meets side ac in point y. That completes the proof.

9.15. Right and left turns and sides of lines. Recall that Right andLeft are define in ECG in terms of IntersectCircles1 and IntersectCircles2 . Ofcourse, these axioms can also be viewed as defining which intersection point ofcircles is which, in terms of the fundamental and intuitive notions of right andleft turns. These notions are in turn axiomatized by other axioms of ECG thatconnect Right and Left to the notions of SameSide and OppositeSide . Theseconnections can be completed with a few lemmas, which we now give.

Lemma 9.89.

B(a, b, e) ∧ Right (a, b, c) → Right (a, e, c)

B(a, e, b) ∧ Right (a, b, c) → Right (a, e, c)

B(e, a, b) ∧ Right (a, b, c) → Right (e, b, c)

and similarly with Left instead of Right .

Proof. Suppose B(a, b, e) ∧ Right (a, b, c). Then by Axiom H8, Right (c, a, b).Then e and b are on the same side of Line (a, c), as witnessed for example bythe reflection x of b in point a, i.e. the point x such that B(x, a, b) and bx = ab.Then B(x, a, e) by Lemma 9.7. Hence by Axiom H4, we have Right (a, e, c).That proves the first assertion of the lemma; the other five assertions are provedsimilarly. We omit the proofs.

The following rather basic lemma seems to require Lemma 9.89:

Lemma 9.90. If c is on Ray (a, b) and c 6= a then

IntersectLines (Line (a, b),K) ∼= IntersectLines (Line (a, c),K)

IntersectLineCircle1 (Line (a, b), C) ∼= IntersectLineCircle1 (Line (a, c), C)

IntersectLineCircle2 (Line (a, b), C) ∼= IntersectLineCircle2 (Line (a, c), C)

Proof. To prove the first assertion, suppose IntersectLines (Line (a, b),K) is de-fined. Then Line (a, b) and K do not coincide, by Axiom I21, and there is a pointm on both Line (a, b) and K, by Axiom I4. Since c and a are on both Line (a, b)and Line (a, c), by Axiom I3, those lines coincide; in particularm is on Line (a, c).Since m is also on K, and since K and Line (a, c) do not coincide, (or else K andLine (a, b) would coincide, which they do not), IntersectLines (Line (a, c),K) isdefined, by Axiom I12, and is equal to m, by Axiom I13. That completes theproof of the first assertion.

Letp = IntersectLineCircle1 (Line (a, b), C)

q = IntersectLineCircle2 (Line (a, b), C)

p′ = IntersectLineCircle1 (Line (a, c), C)

q′ = IntersectLineCircle1 (Line (a, b), C)

Then for points x not on Line (a, b),

Right (x, p, q) ↔ Right (x, a, b) by Axiom Cont 4

↔ Right (x, a, c) by Lemma 9.89

↔ Right (x, p′, q′) by Axioms Cont 4

100 MICHAEL BEESON

But the points p′ and q′ are the two intersection points of the line and circle, so

¬¬ ((p′ = p ∧ q′ = q) ∨ (p′ = q ∧ q′ = p)).

But since x is not on L, the second alternative would violate Axiom H10 andLemma ??. Hence, by the stability of equality, the first alternative holds. Thatcompletes the proof of the lemma.

Lemma 9.91. Let a and b lie on line L, and x and y not lie on line L, andsuppose Right (a, b, x). Then x and y are on the same side of L if and only ifRight (a, b, y), and x and y are on opposite sides of L if and only if Left (a, b, y).Similarly with Right and Left interchanged.

Proof. The left-to-right implication in the first claim is Axiom H4 for Rightand Axiom H5 for Left . Now suppose Right (a, b, x) and Right (a, b, y). Wemust show x and y are on the same side of L. By Lemma 9.83, it suffices toshow that no point of L is between x and y. Suppose B(x, z, y) and z is on L.Then x and y are on opposite sides of L. Then by Axiom H6, Left (a, b, y).Taking p = q = y in Axiom H3, the hypotheses of H3 hold, and we getB(y, IntersectLines (Line (y, y),Line (a, b), y), which false since Line (y, y) is un-defined, and even if it were it would violate Axiom B1-ii. That completes theproof of both directions of the first claim. The second claim is proved similarly,with Right and Left interchanged, appealing to Axiom H7 instead of H6. Thatcompletes the proof.

Lemma 9.92. Left (a, b, p) ∧Right (a, b, p) → on(p,Line (a, b)).

Proof. Let C = Circle (a, p) and K = Circle (b, p). By definition of Left andRight , if Left (a, b, p) or Right (a, b, p), then both intersection points of C and Kare defined. Hence a 6= b, by Axiom I17 or Axiom I18. Then L = Line (a, b) isdefined. Assume, for proof by contradiction , that p does not lie on L. Let M bethe perpendicular from p to L, let x be its intersection point with L, and let q bethe reflection of p in L, i.e. the point with qx = px and B(p, x, q). Then p and qare on opposite sides of L so by Axiom H6, we have Left (a, b, q). Now triangle axpis congruent to triagle axq by SAS, since L is perpendicular to M by Lemma 9.38(or by SAS since all right angles are congruent). Hence ap = aq. Similarlybp = bq. Then C = Circle (a, q) and K = Circle (b, q). Then by definition of Leftand Right , since Left (a, b, q) we have q = IntersectCircles2 (C,K). But then byhypothesis, q = p. Since B(p, x, q) we now have B(p, x, p), contradicting AxiomB1-ii. This contradiction shows ¬¬on(p, L). By Axiom S3 we have on(p, L).That completes the proof of the lemma.

Lemma 9.93. Suppose p = IntersectCircles1 (C,K) = IntersectCircles2 (C,K).Then p is collinear with the centers of C and K.

Proof. Let a and b be the centers of circles C and K, and let L = Line (a, b).By the definitions of Right and Left we have Right (a, b, p) and Left (a, b, p). ByLemma 9.92, p lies on Line (a, b). That completes the proof.

Lemma 9.94. Let C and K be two distinct circles, and suppose x lies onboth C and K. Then x is not different from both IntersectCircles1 (C,K) andIntersectCircles2 (C,K).


Proof. Let p = IntersectCircles1 (C,K) and q = IntersectCircles2 (C,K). Thesepoints exist by circle-circle continuity, since the point c on C is both non-strictlyinside and non-strictly outside of K. Let a and b be the centers of C and Krespectively. Since the circles are distinct and have a point in common, a 6= b.Then L = Line (a, b) is defined. If p 6= q then Line (p, q) is perpedicular to a.Let e on L be the intersection of L with Line(p, q). Since x lies on both circles,we have ax = ap and bx = bp, so triangle abx is congruent to triangle abp. ByLemma 9.68, if x is on the same side of L as p then x = p, and if it is on thesame side of L as q then x = q. Since p and q are on opposite sides of L, by theplane separation theorem if x is on the opposite side of L from q then it is onthe same side as p, so x = p. On the other hand if x is not on the opposite sideof L from q then by Lemma 9.83, it is on the same side as q, so x = q. But

¬¬OppositeSide (x, q, L) ∨ ¬OppositeSide (x, q, L))

is logically valid; hence ¬¬(x = p ∨ x = q). That completes the proof.

Lemma 9.95. Suppose a, b, and c are not collinear. Then

¬¬(Right (a, b, c) ∨ Left (a, b, c)).

Remark. We cannot remove the double negation, since ECG cannot prove anydisjunctive theorem, as we will show in a subsequent section.

Proof. Suppose a, b, and c are not colliear, and suppose ¬Right (a, b, c) and¬Left (a, b, c). We must derive a contradiction. Let C = Circle (a, c) and K =Circle (b, c). Then by definition of Right and Left , c 6= IntersectCircles1 (C,K)and c 6= IntersectCircles2 (C,K). By Lemma 9.94, this is impossible. Thatcompletes the proof.

9.16. Definability of IntersectCirclesSame and IntersectCirclesOpp. Wehave included a function symbol IntersectCirclesSame to produce the intersec-tion point of two circles on the same side of the line joining their centers as agiven point x. In this section we show that it was superfluous to have a primi-tive symbol for that purpose, as a compound term already plays that role. Theterm, however, is rather complicated, so in the interest of supporting Euclidmore directly, it may be handy to have a primitive symbol.

Up until now, we have mentioned IntersectCirclesSame only once, when weneeded it to prove the stability of angle ordering in Lemma 9.72. That lemma inturn is needed for the propositions in the latter part of Euclid Book I, but hasnot been used anywhere in this paper; it was proved only to lay the foundationfor Euclid. In particular none of the constructions we have defined so far usedIntersectCirclesSame, so we may use them in the proof that it is definable.

Theorem 9.96 (Definability of IntersectCirclesSame ). There is a termt(R,K, x) of ECG not containing IntersectCirclesSame or IntersectCirclesOpp ,such that ECG proves

t(R,K, x) = IntersectCirclesSame (R,K, x).

More precisely, the following is provable in ECG without the axioms forIntersectCirclesSame and IntersectCirclesOpp : If circles R and K are not con-centric, and L is the line joining their centers, and x is a point not on L, then

102 MICHAEL BEESON

Figure 21. Definability of IntersectCirclesSame

Lb

bq

b xbw

b

bab

e

bp c

br

t(R,K, x) ↓ if and only if R and K have a point in common, and in that casep = t(R,K, x) is a point of intersection of R and K, and if there are two dis-tinct points of intersection of R and K, then p is on the same side of L as x.Moreover if t(R,K, x) ↓ then the centers of R and K are different and x is noton L.

Remark. If there is only one point of intersection, then by Lemma 9.93 thatpoint is on L.

Proof. Here is the construction script defining the term t:

t(Circle R, Circle K, Point x)

a = center(R)

b = center(K)

L = Line(a,b)

c = IntersectCircles1(R,K)

J = Perp(c,L)

e = IntersectLines(J,L)

M = Perp(x,J)

w = IntersectLines(J,M)

p = IntersectLineCircle2(Line(e,w),Circle(e,c)

return p

Here is the correctness proof of the script. Since R and K are not concentric,L in line 3 is defined. If R and K have no point in common, then c in line4 is not defined, so t(R,K, x) is not defined, as required. We may thereforesuppose that R and K have a point in common. Then c in line 4 of the scriptis defined. (The point c might lie on L if the circles are tangent.) Since Perpis always defined, J in line 4 is defined, whether or not c is on L. By definitionof perpendicular, J meets L, so e is defined. We note that the definition ofPerp in Lemma 9.56 involves IntersectCircles1 and IntersectCircles2 , but notIntersectCirclesSame or IntersectCirclesOpp . Since Perp is always defined, Mis defined, and by definition of perpendicular, w is defined. We claim w is not


on L. Since wz and L are both perpendicular to J , if w were on L, then by theuniqueness of perpendiculars at a point (Lemma 9.80), Line (w, x) and L wouldcoincide, but x is on Line (w, x) but not on L. Hence w is not on L. Hencee 6= w; hence J coincides with Line (e, w). Since J passes through the centerof Circle (e, c), J meets Circle (e, c), so z in the penultimate line of the scriptis defined, and hence t(R,K, x) is defined. Observe that t(R,K, x) ↓ then thetwo circles do intersect, as the subterm of t(R,K, x) defining c must be defined,according to the laws of the logic of partial terms. Also if t(R,K, x) is defined,then Line (e, w) is defined, so e 6= w, which implies x is not on L (since if x ison L then both L and ew are perpendiculars to J from x). That proves the firstclaim of the lemma, that t(R,K, x) is defined if and only if R and K have apoint in common.

Note that although J coincides with Line (e, w), they may possibly have op-posite orientations, so it is important to put Line (e, w) in the penultimate lineof the script and not J .

Now suppose x is not on L and t(R,K, z) is defined, and let p = t(R,K, x). Wemust show that p is on the same side of L as x. ( Fig. 21 shows p on the oppositeside of L from x, because it illustrates the situation in a proof by contradiction;in reality p and q are switched from what is shown in Fig. 21.) Since wx doesnot meet L, by Lemma 9.83 x and w are on the same side of L. Therefore itsuffices to show that w and p are on the same side of L. Suppose w and p arenot on the same side of L. Then by Lemma 9.83, wp meets L, so B(w, e, p). Letq be the other intersection point of the two circles, namely

q = IntersectLineCircle1 (Line (e, w),Circle (e, c))

Now suppose p 6= q. Observe that triangles abp and abq are congruent. ByLemma 9.68 p and q are not on the same side of L. By the plane separationtheorem p and q are opposite sides of L. Then B(p, e, q).

Let r be any point not on J . By Axioms Cont 4 and Cont4, we have Right (r, q, p)if and only if Right (r, e, w), and similarly with Left in place of Right. SupposeRight (r, q, p). Then Right (r, e, w). We have

Right (q, p, r) by Axiom H8 since Right (r, q, p)

Right (e, w, r) by Axiom H8 since Right (r, e, w)

Right (e, p, r) by Lemma 9.89 since Right (r, q, p) and B(q, e, p)

Now since B(w, e, p) and B(q, e, p), we have by Axiom B3

¬¬(B(e, q, w) ∨ B(e, w, q) ∨ q = w).

If q = w we have Right (w, p, r) since Right (q, p, r). In the other two cases wehave Right (w, p, r) since Right (q, p, r), by Lemma 9.89. Hence ¬¬Right (w, p, r).But Right is stable (since it is defined by an equality); hence Right (w, p, r). Wethen have

Right (w, e, r) by Lemma 9.89 since Right (w, p, r) and B(w, e, p)

Left (e, w, r) by Axioms H8 and H10

104 MICHAEL BEESON

Now we have Right (e, w, r) and Left (e, w, r). But then r lies on Line (e, w), byLemma 9.92. But since J coincides with Line (e, w), that contradicts our choiceof r. This contradiction shows that the assumption ¬SameSide (w, p, L) cannothold; that is, we have proved ¬¬SameSide (w, p, L). But SameSide is stable, byLemma 9.85. Hence SameSide (w, p, L). That completes the proof.

Corollary 9.97. There is a term s(R,K, x) of ECG not containingIntersectCirclesSame or IntersectCirclesOpp such that ECG proves

s(R,K, x) = IntersectCirclesOpp (R,K, x).

Remarks. This theorem, like the previous one, is to be understood as meaningthat the defining property of s can be proved without using the axioms forIntersectCirclesSame or IntersectCirclesOpp .

Proof. If x is not on L then we can reflect x in L, by dropping a perpendicularfrom x to L , with foot u, and extending the non-null segment xu by segment xuto point z. Then B(x, u, z) with z on L, so z is on the opposite side of L fromx. Now let s(R,K, x) be t(R,K, z). Then (provided R and K are not concentricand do meet) s(R,K, x) is an intersection point of R and K on the same sideof L as z. But then by the plane separation theorem, since z is on the oppositeside of L from x, s(R,K, x) is on the opposite side of L from x. That completesthe proof.

9.17. Rotation and Uniform Reflection. In this section, we define a con-struction that can rotate a given point through a given angle (without a casedistinction whether the point lies on the vertex of the angle or not), and apply itto define a construction that reflects a point in a line, without a case distinctionwhether the point is on the line or not. These “uniform” constructions, alongwith the uniform perpendicular construction Perp of Lemma 9.56, are importantin constructive geometry, and their definitions are far from obvious.

Lemma 9.98 (Construction of angle bisector). There is a term Bisect(a, b, c)such that if a, b, c are distinct points, and e = Bisect(a, b, c), then e 6= b andangle abe is congruent to angle cbe, and e is between two points p and q onRay (b, a) and Ray (b, c) respectively.

Proof sketch. LetMidpoint be the construction given in the proof of Lemma 9.40.Here is a construction script for the bisector:

Bisect(Point a, Point b, Point c)

C = Circle(a,b)

e = IntersectLineCircle2(Line(b,c),C)

d = Midpoint(a,e)

return d

Then as in the proof of Lemma 9.40, triangle abd is congruent to triangle ebd,so angle abd is congruent to angle ebd as required. That completes the proof.

Lemma 9.99 (Uniqueness of angle bisector). Given three non-collinear pointsa, b, and c, and points d and e such that angles abd and cbd are congruent, andangles abe and cbe are congruent. Then b, e, and d are collinear.


Proof. Let C be a point on Ray (b, c) with aC = ab. Since a, b, and c are non-collinear, L = Line (a, C) is defined. By the crossbar theorem (Lemma 9.61),L meets Line (b, d) in a point D, and L meets Line (b, e) in a point E, withB(a,D,C) and B(a,E,C). Then by SAS, triangles abD and DbC are congru-ent. Hence L is perpendicular to Line (b,D). Similarly L is perpendicular toLine (b, E). Suppose Line (b,D) and Line (b, E) do not coincide. Then they areparallel by Lemma 9.48. But b lies on both those lines, contradiction. Hencethose lines cannot fail to coincide. Then b, E, and D cannot fail to be collinear.But collinearity is defined by a negation, so it is stable. Hence b, E, and D arecollinear. Hence b, e, and d are collinear, as claimed. That completes the proof.

Definition 9.100. Let Rotate be the term represented by the following con-struction script:

Rotate(P,O,Q,A)

E = Bisect(P,O,Q)

K = Perp(A,Line(O,P))

B = IntersectLines(K,Line(O,E))

Z = project(B,Line(O,Q))

return Z

Lemma 9.101. There are terms of ECG which we denote Project, Para, andRotate , such that the properties of these terms used in Section 5 are provable inECG. Specifically,

(i) Project(x, L) is a (and in fact the unique) point on L such that a lineperpendicular to L at x passes through x, and

(ii) If P , O, and Q are not collinear, and A lies on Line (O,P ) (but notnecessarily on Ray (O,P ), then the value Z = Rotate (P,O,Q,A) is defined andlies on Line (O,Q).

(iii) OA = OZ, and if A 6= O, then POQ is a right turn if and only if AOZis a right turn, and POQ is a left turn if and only if AOZ is a left turn; and ifA = O then Z = O.

Proof. The uniform construction of the perpendicular through point x to line L,which is Perp, was exhibited and proved correct in Lemma 9.56. The projectionProject(x, L) of x on L is the foot of that perpendicular, i.e.

Project(x, L) = IntersectLines (L,Perp(x, L)).

The uniform parallel construction, called Para, is defined by

Para(x, L) = Perp(x,Perp(x, L)).

It produces a line parallel to L through x if x is not on L, and a line that coincideswith L if x is on L, but without requiring a case distinction.

Now we turn to (ii). Let Bisect be as in Lemma 9.98. Then we can defineRotate as in Section 5, namely, by a term corresponding to the script in Def-inition 9.100 (see Fig. 2). We now argue for (ii). Suppose P , O, and Q arenot collinear, and A lies on Line (O,P ). Then e = Bisect (P,O,Q) is defined,and K = Perp(A,Line (O,P ) is defined. Since P , O, and Q are not collinear,Ray(b, e) is the angle bisector of angle POQ, which means that angle POB

106 MICHAEL BEESON

is congruent to QOE. According to Lemma 9.98, E is between two points onRay (O,Q) and Ray (O,P ) and not equal to O. Since these rays do not lineon the same line (because O, P , and Q are not collinear), E does not lie onLine (O,P ). Then A does not lie on Line (O,E). Hence K and Line (O,E) donot coincide. Hence B, the intersection point of these two lines, is defined. SinceO 6= Q, Line (O,Q) is defined and hence Z, the projection of B on Line (O,Q),is defined. Note that in spite of the appearance of Fig. 1, we did not assumeanything about the position of A on Line (O,P ). That is, Z is defined, no matterwhether A lies on Ray (O,P ) or the opposite ray. That completes the proof of(ii).

Turning to (iii), first suppose A = O. Then B = O and Z = 0, as claimed, bythe properties of Project. In particular OA = OZ as both segments are OO.

Now suppose A is on Ray (O,P ). Then Z is on Ray (O,Q). But angle AOB isthe same angle as angle POE, and angle QOE is the same angle as angle ZOB;hence angle ZOB is congruent to angle AOB. Angles BAO and BZO are rightangles, by the property of Perp, which was used to construct points B and Z.Since all right angles are congruent, angle BAO is congruent to angle BZO.Side OB is congruent to itself. Hence triangles BAO and BZO are congruent,by SAA (Lemma 9.69). Hence OA = OZ as corresponding sides of congruenttriangles.

On the other hand, if A is not on Ray (O,P ), then since A is on Line (O,P )we have B(A,O, P ). Let P ′ and Q′ be the reflections of P and Q in point O.Let E′ be the reflection of E in O. By the congruence of vertical angles, OE′

bisects angle P ′OQ′. Let E′′ = Bisect (P ′, O,Q′). Then both OE′′ and OE′bisect angle P ′OQ′, and OE = OE′ = OE′′, so E′ = E′′, by Lemma 9.99. Thusthe situation is exactly the reflection in O of Fig. 2; arguing as above with primeson P , E, and Q, conclude OA = OZ.

Now we have proved OA = OZ in three cases that are classically mutuallyexclusive. Hence ¬¬OA = OZ. Then by the stability of equality, OA = OZ.

Now suppose A 6= O and A lies on Ray (O,P ). We will show in this case POQis a right turn if and only if AOZ is a right turn.

We claim Z lies on Ray (O,Q). Since lying on a ray is defined by a negation,it is stable, so we may prove this by contradiction. Suppose Z does not lie onRay (O,Q). Then B(Z,O,Q). Then Z is on the opposite side of L = Line (O,E)from Q. But A is on the same side of L as P , by Lemma 9.83, since AP meets Lonly at O, and B(O,A, P ). Since P and Q are on opposite sides of L, then A ison the opposite side of L from Q, by the plane separation property. Now A andZ are both on the opposite side of L from Q; hence they are on the same side ofL. But then triangles OBA and OBZ are two congruent triangles on segmentOB with third vertex on the same side of L, so by Lemma 9.68 we have Z = A.But then Q and P lie on Line (O,Z), contradicting the assumption that P , Q,and O are not collinear. Hence Z does lie on Ray (O,Q). We have Z 6= O since ifZ = O then B = O and then A = O, contradiction. Now by Lemma 9.89, POQis a right turn, if and only if AOQ is a right turn, and since Z lies on Ray (O,Q)by Lemma 9.89 again, if and only if AOZ is a right turn.

Similarly, assuming A lies on Ray (O,P ) and A 6= O, POQ is a left turn ifand only if AOZ is a left turn; and similarly, we reach the desired conclusions


if B(A,O, P . (Alternately we could prove as a lemma that Right and Left arepreserved under reflection in a point, and apply that lemma.)

Since these cases are classically mutually exclusive, we have proved

A 6= 0 → ¬¬(Right (P,O,Q)↔Right (A,O,Z))∧(Left (P,O,Q)↔Left (A,O,Z))

Right and Left are defined by the equality of certain terms (see Definition 8.8),so they are stable (implied by their double negations) by the stability of equality,Axiom S1. Hence we can push the double negation inwards, and drop it in frontof Left and Right , obtaining the desired conclusion:

A 6= 0 → (Right (P,O,Q)↔Right (A,O,Z)) ∧ (Left (P,O,Q)↔Left (A,O,Z))


Lemma 9.102 (Inverse rotation). If Z = Rotate (P,O,C,A) thenZ = Rotate (C,O, P, Z).

Proof. The bisector of angle QOP is the same as the bisector of angle POQ, byLemma 9.99. Then when, in the second rotation, we erect the perpendicular toOQ at Z, it intersects OL at B because of the uniqueness of the perpendicularat Z (Lemma 9.80). Then when we drop the perpendicular to OP , it meetsLine (O,P ) at A because of the uniqueness of the projected point, proved inLemma 9.101. That completes the proof.

Definition 9.103. The term I is given by the following construction script.(The script takes no arguments, which means that the term I has no variables.)

I()

L = Line(0,1)

J = Perp(0,L)

K = Perp(gamma,J)

p = IntersectLines(J,K)

return IntersectLineCircle2(Line(0,p),Circle(0,1))

The following lemma states an “obvious fact” whose proof is surprisingly in-tricate.

Lemma 9.104. Right (1, 0, I).

Proof. Since 0 is another name for α and 1 is another name for β, and αβγ isa left turn by Axiom H12, 10γ is a right turn by Axiom H9. Then we definep = IntersectLines (J,Perp(γ, J)). Then p 6= 0 since if p = 0, K and L coincide,since both are perpendicular to J at 0, but then α, β, and γ would be collinear,which they are not. Since J and Perp(γ, J) meet only at p, we conclude that10p is a right turn, by Lemma 9.89.

Letq = IntersectLineCircle2 (Line (0, p),Circle (0, 1)).

(This is the point you might call −I, on the y-axis J opposite I.) By Axiom Cont4, for all points x not on Line (0, p) (which coincides with J , but might have thesame or opposite orientation), we have Right (x, q, I) if and only if Right (x, 0, p).

108 MICHAEL BEESON

Taking x to be 1, we have Right (1, q, I) if and only if Right (1, 0, p). But we haveproved Right (1, 0, p). Hence Right (1, q, I). By Lemma 9.12 we have B(q, 0, I).Hence by Lemma 9.89, we have Right (1, 0, I). That completes the proof of thelemma.

We now turn to uniform reflection; by that we mean a construction to reflectpoint x in line L without a case distinction as to whether x is on L or not. By theuniform perpendicular construction we can get a line through x perpendicularto L, meeting L at a point e, but since only non-null segments can be extended,we cannot extend segment xe as we desire to do, unless x is not on L. Thekey to avoiding this case distinction is to rotate x twice by ninety degrees! Theconstruction is illlustrated in Fig. 22. Point x is rotated twice, first to point yand then to the desired answer, point z. The dashed lines show the constructionused to perform the rotation, according to the Rotate construction, using twoangle bisectors and Perp to construct points c and a. Here a, b, and c are pointsequidistant from e and definitely different than e, while we do not assume thatx 6= e.

Figure 22. Uniform Reflection

Lb

eb

zb

x

b

y

b

a

bc

b

b

b b

Lemma 9.105 (Uniform reflection). There is a term Reflect (x, L) that pro-duces the reflection of x in line L, without a case distinction whether x is on Lor not. That is, if z = Reflect (x, L) and K perpendicular to L contains x andmeets L at e, then z lies on K with ez = ex, and unless x lies on L, B(x, e, z).

Proof. Here is the construction script illustrated in Fig. 22.

Reflect(Point x, Line L)

K = Perp(x,L)

e = IntersectLines(K,L)

C = Circle(e,


alpha, beta

)

a = IntersectLineCircle1(L,C)

b = IntersectLineCircle1(K,C)

c = IntersectLineCircle2(L,C)

y = Rotate(a,e,b,x)

z = Rotate(b,e,c,y)

Return z;

Since Perp is always defined, K is defined, and since K is perpendicular to L,e in line 2 is defined. The constants α and β are used in line 3 to ensure thatC has a nonzero radius. Since L and K both pass through the center of C,the points a, b and c are defined. Since these three points are different frome (because C is not a degenerate circle), y and z are defined by Lemma 9.101.Hence Reflect (x, L) is defined (regardless of whether x is or is not on L).

By the properties of Rotate , we have ex = ey and ey = ez. Hence ex = ez. Itremains to show that if x does not lie on L (i.e. x 6= e) then B(x, e, z). Supposex 6= e. Then x and z both lie on circle C and line K. Since K is a diameterof C, its center lies between its endpoints, by Lemma 9.12. That completes theproof of the lemma.

Lemma 9.106 (Reflection in a diameter). Let line L pass through the centerof circle C and let p lie on C. Then Reflect (p) lies on C.

Proof. Let L pass through the center e of C and let p lie on L. Let M =Perp(p, L). Let x be the intersection point of M and L. Let q = Reflect (p, L).Then q lies on M and xp = xq, by Lemma 9.105. We must show q lies on C,which will follow from qe = pe. By Axiom S4, the stability of On, we can argueby cases.

Case 1, e = p. Then circle C is degenerate and so q = p and we are finished.Case 2, p does not lie on L and e does not lie on M . Then e, p, q, and x are all

distinct, and since px = qx and all right angles are congruent, we have trianglepxe congruent to triangle qxe by SAS. Hence ep = eq, completing Case 2.

Case 3, p lies on L. Then p = q by Lemma 9.105, so pe = qe, completing Case3.

Case 4, e lies on M . Then Line (e, p) coincides with M , so q lies on Line (e, p).Then x = e by the uniqueness of dropped perpendiculars. Since px = qx byLemma 9.105, we have pe = qe. That completes Case 4, and with it, the proofof the lemma, since these cases are clasically mutually exclusive.

Lemma 9.107 (Reflection in a point). There is a construction Reflect (a, x, L)such that if a and x are points on line L then Reflect (a, x, L) is defined and if p =Reflect (a, x, L) then pa = ax and T(p, a, x), where T is non-strict betweenness.

Remarks. This trivial-sounding lemma is not trivial to prove, because we cannotmake a case distinction whether x = a or not, so we cannot form Line (x, a) orextend the segment xa. Once constructed, we say “p is the reflection of x in a”if we know that x 6= a; in that case it is not necessary to mention L since L

110 MICHAEL BEESON

necessarily coincides with Line (a, x); but being given L is necessary in order toconstruct the reflection without a case distinction.

Proof. Here is the construction script. It calls on the construction of Lemma 9.105;we use Reflect for both terms; since one has more arguments than the other, theycan be distinguished.

Reflect(point a, point x, Line L)

K = Perp(a,L);

return Reflect(a,K)

The claimed properties of this term follow immediately from Lemma 9.105.That completes the proof of the lemma.

9.18. The other intersection point. Many Euclidean constructions in-volve constructing one intersection point p of a line L = Line (a, b) and a circleC, and then we say “Let q be the other intersection point of L and C”. Thequestion is, whether in ECG we can give a uniform method of constructing Qfrom p, L, and C. Yes, we can, as we will now show.

Figure 23. Definability of “the other intersection point”.

L

b a

b

pb

qb

x

C

Theorem 9.108. There is a term Other(p, L,C) such that, if point p lies online L and circle C, then Other(p, L,C) is defined and lies on both L and C,and if z 6= p and z lies on both L and C then Other(p, L,C) = z.

Proof. Here is the construction script for Other:

Other(Point P, Line L, Circle C)

a = center(C)

K = Perp(a,L)

q = Reflect(p,K);


return q;

Since Perp is always defined, K in line 2 is defined. Let x = IntersectLines(K,L).By Lemma 9.105, qx = px and q lies on L. We must prove q lies on C. If adoes not lie on L then since K and L are perpendicular at x we have triangleaxp congruent to triangle axq. Hence corresponding sides ap and aq are equal.If a does lie on L then since a also lies on K we have a = x, hence ap = aq. Butwe have ¬¬ (on(a, L) ∨ ¬ on(a, L)). Hence ¬¬ap = aq. Hence ap = aq by thestability of equality. Hence q lies on C.

Now suppose z is any point different from p that lies on both L and C. Wemust show z = q. By Lemma 9.52 there are exactly two points on both L andC. But p and q are two such points, and z 6= p. Hence ¬¬z = q. By the stabilityof equality then z = q. That completes the proof.

Remark. In case one thinks, “this is not what Euclid had in mind!” (when hespeaks of the other intersection point), that is of course true; but Euclid nevertried to give uniform constructions; and in general it takes much more care toproduce uniform constructions to avoid case distinctions, as we have seen alreadywith uniform perpendiculars and uniform midpoints.

We next want to show that similarly, one can construct the “other intersectionpoint” of two circles.

Theorem 9.109 (The other intersection point of two circles). There is a termOther2(p, C,K) such that, if point p lies on circle C and on circle K, and C

and K are not coincident, then Other2(p, C,K) is defined and lies on both Cand K, and if z 6= p lies on both C and K then Other2(p, C,K) = z.

Proof. Here is the construction script:

Other2(Point p, Circle C, Circle K)

a = center(C);

b = center(K);

L = Line(a,b);

q = Reflect(p,L);

return q;

Here is the correctness proof of the script. Then C and K are not concentric,since if a = b then C and K coincide (because then ap = bp, so for all z, On(z, C)if and only if za = ap if and only if ab = bp if and only if On(z,K)). But byhypothesis, C and K do not coincide. Hence a 6= b. Therefore L in line 3 isdefined. Since Reflect is always defined, by Lemma 9.105, q is defined, and byLemma 9.106, q lies on circles C and K. That completes the proof.

9.19. Defining the order of points on a line. We have earlier (see section9.3) defined an ordering on segments. That corresponds to arithmetic on positivenumbers. We now define an ordering on the points of a line, corresonding topositive and negative numbers.

112 MICHAEL BEESON

Let 0 be another name for β and let 1 be another name for α, and letL = Line (α, β) be called the x-axis. Let K be perpendicular to L at 0. ByLemma 9.56, there exists a line M containing γ and perpendicular to K (we donot need to know whether γ is on K or not). Let z on K be the foot of thisperpendicular. By Lemma 9.48, M is parallel to L. On Ray (0, z) construct apoint I such that 0I = 01. Then I is on the same side of L as γ, since M isparallel to L by Lemma 9.48. (Note, this does not use the parallel axiom; thatlemma was derived in neutral geometry.) Since αβγ is a right turn (by AxiomsH8, H10 and H12), and I is on the same side of L as γ, then αβγ, which canalso be written 10I, is also a right turn, by Axiom H4.

Definition 9.110. For points x and y on L:

x < y := Left (x, y, I)

x > y := Right (x, y, I)

x ≤ y := ¬y < x

x ≥ y := y ≤ x

Note that Left (x, y, I) already implies x 6= y, so it is not necessary to stipulatex 6= y.

Lemma 9.111. With notation as above,

0 6< x ∧ x 6< 0 → x = 0.

Proof. Suppose 0 6< x and x 6< 0. Then 0xI is not a left turn and x0I is not aleft turn. But that contradicts Lemma 9.95. That completes the proof.

Note that ECG cannot prove a < b → a < x ∨ x < b, since the decisionas to which alternative holds cannot be made continuously in x. This principle,called “apartness”, is not provable in ECG.

For points on L, we say that x is on the same side of y as z if y is not betweenx and z and neither x nor z is equal to y. Note that this is equivalent to sayingthat x and y are on the same side of the line perpendicular to L at y. Then wehave

Lemma 9.112. With notation as above, fix a point 1 > 0 on L. Then forpoints x on L, x > 0 if and only if x is on the same side of 0 as 1; and 0 < x ifand only if ¬B(x, 0, 1).

Remark. This lemma connects the definition of order in terms of Right and Leftwith the primitive notion of betweenness.Proof. Let K be the line perpendicular to L at 0. Then we must show that x > 0if and only if x is on the same side of K as 1. Suppose x > 0; we must prove xand 1 are on the same side of K. Since x > 0, we have x 6= 0 and 0xI is a leftturn. Similarly 01I is a left turn.

By Axiom H9 (applied twice) I0x is a left turn. Since 0 < 1, we have by thedefinition of < that 01I is a left turn. Hence by Axiom H9, I01 is a left turn.Hence by Axiom H2, x and 0 are on the same side of Line (0, I), which has thesame points (and hence the same sides) as line K. Hence x and 0 are on thesame side of K, which was what was to be proved.


Conversely, suppose x and 1 are on the same side of K. Since 0 < 1, as aboveI01 is a left turn. By Axiom H5 then, I0x is a left turn. By Axiom H10, Ix0 is aright turn, and then by Axiom H8, x0I is a right turn; then x > 0 by Definition9.110. That completes the proof of the first claim of the lemma.

The second claim can be derived from the first, but it can also be proveddirectly, since (for points x on L), if x is on the same side of 0 as 1 then bythe plane separation theorem ¬B(x, 0, 1), so by Lemma 9.89, and the fact thatRight (1, 0, I), we have Right (x, 0, I), hence Left (0, x, I), which is the definitionof 0 < x. Conversely, to prove 0 < x → ¬B(x, 0, 1), suppose 0 < x andB(x, 0, 1). Let −1 be the extension of segment 10 by 10. Then B(−1, 0, 1), soby definition of SameSide , −1 and x are on the same side of K. By definitionof 0 < x we have Left (0, x, I); by Lemma 9.89 we have Left (0,−1, I). HenceRight (−1, 0, I). By Lemma 9.89 we have Right (−1, 1, I) and then Right (0, 1, I).But since 0 < 1 we have Left (0, 1, I), contradicting Lemma 9.92. That completesthe proof of the lemma.

§10. The parallel postulate. We will introduce three versions of the par-allel postulate: Euclid’s own version (Euclid 5), which says that under certainconditions, two lines will meet in a point; a version we call the strong parallelpostulate, which weakens Euclid’s conditions for the two lines to meet; and Play-fair’s version, which says two parallels to the same line through the same pointmust coincide (and makes no assertion at all about the existence of an intersec-tion point). We then start to consider the relations of implication between theseversions (relative to neutral ECG), and finally draw some consequences of thestrong parallel axiom.

10.1. Alternate Interior Angles. In connection with parallel lines, thefollowing terminology is traditional: If L and K are two lines, and p lies onL and q lies on K but not on L, then pq is a transversal of L and K. ThenLine (p, q) makes four angles with L and four angles with K; certain pairs ofthem are called alternate interior angles or adjacent interior angles. In ECG,there is no primitive concept of “angle”; angles are treated as triples of points,and angle congruence is a defined concept (it is defined in Definition 9.21). Inthe next section we will discuss Euclid’s parallel postulate, and we will need toexpress it without mentioning angles directly. We therefore discuss how to dothat now.

Angle congruence is defined in terms of triangle congruence, which is in turneddefined by the SSS criterion: the three sides are pairwise congruent.

We will now unwind these definitions to express the concept of alternate inte-rior angles being equal, without mentioning angles. The following definition isillustrated in Fig. 24.

Definition 10.1. “pq makes alternate interior angles equal with Kand L”, written AI(p, q, L,K), means

on (p, L) ∧ on (q,K) ∧ ¬on (q, L) ∧ p 6= q

∧ ∃r, s, t (on(r,K) ∧ on (s, L) ∧ B(r, t, s) ∧ B(p, t, q) ∧ r 6= p ∧ q 6= s

∧ pt = qt ∧ rt = st)

114 MICHAEL BEESON

Figure 24. Transversal pq makes alternate interior anglesequal with L and K, if pt = tq and rt = st.

b

p

b

qb

s

bt

b

r

L

K

This can also be expressed as “K and L make interior angles equal withpq”.

The reason for requiring B(r, s, t) is that r and t need to be on opposite sidesof Line (p, q) in order that the angles rpt and qst are alternate interior angles. Atthe cost of introducing two more points into the definition, we could have madeit easier to verify; any two congruent triangles with those angles as a vertex couldreplace prt and qst, as illustrated in Fig. 25. As the lemmas given after anglecongruence is defined in Definition 9.21 show, if we can produce x and y so thatFig. 25 is correct, then AI(p, q, L,K) will hold.

A careful reader might have noticed that we omitted the congruence rp = qs inthe definition, which would be required to express the congruence of the shadedtriangles according to the definition of triangle congruence. This was not anaccident. In case the lines K and L are as shown in the diagram, the trianglesare congruent anyway by SAS, since the angles at t are vertical angles, and (aswe prove in Lemma 9.33), vertical angles are congruent in neutral ECG. Butnot only is rp = qs not necessary, there is a good reason to omit it: it allows thedefinition to make sense even when K and L coincide! This is useful because itenables us to avoid making case distinctions according as K and L do or do notcoincide.

Figure 25. Another way to say that transversal pq makes alter-nate interior angles equal with L and K. The shaded trianglesare congruent.

b

p

b

qb

s

bt

bx

b y

b

r

L

K


On the other hand, instead of introducing more points using ∃, we can get ridof ∃ entirely by constructing specific points r, s, and t:

r = IntersectLineCircle2 (K,Circle (p, α, β))

t = Midpoint (p, q)

s = IntersectLineCircle2 (Line (r, t)Circle (t, r))

These points will do for r, t, and s in AI(p, q, L,K) if any points will do. In thisway AI(p, q, L,K) can be expressed by a quantifier-free formula. Similarly x andy can be calculated in Fig. 25, by requiring xp = rp and yq = sq. That is actuallywhat we would get by directly expanding the definition of angle congruence.

10.2. Euclid’s parallel postulate. Euclid’s postulate 5 is

If a straight line falling on two straight lines make the interior angleson the same side less than two right angles, the two straight lines, ifproduced indefinitely, meet on that side on which are the angles lessthan the two right angles.

We consider the formal expression of Euclid’s parallel axiom. Let L and Mbe two straight lines, and let pq be the “straight line falling on” L and M , withp on M and q on L. We think that what Euclid meant by “makes the interiorangles on the same side less than two right angles” was that, if K is another lineK through p, making the interior angles with pq equal to two right angles, thenM would lie in the interior of one of those angles (see Fig. 26). For K to makethe interior angles on the same side of pq equal to two right angles, is the sameas for K to make the alternate interior angles equal. In the previous section wediscussed the formal expression of that concept. Once we use this criterion toexpress that L and K make interior angles on the same side equal to two rightangles, we can then use three more points to “witness” that one ray of line Memanating from p lies in the interior of one of the interior angles made by M .Fig. 26 illustrates the situation.

Figure 26. Euclid 5: M and L must meet on the right side,provided B(q, a, r) and pq makes alternate interior angles equalwith K and L.

b

p

b

a

b

q

b

r

L

K

M

Here is a formal version of Euclid’s parallel axiom, using the formula AIE fromthe previous section to express that pq makes alternate interior angles equal withK and L:

116 MICHAEL BEESON

on(p,K) ∧ on(p,M) ∧ on(a,M) ∧ on(r,K) ∧ on(q, L) ∧ B(q, a, r)∧ AIE(p, q,K,L) → B(p, a, IntersectLines (L,M))

The logical axioms of LPT make it superfluous to state in the conclusion thatIntersectLines (L,M) is defined. That follows automatically.

We now write out Euclid 5 in the primitive syntax of ECG, eliminating thedefined concept AIE. The result, illustrated in Fig. 27, is the official version ofEuclid 5; that is, when we refer to Euclid 5 in subsequent sections, this is whatwe mean.

Figure 27. Euclid 5: M and L must meet on the right side,provided B(q, a, r) and pt = qt and rt = st.

b

p

b

a

b

qb

s

b

r

b

t

L

K

M

on(p,K) ∧ on(p,M) ∧ on(a,M) ∧ on(r,K) ∧ on(q, L) ∧ B(q, a, r) (Euclid 5)∧ on (s, L) ∧ B(p, t, q) ∧ pr = qs ∧ pt = qt ∧ rt = st ∧ ¬ on (p, L)

→ B(p, a, IntersectLines (L,M))

We consider strengthening this axiom by removing the hypothesis ¬ on (p, L).Classically it makes no difference, since if p is on L then the conclusion that Mmeets L is already satisfied; but the relation between the two versions is not soobvious with constructive logic. If on (p, L), then because B(p, t, q) and on (q, L),we have on (t, L); hence also on (r, L). But then we have on (q, L) and on (r, L)and B(r, a, q), so on (a, L). Similarly if on (a, L), there is a similar “collapse”,and one can show on (p, L). So the hypothesis ¬ on (p, L) could be replaced by¬ on (a, L). We do not know if the axiom really becomes stronger if we remove¬ on(p, L), but we leave that hypothesis there, because it seems to be whatEuclid intended.

10.3. The strong parallel postulate. Although we have finally arrived ata satisfactory formulation of Euclid 5, that formulation is satisfactory only in thesense that it accurately expresses what Euclid said. It turns out that this axiomis not satisfactory as a parallel postulate for ECG. The main reason is that it isinadequate to define division geometrically. The details will be made clear whenwe define a geometric construction Reciprocal geometrically that constructs asegment of length 1/x from a segment of length x. For now, it is enough toexplain that as x gets nearer and nearer to 0, the number 1/x requires a line ofsmaller and smaller slope to meet a certain horizontal line. If x passes throughzero, this intersection point “goes to infinity”, then is undefined when x = 0,but then “reappears on the other side”, coming in from minus infinity. Without


knowing the sign of x, we will not know on which side of the transversal pq thetwo adjacent interior angles will make less than two right angles. In other words,with Euclid 5, we will only be able to divide by a number whose sign we know;and the principle x 6= 0 → x < 0 ∨ x > 0 is not an axiom (or theorem) ofECG. The conclusion is that if we want to divide by nonzero numbers, we needto strengthen Euclid’s parallel axiom.

We make three changes in Euclid 5 to get what we call the strong parallelpostulate:

(i) We change the hypothesis B(q, a, r) in Euclid’s axiom to ¬on(a,K). Inother words, we require that the two adjacent interior angles do not make exactlytwo right angles, instead of requiring that they make less than two right angles.

(ii) We change the conclusion to state only that M meets L, without specifyingon which side of the transversal pq the intersection lies.

(iii) We drop the hypothesis ¬ on (p, L).

Here is the strong parallel postulate, which we take as an axiom of ECG:

on(p,K) ∧ on(p,M) ∧ on(r,K) ∧ on(q, L) (Strong Parallel Postulate, SPP)∧ on(a,M) ∧ ¬ on (a,K)∧ on (s, L) ∧ B(p, t, q) ∧ pr = qs ∧ pt = qt ∧ rt = st

→ IntersectLines (L,M) ↓

Figure 28. Strong Parallel Postulate: M and L must meet(somewhere) provided a is not on K and pt = qt and rt = st.

b

p

b a

b

qb

s

b

r

b

t

L

K

M

The strong parallel axiom differs from Euclid’s version in that we are notrequired to know in what direction M passes through P ; but also the conclusionis weaker, in that it does not specify where M must meet L. In other words,the betweenness hypothesis of Euclid 5 is removed, and so is the betweennessconclusion. Since both the hypothesis and conclusion have been changed, it isnot immediate whether this new postulate is stronger the Euclid 5, or equivalent,or possibly even weaker, but it turns out to be stronger–hence the name.

Also we call attention to the fact that nothing in the postulate prevents p fromlying on L. In that case the “triangles” (shaded in the diagram) are no longertriangles, but the statement that their sides are pairwise congruent still makessense. The condition B(t, p, q) will force L and K to coincide if p lies on L. We

118 MICHAEL BEESON

show below, in Corollary 10.13, that it does not matter if we add the hypotheses¬on (p, L) and ¬on (a, L) to the strong parallel postulate–the modified axiom isequivalent to the one given above. In particular, change (iii) that we made inpassing from Euclid 5 to the SPP is insignificant.

10.4. Playfair’s axiom. The astute reader may have observed that we havenow discussed two “parallel postulates” in detail, yet have never defined oreven mentioned the word “parallel”. Euclid did not give his Postulate 5 thename “parallel postulate” (or any other name). A case can be made that it ismore of a “triangle construction postulate.” Be that as it may, we now defineParallel (L,K) for lines L and K to mean that the lines do not meet:

Definition 10.2.

Parallel (L,K)↔∀x¬ (on(x, L) ∧ on(x,K)).

This requires a universal quantifier. Compare it to the quantifier-free formula

¬ IntersectLines (K,L) ↓ .If L and K are parallel, then IntersectLines (K,L) is undefined (by Axiom I4);but it will also be undefined if L and K coincide, by Axiom I21. The relationdefined by ¬IntersectLines (L,K) ↓ can be referred to informally as “weaklyparallel”; classically it means “parallel or coincident”, but constructively, wegenerally cannot make the decision of two weakly parallel lines whether theyare coincident or parallel. Instead, weakly parallel means “not not parallel orcoincident.” (Consider, for example, the x-axis and the line y = a; these linesare parallel if a 6= 0 and coincident if y = 0, so if we could decide whether twoweakly parallel lines are coincident or parallel, we could decide y = 0 ∨ y 6= 0.)

That this concept does have something to do with Euclid’s parallel postulateis hinted at by the following lemma:

Lemma 10.3. Let K and L be two lines, with point p on K but not on L, andlet traversal pq make equal alternate interior angles with K and L. Then K andL are parallel.

Proof. Refer to Fig. 24 for an illustration; let points p, r, s, q, and t be as shownin the figure with triangle prt congruent to triangle qst. Suppose, for proof bycontradiction, that K and L meet in point x. (Since Parallel is defined as notmeeting, a constructive proof of it consists in deriving a contradiction.) Considertriangle pqx. If x is on Ray (p, r), then the angles of triangle pqx are pxq, qpx,and pqx. The two adjacent interior angles on pq make two right angles together,but in neutral geometry one can prove that two angles of a triangle are less thantwo right angles. (This is Euclid I.17, proved in two lines from Euclid I.16, whichwe prove constructively in Lemma 9.73 below.) This contradiction shows thatK and L cannot meet in a point x on Ray (p, r). Similarly, they cannot meet ina point x on Ray (r, p). But, if they meet in a point x at all, then by Axiom B3,

¬¬ (B(x, p, r) ∨ x = p ∨ B(p, x, r) ∨ x = r ∨ B(p, r, x)).

In each of those cases, x lies on Ray (r, p) or on Ray (p, r); so we have a contra-diction, and K cannot meet L at all. Hence K is parallel to L as claimed. Thatcompletes the proof of the lemma.


Most modern treatments of geometry formulate the parallel axiom in a differ-ent way:

If two lines K and M are parallel to L through point p, then K = M .

We call this the “Playfair’s postulate”, or for short just “Playfair”, after JohnPlayfair, who published it in 1795, although (according to Greenberg [12], p.19) it was referred to by Proclus. Since we can always construct one parallel, itwould be equivalent to assert that, given a line L and a point p not on L, thereexists exactly one parallel to L through p.

Figure 29. Playfair: M and L can’t fail to meet.

b

p

L

K

M

The formal expression of Playfair’s postulate says that if there are two linesthrough point p that are weakly parallel to L then those two lines coincide.Classically, “weakly parallel” means “parallel or coincident”, but constructively,it just means that IntersectLines (K,L) is not defined (without specifying whynot).

¬ IntersectLines (K,L) ↓ ∧¬ IntersectLines (M,L) ↓ ( Playfair’s Postulate)∧on(p,K) ∧ on(p,M) → on(q,K)↔on(q,M)

We do not say L = M since equality between lines is intensional; instead weassert that the same points are on L as are on M . Playfair’s postulate is classi-cally equivalent to the more usual formulation, in which the hypothesis mentionsParallel instead of “weakly parallel”:

Parallel (K,L) ∧ Parallel (M,L) (weak Playfair)∧on(p,K) ∧ on(p,M) → on(q,K)↔on(q,M)

Neither version of Playfair’s axiom makes any existential assertion at all; tosee this just recall that the definition of Parallel only uses a universal quanti-fier. They are therefore the weakest versions of the parallel postulate that wewill consider. It is an empirical fact that Playfair’s axiom suffices to prove allthe consequences of the parallel postulate that do not themselves make exis-tential assertions. “Weakly parallel” means we do not make a case distinctionwhether the lines coincide or not. Of course if they do coincide the conclusionis trivial. Since we need to deal with weakly parallel lines, in order to supportcoordinatization without case distinction, Playfair’s axiom is more useful thanweak Playfair. Another relevant fact is that Playfair is equivalent to its doublenegation, which one cannot say about weak Playfair (since Parallel involves auniversal quantifier and ¬¬, ∀ is not the same as ∀¬¬).

120 MICHAEL BEESON

Lemma 10.4. Playfair’s axiom implies (in neutral ECG) that weakly parallellines make equal alternate interior angles with any transversal. Similarly, weakPlayfair’s axiom implies that parallel lines make equal alternate interior angleswith any transversal.

Proof. Let L and K be weakly parallel lines, and let pq be a transversal, with pon K and q on L and p 6= q. Let s be any point on L not equal to q. Let t bethe midpoint of pq, and extend segment st to point r with B(s, t, r) and tr = ts.Let M = Line (p, r). Then by Definition ??, pq makes equal alternate interiorangles with L and M .

If p does not lie on L, then by Lemma 10.3, M and L are parallel. Butthen M and K are two weakly parallel lines to L passing through p. Hence byPlayfair’s axiom, they coincide. Hence pq makes alternate interior angles withL and K. On the other hand, if p does lie on L, then M and K coincide andwe are finished in that case too. But by Axiom S3, the stability of on , we have¬¬ (on (p, L) ∨ ¬on (p, L)), so it is constructively legal to argue by these twocases. That completes the proof.

The next lemma will be needed below.

Lemma 10.5. Given a point p and a line L, we can construct a point on Ldifferent from p.

Remark. Of course the lemma is classically trivial. But notice that the similar-sounding lemma, given lines K and L that do not coincide, there is a point onL that is not on K, is not constructively correct, since such a point cannot befound continuously in K and L.

Proof. Let a = pointOn1 (L) and b = pointOn2 (L). Then a 6= b. Extendsegment ab by ap to point z; then extend segment az by ab to point t. Thenaz > ap (by Definition 9.64). Let C = Circle (a, t). Since ap = az < ap, pointp does not lie on C (by Lemma 9.65 and Axiom C5). Hence p 6= t, so point tmeets the conditions of the lemma. That completes the proof of the lemma.

10.5. Another version of the strong parallel postulate. We now con-sider yet another expression of the parallel postulate, which we call SPPE for“strong parallel postulate equivalent.” We show that this version is closely re-lated to the stability principle ¬¬t↓ → t ↓.¬ IntersectLines (K,L) ↓ ∧ on(p,K) ∧ on(p,M) (SPPE)∧ on(a,M) ∧ ¬ on(a,K) → IntersectLines (L,M) ↓

In other words, if K is weakly parallel to L through p, then any other line Mthrough p must meet L.

Lemma 10.6. Playfair’s postulate proves that SPPE is equivalent to the strongparallel postulate SPP.

Proof. We show first that the strong parallel postulate implies SPPE (withPlayfair). Let K and L be weakly parallel, i.e. IntersectLines (K,L) is notdefined, and let p be on K, and let M be another line through p. By Lemma 10.5,we can find a point q on L different from p. By Playfair’s axiom and Lemma 10.4,pq makes alternate interior angles equal with K and L. Then by the strongparallel axiom, M meets L, but that is the conclusion of SPPE.


Figure 30. SPPE: K and L are parallel, so M and L mustmeet, but we don’t say where, and we assume nothing about aexcept that it is not on K.

b

p

b

a

L

K

M

Conversely, suppose SPPE, and let K and L be lines, and pq a transversalmaking equal alternate interior angles withK and L, andM another line throughp, with p on K, and a a point of M that is not on K. By Lemma 10.3, K and Lare parallel, so by SPPE, M meets L. That completes the proof of the lemma.

10.6. Coordinates. In this section, we show that the strong parallel axiomis sufficient to introduce the usual Cartesian coordinates.

The “coordinates” of a point b are given by terms X(b) and Y (b). These termsare defined by the following construction scripts. The term I in the second scriptis defined in Definition 9.103.

Definition 10.7. X(Point b)

L = Line(0,1)

J = Perp(0,L)

H = Perp(b,L)

return IntersectLines(H,L)

Y(Point b)

L = Line(0,1)

J = Perp(0,L)

M = Perp(b,J)

P = IntersectLines(M,J)

return Rotate(I,0,1,P)

Note that Y (b) returns a point on the x-axis L, not a point on the Y -axis J .Coordinatization can be done without any parallel axiom. To “uncoordinatize”

we define a term MakePoint (x, y). This term takes two points on Line (0, 1)and makes a point whose coordinates are the given points. A related termMakePoint2 starts with a point on the x-axis and another point on the y-axis.For technical reasons we define MakePoint slightly more generally, so that itworks with any points x and y and first projects them on Line (0, 1); this stepdoes nothing if the points are on Line (0, 1) to begin with.

122 MICHAEL BEESON

Definition 10.8. The term MakePoint (x, y) is defined by the following con-struction script, which is illustrated in Fig. 31.

MakePoint(Point x, Point y)

L = Line(0,1)

J = Perp(0,L)

U = Perp(x,L)

W = Perp(y,L)

z = Rotate(1,0,I,\IntersectLines(W,L))

V = Perp(z,J)

return IntersectLines(U,V)

We already proved in Lemma 9.104 that Right (1, 0, I). A similar fact, whichcould not be stated until coordinates were defined, is the following:

Lemma 10.9. Right (1, 0, Y (γ)).

Proof. By definition of Y , Y (γ) = Rotate (I, 0, 1, p). By Lemma 9.101, Y (γ) is onRay (0, 1). Hence by Lemma 9.89, Right (Y (γ), 0, I) if and only if Right (1, 0, I),which we proved in Lemma 9.104. Hence Right (Y (γ), 0, I). That completes theproof of the lemma.

Next we prove the basic properties of MakePoint , and the coordinate functionsX and Y .

Figure 31. MakePoint constructs a point with coordinates xand y, shown by the small open circle.

L

J

bz

b

0b

yb

x

U W

Lemma 10.10 (Uncoordinatization). If b = MakePoint (x, y) is defined, thenx = X(b) and y = Y (b). The strong parallel postulate implies that for every pairof points (x, y) on L, MakePoint (x, y) is defined.

Proof. Let U = Perp(x, L) and V = Perp(z, J) as in the script for MakePoint (x, y).If b = IntersectLines (U, V ) is defined thenX(b) = x and Y (b) = y, by Lemma 9.80


(the uniqueness of perpendiculars at a point). To prove the second claim, assumethe strong parallel postulate. We only have to prove that U and V do intersect.Since V and L are both perpendicular to J , they are weakly parallel, that is,IntersectLines (V, L) is not defined, since if it were defined then by Axiom I14,p = IntersectLines (V, L) would be on both V and L, and hence by Lemma 9.81,V and L would coincide, contradicting Axiom I21. (Note that we needed theuniform version of uniqueness of perpendiculars, since we do not know whether pis on J or not.) U does not coincide with L, and in fact we can construct a pointa on U that is not on L, for example a = IntersectLineCircle1 (U,Circle (x, 0, 1)).That point is not on L since it is different from x, and if it were on L, then Land U would coincide, but perpendicular lines cannot coincide, as remarked afterDefinition 9.35. Now L is weakly parallel to V , and U and L have a commonpoint x and point a lies on U but not on L, so SPPE (which is equivalent tothe strong parallel postulate) applies, with (L,K,M, p, a) in SPPE taken to be(V, L, U, x, a), and the conclusion is that IntersectLines (V, U) ↓. That completesthe proof.

Remark. If SPPE is weakened by changing the hypothesis

¬ IntersectLines (K,L) ↓to Parallel (K,L), then we need a case distinction y = 0 ∨ y 6= 0. In either casewe can conclude IntersectLines (U, V ) ↓, but since the stability of t ↓ is not anaxiom of ECG, we cannot complete the argument by double-negating the casedistinction.

10.7. The strong parallel postulate and stability of IntersectLines. Re-call that the stability axioms omitted the axiom ¬¬ t ↓ → t ↓, because it wasprovable except in the case when t involves IntersectLines . We now show that inthat case, it is relatively strong! If we assume that stability axiom, then we canno longer distinguish the various versions of the parallel postulate. In a sense,that simply argues for accepting both SPP and the stability postulate, which iswhat we actually do in ECG, but we also wish to investigate the constructiverelations between the different forms of the parallel postulate relative to someweaker theory, which we take to be Playfair plus neutral ECG.

Lemma 10.11. Let L and K be lines that do not coincide and are not parallel.Then there exists a point on K that is not on L.

Remark. Of course classically the lemma is trivial and does not even need the hy-pothesis that the lines are not parallel. Constructively that hypothesis is crucial,since without it, a point on M that is not on L cannot be found continuously asM varies (ever so slightly from a parallel).

Proof. By making a rotation and translation, we can assume that L is Line (0, 1).Let J = Perp(0, L). Then from each point b of K, we can drop a perpendicularM(b) to J . We claim that M(b) meets K only in b. To prove this, suppose thatM(b) meetsK at another point; thenM(b) andK coincide, soK is perpendicularto J ; then K is parallel to L by Lemma 9.48), but that contradicts the hypothesisthat K is not parallel to L. Hence, as claimed, M(b) meets K only in b. Let Y (b)be the intersection point ofM(b) with J , which exists sinceM(b) is perpendicular

124 MICHAEL BEESON

to J . We only need to show that Y (b) takes on a nonzero value for some b, sinceif b is on L, then H(b) coincides with L, so H(b) = 0. As usual let I be a pointon J with 0I = 01. By Lemma 10.10, using the strong parallel axiom there is apoint b such that X(b) = x and Y (b) = I. That b lies on M but not on L. Thatcompletes the proof.

Lemma 10.12. SPP (and SPPE) imply the stability principle:

¬¬ IntersectLines (L,M) ↓ → IntersectLines (L,M) ↓ .

Conversely, that stability principle plus Playfair implies SPP (and SPPE).

Proof. First we show that SPPE implies the stability principle for IntersectLines .¬ IntersectLines (L,M) ↓ is the definition of “weakly parallel.” Suppose thenthat L and M are not weakly parallel. Then they do not coincide and are notparallel. Let p be a point on M (by Lemma 10.11 we can even choose p sothat p is not on L). Let K be weakly parallel to L through p (for exampleK = Perp(p,Perp(p, L)) will do). Let a be a point on M different from p (forexample

IntersectLineCircle1 (M,Circle (a, α, β))

will do). Then a is not on K since if it were, M and K would coincide, but M isnot weakly parallel to L and K is weakly parallel to L. Then by SPPE, L andM meet. That completes the proof that SPPE implies the stability principle forIntersectLines .

Conversely, suppose that stability principle holds. We will prove SPPE. Let Kbe weakly parallel to L through p, and let M be another line through p, contain-ing a point a that is not onK. By S6 it suffices to prove ¬¬ IntersectLines (L,M) ↓.Suppose then that ¬ IntersectLines (L,M) ↓. Since L and M do not coincide,they are parallel. But that contradicts strong Playfair. That completes the proofof the lemma.

Corollary 10.13. If, in the strong parallel postulate or in SPPE, we add

¬on (a, L) ∧ ¬on (p, L)

to the hypothesis (so the lines L and K are required to be parallel in SPPE, notjust weakly parallel), and the case of L coinciding with M in the strong parallelpostulate is ruled out, then the resulting principles are equivalent to the principlesas formulated, and to the stability principle for IntersectLines .

Proof. The principles with one or two additions to the hypothesis become primafacie weaker, so we only have to show that the modified principles still imply thestability principle for IntersectLines . The proof of Lemma 10.12, by appealingto Lemma 10.11, works just as well for the modified SPPE; hence the modifiedSPPE implies the stability principle. On the other hand the modified strongparallel postulate implies the modified SPP, by the proof of Lemma 10.6, sincethe additional hypotheses in question are the same in both principles. Thatcompletes the proof.


10.8. Stability of definedness and ∼=. When listing the stability axiomsof ECG, we discussed the principle S6, namely ¬¬t ↓ → t ↓, which we didnot take as an axiom. In this section we will prove that principle in ECG,and show that the strong parallel postulate is needed only in case t containsIntersectLines . The key to this theorem is to establish geometric conditions fort to be defined. To that end we begin with lemmas. The first one is used to givegeometric conditions for a line and circle to intersect.

Lemma 10.14. Let C be a circle with center a and L a line. Let K be theperpendicular to L through a. Let p be the intersection point of K and L. If anypoint of L is inside C then p is inside C; if any point of L is strictly inside Cthen p is strictly inside C.

Proof. If any point of L is inside C, then by the line-circle continuity axiom, Lmeets C at one or two points u, v.

Case 1: If u 6= v then triangles upa and vpa are congruent by Lemma 9.50,since they are right angles sharing a leg and with equal hypotenuses ua = va.Hence up = vp. Hence by Lemma 9.5, u and v do not lie on the same side of p.Hence by Markov’s principle for betweenness, B(u, p, v). Hence by Lemma 9.75,ap < ab, since ab is the radius of C. Then p is inside C, by definition of “inside.”

Case 2: If u = v then by the definition of “tangent”, L is tangent to C atv, so by Lemma 9.54, Line (a, u) is perpendicular to L; then K and Line (a, u)are both perpendicular to L from a, so by the uniqueness of perpendiculars,p = u = v. Then ap = au = ab, so p is inside C in that case too.

Now we justify an argument by cases. Since ¬¬(u = v ∨ u 6= v), we have¬¬ap ≤ ab; hence by the stability of segment ordering (Lemma 9.15) we haveap ≤ ab, so p is inside C. That completes the proof of the lemma.

The following lemma is used to give geometric conditions for two circles tointersect.

Lemma 10.15. Let C and K be circles with centers a and b, respectively, andsuppose a 6= b. Let L = Line (a, b) and let p = IntersectLineCircle1 (L,K) andq = IntersectLineCircle2 (L,K). Then if any point of K is inside C, p is insideC; if any point of K is strictly inside C, p is strictly inside C; if any point ofK is outside C, q is outside C, and if any point of K is strictly outside C, q isstrictly outside C.

Proof. Let u = IntersectLineCircle1 (L,C) and v = IntersectLineCircle2 (L,C).According to the definitions of “inside” and “outside” (Definition 8.11), what wehave to prove is that if x is any point on K, then

xa < au → pa < au

xa ≤ au → pa ≤ au

xa > au → qa > au

xa ≥ au → qz ≥ au

It will suffice to show pu ≤ au with equality only when x = p, and ua ≤ quwith equality only when x = q. By Lemma 9.72 (the stability of angle ordering),these statements are stable; so we can argue by cases on the possible betweenness

126 MICHAEL BEESON

relations of the six points in question. In each case we assume the previous casesdo not hold.

Case 1, p = q. Then K is a degenerate circle and x = p. Then the conclusionsare all immediate, completing this case.

Then p 6= q, so by Lemma 9.12, we have B(p, b, q), and by Lemma 9.91, p is onRay (b, a) and q is on the opposite ray. We have q 6= a, since q is on the oppositeray to Ray (b, a), but a is not on that ray since a 6= b.

Case 2, C is a degenerate circle, i.e. u = v = a. Then if any point x of K isinside C, we have x = a, and so x lies on L. Then by Lemma 9.52, x cannot bedifferent from both p and q. But x 6= q since x = a and a 6= q. Hence ¬¬x = p.Hence x = p by the stability of equality. Hence a = p. Hence p is inside x. Sinceq 6= a, q is certainly outside a. That completes this case.

Now neither K nor C is a degenerate circle; hence by Lemma 9.12 we haveB(u, a, v).

Assume q is strictly inside C. Then

B(p, b, q) by Lemma 9.12

B(a, b, q) ∧ B(u, a, b) by Lemma 9.91

B(p, a, b) by inner transitivity of betweenness

B(u, q, v) since q is strictly inside C

We also have

B(u, a, b) derived above

B(a, b, q) derived above

B(u, a, q) inner transitivity

B(u, q, v) derived above

B(a, q, v) inner transitivity

We will now show that p is inside C. Since B(p, a, b) we have

ap < pb by definition of <

= bq since p and q are on K with center b

< aq since B(a, b, q)

< av since B(a, q, v)

= au since u and v are on C with center a

Hence ap < au, so p is strictly inside C. We have now proved that if q is strictlyinside C, so is p. Now suppose x is any point of K that is inside C. If p is strictlyinside C, then the first inequality is satisfied. Hence if q or p is strictly insideC then the first inequality is satisfied. Then, by the stability of congruence, wemay assume that both q and p are outside C. Then B(v, p, b). Let x be anypoint on K, and let m be the foot of the perpendicular from x to L. Then wehave xa > am since the leg of a right triangle is less than the hypotenuse byLemma 9.74; and B(p,m, b); so since B(a, v, p) and B(v, p,m) we have B(a, v,m)and hence av < am. Hence au = av < am < ax, so x is strictly outside K.Hence the first inequality holds.


The second inequality is proved the same way, using “inside” instead of “strictlyinside”; we need to first prove that if q is inside C, so is p, which is done thesame way as for “strictly inside.”

Now for the third inequality. Assume x is strictly outside C. Assume q isinside C. Then, as shown above, p is inside C also. Then

bx = bq since x and q lie on K

< aq since B(a, b, q)

< av since B(a, q, v) since q is inside C

= au since u and v both lie on C

Hence x does not lie outside C, contradiction. Hence q does not lie inside C.Hence, by the stability of congruence, q lies outside C, which establishes thethird inequality. The fourth inequality is proved the same way, using “outside”instead of “strictly outside.” That completes the proof of the lemma.

Theorem 10.16. (i) Let t be a term not containing IntersectLines . Then

¬¬ t ↓ → t ↓is provable in neutral ECG, i.e. without any parallel axiom.

(ii) The strong parallel axiom proves every instance of the schema

¬¬ t ↓ → t ↓ .Remark. This theorem shows that the principle S6, namely ¬¬ t ↓ → t ↓,

that we did not take as an axiom of ECG, turns out to be the strong parallelpostulate in disguise. Principle S6 seems quite natural in connection with theother stability axioms, because it looks like another form of Markov’s principle–if two lines cannot fail to meet then they meet. This might be considered as anargument in favor of the strong parallel axiom. At the very least it shows that ifwe want to study the logical relationships of various forms of the parallel axiom,the base theory had better not include S6.

Proof. It suffices to prove the lemma for terms t that do not containIntersectCirclesSame or IntersectCirclesOpp , by Theorem 9.96. We proceedby induction on the complexity of such terms t. The basis case is when t is aterm consisting of one function symbol with variables for arguments. We have toconsider each of the function symbols of ECG separately. Several of the functionsymbols are total, i.e. f(x) ↓ is provable, immediately from the axioms. Theseinclude pointOn1 , pointOn2 , pointOnCircle , and center. For example, givencircle C, we have by Axiom CA6, Circle (center(C), pointOnCircle (C)) = C,from which it follows that center(C) ↓ and pointOnCircle (C) ↓. Similarly fromAxiom CA1 we have pointOn1 (L) ↓ and pointOn2 (L) ↓.

When f is IntersectLineCircle1 or IntersectLineCircle2 , we appeal to the line-circle continuity axiom Cont 1. Suppose IntersectLineCircle1 (L,C) ↓, and letz = IntersectLineCircle1 (L,C). Then by Axiom I6, z lies on both L and C.Then with b = a = x = p, the hypothesis of Axiom Cont 1 is satisfied. That is,IntersectLineCircle1 (L,Circle (a, b)) ↓ if and only if

∃p, x(on(p, L) ∧ ax = ap ∧ T(a, x, b).

128 MICHAEL BEESON

We now claim that this last condition is stable. That is,

¬¬∃p, x(on(p, L)∧ax = ap∧T(a, x, b)) → ∃p, x(on(p, L)∧ax = ap∧T(a, x, b))

Suppose ¬¬∃p, x (on(p, L) ∧ ax = ap ∧ T(a, x, b)). The hypothesis is the doublenegation of “some point on L is inside C”. We will construct p and x as required.LetK =⊥ (a, L) and let p be the intersection point ofK and L. By Lemma 10.14,if any point of L is inside C, then this point p is inside C. Let x be a point onRay (a, b) such that ax = ap. Then on(p, L) by construction of p and ax = ap byconstruction of x. Then by the definition of “inside”, if any point of L is inside C,then T(a, x, b). But we have assumed that not not some point of L is inside C;hence ¬¬T(a, x, b). Since T is defined by a negated formula, and triple negationis the same as single negation, we have T(a, x, b). That completes the proof ofthe stability of IntersectLineCircle1 (L,C) ↓. The case of IntersectLineCircle2is treated the same way.

We next consider the case when f is IntersectCircles1 or IntersectCircles2 .By Axiom I18, if C and K are concentric, then IntersectCircles1 (C,K) andIntersectCircles2 (C,K) are not defined. Let C and K be two circles thatare not concentric. By the circle-circle continuity axiom, C and K intersectif there is a point of K inside C and a point of K outside C. In fact theyintersect if and only if this condition holds, since if they do intersect, the in-tersection points are both inside and outside C. Let a = center(C) and b =center(K). Let L = Line (b, a), which is defined since C and K are not concen-tric. Let p = IntersectLineCircle1 (L,K) and q = IntersectLineCircle2 (L,K).By Lemma 10.15, if K has a point inside C and another point outside C, thenp is inside C and q is outside C. Hence K and C intersect if and only if p isinside C and q is outside C. Let c = pointOnCircle (C). Then K and C intersectif and only if ap ≤ ac and ac ≤ aq. But since segment ordering is stable byLemma 9.15, this last condition is stable. Hence “non-concentric circles C andK intersect” is stable.

Now suppose ¬¬ IntersectCircles1 (C,K) ↓. Then by Axiom I17 and the sta-bility of equality, C and K are not concentric, and by Axiom I8, not not thereis a point on both circles. But as just shown, this condition is stable, so there isindeed a point on both circles. Hence by Axiom I16, IntersectCircles1 (C,K) ↓.That completes the proof in the case f = IntersectCircles1 (C,K). The case ofIntersectCircles2 (C,K) is treated similarly, appealing to Axiom I19, I9, and I17instead of I18, I8, and I16.

That completes the basis case of the induction on the complexity of t for claim(i), since we have considered all the function symbols except the two excluded inthe first sentence of the proof. For claim (ii), we must also treat the case whent = IntersectLines (r, s). By Lemma 10.12, with the strong parallel postulate wehave

¬¬ IntersectLines (L,M) ↓ → IntersectLines (L,M) ↓ .Suppose ¬¬ t ↓. Then by the strictness axiom of LPT, ¬¬ r ↓ and ¬¬ s ↓. Thenby the induction hypothesis r ↓ and s ↓. Then we can substitute r and s for L andM in the stability principle for IntersectLines , obtaining IntersectLines (r, s) ↓;but that is t ↓. That completes the proof of the lemma.


Here is the induction step: Consider a term t = f(s1, . . . , sn), where f is afunction symbol and the sk are terms. Suppose ¬¬t ↓. Then for each k, ¬¬sk ↓.By the induction hypothesis, there is a proof of sk ↓ (in neutral ECG unless tcontains IntersectLines , and in ECG if it does contain IntersectLines ). By thebasis case we have

¬¬f(x1, . . . , xn) ↓ → f(x1, . . . , xn) ↓ .Since we have proved sk ↓ for each k = 1, 2, . . . , n, we can substitute sk for xk

and conclude

¬¬f(s1, . . . , sn) ↓ → f(s1, . . . , sn) ↓ .That is the desired conclusion ¬¬t ↓ → t ↓. That completes the proof.

We now ask whether t ∼= q is stable. The answer is, that the strong parallelaxiom implies that t ∼= q is stable, but we really need the strong parallel axiomto prove that.

Lemma 10.17. The strong parallel axiom implies ¬¬ t ∼= q → t ∼= q.

Proof. We have

¬¬t ∼= s↔¬¬ ((t ↓ → s ↓ ∧ t = s) ∧ s ↓ → t ↓ ∧ t = s)

↔ (¬¬ t ↓ → ¬¬ s ↓ ∧¬¬ t = s) ∧ (¬¬ s ↓ → ¬¬t ↓ ∧ ¬¬ t = s)

↔ (¬¬ t ↓ → ¬¬ s ↓ ∧t = s) ∧ (¬¬ t ↓ → ¬¬ s ↓ ∧ t = s)

We will be able to complete the proof if we know that ¬¬t ↓ and ¬¬s ↓ arestable. In ECG, by the stability axioms, this is true as long as t and s do notcontain the function symbol IntersectLines ; to prove the stability of definednessof terms involving IntersectLines , we need the strong parallel axiom. However,with the aid of the strong parallel axiom we do have that stability, by Theorem10.16. That completes the proof.

10.9. ECG proves Euclid’s parallel postulate. In this section, we givea direct proof of Euclid 5 in ECG. Later we give a more informative proof, bycharacterizing the models of each theory as different types of Euclidean fields.But it is still instructive to have a direct proof.

The strong parallel postulate SPP has a weaker conclusion than Euclid 5,because it does not specify on which side of P the intersection point will lie. Onthe other hand, SPP also has a weaker hypothesis than Euclid 5, so its exactrelationship to Euclid 5 is not immediately clear. One direction is settled by thefollowing theorem:

Theorem 10.18. ECG proves Euclid’s Postulate 5.

Proof. Since we proved SPPE is equivalent to SPP using Playfair, we can useSPPE. Let L be a line, p a point not on L, K parallel to L through P , M anotherline through P , q a point on L, q a point on M not on pq, and r the intersectionof QA with K. Suppose that the interior angles made by L, M , and PQ makeless than two right angles, which formally means that a is between q and r. SeeFig. 26 for an illustration. By SPPE, since M and K do not coincide, M doesmeet L at some point e (indicated by the open circle in the diagram). It remainsto show that a is between p and e. By Markov’s principle for betweenness, it

130 MICHAEL BEESON

suffices to prove that p is not between e and a, and e is not between p and a.Suppose first that p is between e and a. Then e is on the opposite side of lineK from a. Segment eq lies on line L, and line L does not meet K, so segmenteq does not meet K. Hence by Lemma 9.83, e and q are on the same side ofline K. Since e is on the opposite side of K from a, it follows that a and q areon opposite sides of line K. Hence point r, the intersection of aq with K, mustbe between a and q. But that contradicts the fact that a is between q and r.Hence the assumption that p is between e and a has led to a contradiction. Nowsuppose instead that e is between p and a. Then a and p are on opposite sidesof L.

But rp does not meet L, since rp lies on K, and K does not meet L. Henceby Lemma 9.83, r and p are on the same side of line L. Hence a and r are onopposite sides of L. But then the intersection point of ar and L, which is q, liesbetween a and e, contradicting the fact that a lies between q and r. Hence theassumption that e is between p and a has also led to a contradiction. Hence, asnoted already, by Markov’s principle for betweenness, a is between p and e. Thatcompletes the proof of Euclid’s Postulate 5 from the strong parallel postulate.

Now we consider the converse problem, of deriving Axiom 58 from Euclid’sPostulate 5. The obvious proof attempt works only if we assume an additionalaxiom. A convenient formulation of what is required is “of two unequal segmentsone is longer”:

ab 6= cd → ab < cd ∨ cd < ab (Segment Comparison Principle)

Note that, unlike any axioms of ECG this principle involves a disjunction. It isequivalent to the ordering principle

x 6= 0 → x > 0 ∨ x < 0

We shall discuss the constructive acceptability (or non-acceptability) of thisprinciple far below. For now, we just note that, unlike the axioms of ECG, itinvolves a disjunction.

Lemma 10.19. Let T be the theory ECG without any parallel postulate. Thenin T, the Segment Comparison Principle plus Euclid 5 implies the strong parallelpostulate.

Proof. Assume Euclid’s postulate and let p be a point not on line L, and Kparallel to L through p, and line M another line through p as in Axiom 58. Letq be the point on L at the foot of the perpendicular to L from p. Let A on lineM and B on line L be on the same side of pq. Then K makes the interior anglesApq and pqB equal to two right angles. Hence M does not, or it would coincidewith K. If angle Apq is less than a right angle, then by Euclid’s Postulate 5,M meets L. Similarly, if angle Apq is more than a right angle, we can show byEuclid’s Postulate that M meets L (on the other side of q from B). By Axiom75, one of these alternatives must hold. More formally, let S be the intersectionpoint of Line (A,B) with K. Then angle Apq is less than a right angle if A isbetween S and B, and more than a right angle if S is between A and B. Bythe Segment Comparison Principle, one of these alternatives holds. Hence byEuclid’s Proposition 5, M meets L. That completes the proof of the lemma.


10.10. Various forms of the parallel axiom. Within neutral geometry,we can consider the logical relations between various forms of the parallel axiom.In the introduction we have already listed the Playfair axiom, Euclid 5, and thestrong parallel axiom of ECG. There are also a number of other propositionsthat are classically equivalent to Euclid 5. Lemmas in this section continue tobe proved in neutral geometry, i.e. in ECG without any parallel postulate.

The first of these is the “triangle circumcription principle”, which says thatany three non-collinear points lie on a circle. This makes sense constructively,but it also has a stronger version, the “strong triangle circumscription principle”,which says that there is a construction (term of ECG) t(a, b, c, L) such that ifa and b lie on L, and c does not lie on L, then t(a, b, c, L) is a circle passingthrough a, b, and c, and if a = b then t(a, b, c, L) is tangent to L.

We have proved that a line intersects a circle in at most two points, and fromthat it is not difficult to prove Euclid III.1, that a chord of a circle lies insidethe circle. Hence the wording, “triangle circumscription” is appropriate; once acircle passes through the vertices of a triangle, the whole triangle lies inside thecircle.

The triangle circumscription principle is a convenient form for a theory withonly point variables, since it can be naturally stated without mentioning circlesor lines: given three non-collinear points, there is a fourth point equidistant fromall three. We show below that it is constructively equivalent to the strong parallelpostulate, so we could have taken it for the parallel postulate in ECG. Some mayfind it a more natural formulation. According to [30], p. 190, Szmielew took itfor her parallel axiom in her influential manuscript that formed the basis for [26],but Schwabhauser adopted another form of the parallel axiom for publication.

Figure 32. Parallel axiom implies triangle circumscription.The center is where non-parallel lines M and L meet.

L

K

b

x

bc

ba

bb

M

132 MICHAEL BEESON

Lemma 10.20. The strong parallel axiom implies the strong triangle circum-scription principle.

Proof. Suppose a and b lie on line L, and c does not lie on L. Then byLemma 12.4, we can construct a line K perpendicular to L that is the per-pendicular bisector of ab if a 6= b, and contains a if a = b. Let M be theperpendicular bisector of ac; since c is not on L, a 6= c, so this perpendicularbisector exists. Let x be the intersection point of L and K (so x is the midpointof ab if a 6= b, and is a if a = b. Now if K and M meet, that intersectionpoint will be the center of the desired circle, as is easily proved, at least whena 6= b∨a = b; but once the intersection point e of K and M exists, we can argueby cases that ae = be = ce, since equality is stable. The crux of the matter is toprove that K and M do meet.

By Lemma 9.56 there is a line N perpendicular to M containing x. Let y onM be the intersection point of N with M . Let J be perpendicular to N at x.Then N is perpendicular to both J and M ; so by Lemma 9.48, J and M areparallel. Now assume that K and M are parallel, and both contain x. Henceby Playfair (which is a consequence of the strong parallel axiom), K and Mcoincide. Now K is perpendicular to N and to L at x; by the uniqueness of theperpendicular, L and N coincide. But then c lies on L, contrary to hypothesis.Hence K and M are not parallel. Then by the strong parallel axiom, they meet.That completes the proof.

Lemma 10.21. The triangle circumscription principle implies the strong par-allel axiom.

Proof. Suppose L is a line and p is a point not on L; let pw be perpendicularto L at point w on L and let K be perpendicular to Line (p, w) at p. Then Kis parallel to L. Let M be a line through p parallel to L that does not coincidewith K. Let J be perpendicular to M at p and let x be a point on J noton M . We can choose x so that x is not on L (For example, by dropping aperpendicular from p to L, finding its midpoint m, and choosing px = pm.) ByLemma 9.56, there exists a line N perpendicular to L and containing x. Let ybe the reflection of x in M (so x 6= y since x is not on M , and x is on J and J isthe perpendicular bisector of xy. Let z be the reflection of x in L; since x is noton L, z 6= x. If x, y, and z are all collinear, then lines J and N coincide, sincex and y lie on J , and x and z lie on N . Then J is perpendicular to both L andM . Hence M and K are both perpedicular to J at p. By the uniqueness of theperpendicular, M and K coincide, contradicting out assumption that M and Kdo not coincide. That proves that x, y, and z are not all collinear. Then by thetriangle circumscription principle, there is a circle C through those three points.Let its center be e. Then since xy is a chord of C, its perpendicular bisector Mpasses through e. Since xz is a chord of C, its perpendicular bisector L passesthrough C. Hence L and M meet in point e. This contradicts the assumptionthat M is parallel to L. Hence we have proved that every line M parallel to pthrough L coincides with K. In particular any two such lines coincide with eachother. That completes the proof.


Corollary 10.22. The triangle circumscription principle implies the strongtriangle circumscription principle.

Proof. By the previous two lemmas.

Remark. The “uncertainty” in the strong parallel axiom about which sideof the transversal has angles making less than two right angles corresponds toa similar uncertainty about which side of Line (a, b) the noncollinear point cmight lie on. If we weaken the triangle construction principle by strengtheningits hypothesis to say that c lies on the same side of Line (a, b) as a given pointx, then the resulting “weak triangle circumscription principle” is equivalent toEuclid 5.

Lemma 10.23. Playfair’s axiom implies the alternate interior angle theorem,that any line traversing a pair of parallel lines makes alternate interior anglesequal.

Proof. Since ordering of angles is stable, we can argue by contradiction. Hencethe usual classical proof of the theorem applies.

Definition 10.24. A rectangle is a quadrilateral with four right angles.

Lemma 10.25. Playfair’s axiom implies that if a quadrilateral has congruentdiagonals, then it is a rectangle.

Proof. “Angle ABC is a right angle” is stable by Lemma 9.39 and Definition9.42. Hence the classical proof applies.

Lemma 10.26. The opposite sides of a rectangle are equal.

Proof. In quadrilateral ABCD let AC = BD. Then triangle ABC is congruentto triangle ABD since they share side AB and AC = BD and the two rightangles are congruent by Lemma 9.44. Hence AD = BC. Similarly the other twosides are congruent. That completes the proof.

Lemma 10.27. The existence of any rectangle implies Playfair’s axiom.

Proof. FINISH THIS (why did I need this anyway?)

As mentioned above, Tarski [29] and later [26] took a different form of theparallel postulate, illustrated in Fig. 33.28 The following axiom is similar toTarki’s axiom, and we give it his name, but his axiom used non-strict betweennessand did not include the hypothesis that a, b, and c are not collinear. Thedegenerate cases are trivial: if a, b, and c are collinear, then we can (classically,or with more work also constructively) find x and y without any parallel axiom,and if (say) d = b then we can take x = t and y = c, etc. Hence the followingaxiom is classically equivalent in neutral geometry to the one used by Tarski:

B(a, d, t) ∧ B(b, d, c) ∧ a 6= d (Tarski parallel axiom)∧ (¬B(a, b, c) ∧ ¬B(b, c, a) ∧ ¬B(c, a, b))

→ ∃x∃y (B(a, b, x) ∧ B(a, c, y) ∧B(x, t, y))

28Technically, according to [30], the axioms taken op. cit differed in the order of argumentsto the last betweenness statement, but that is of no consequence.

134 MICHAEL BEESON

Figure 33. Tarski’s parallel axiom

x y

ba

bb

bc

bt

bd

Something like this axiom was first considered by Legendre (see [12], p. 223),but he required angle xay to be acute, so Legendre’s axiom is not quite the sameas Tarski’s parallel axiom.

Theorem 10.28. Euclid 5 implies Tarski’s parallel axiom in neutral ECG.

Figure 34. Constructive proof of Tarski’s parallel axiom fromEuclid 5. M is constructed parallel to Line (b, c) and cd = ceand bd = bg. Then x and y exist by Euclid 5.

x

ba

bb

bc

b

t

bd

y

M

b e

b

g

b f

b

h

Proof. Assume the hypothesis of Tarski’s parallel axiom. Construct line Mparallel to Line (b, c) through t. Construct point e by extending segment dc bydc; then ec = dc and B(d, c, e), as illustrated in Fig. 34. Let L be Line (a, c).Then Line (d, t) meets L at a, and does not coincide with L since, if it didcoincide with L, then points d and c would be on L, and hence point b, whichis on Line (b, c), would lie on L by Axiom I3; but that would contradict thehypothesis that a, b, and c are not collinear. Hence Line (d, t) meets L only ata, by Axiom I3. Hence segment dt does not meet L. Hence, by Lemma 9.83,d and t are on the same side of L. Since B(d, c, e), d and e are on oppositesides of L. Hence, by the plane separation theorem, e and t are on opposite


sides of L. Hence, there is a point f on L with B(e, f, t). Now we apply Euclid5; the two parallel lines are Line (b, c) and M , and the conclusion is that Lmeets M in some point, which we call y. Specifically we match the variables(L,K,M, p, r, a, q) in Euclid 5 to the following terms in the present situation:(M,Line (b, c),Line (a, c), c, e, f, t). Then all the hypotheses have been proved,except that we have B(e, f, t) while what is required is B(t, f, e); but those areequivalent by the symmetry of betweenness. Hence Euclid 5 is indeed applicableand we have proved the existence of point y on M and L.

Now, we do the same thing on the other side of angle bac, extending segment dbto point g with db = bg and B(g, d, b), and using the plane separation propertyto show that gt meets Line (a, b) in a point h with B(g, h, t). Then Euclid 5applies to give us a point x on M and Line (a, b).

It only remains to prove B(x, t, y). By the crossbar theorem (Lemma 9.61),Ray (a, d) meets Line (x, y) in a point between x and y; but since a is not onLine (x, y) (because a, b, and c are not collinear), there is only one point ofintersection of Ray (a, d) and Line (x, y), and that point is t; hence B(x, t, y).That completes the proof of the theorem.

We next show that Tarski’s parallel axiom implies Euclid 5 in neutral ECG.Tarski proved29 that it implies Playfair’s axiom (see [26], Satz 12.11, p. 123),and we constructivize his proof here to prove Euclid 5. The proof is short andbeautiful (and the reference is in German and out of print), and we need to checkits constructivity, so we repeat it here.

Figure 35. Tarski’s parallel axiom implies Euclid 5. C is par-allel to L. The points x and y produced by Tarski’s axiom showthat B is not parallel to L.

Lb

t

y

Cbe

ba

bc

x

bb

B

b

d

Theorem 10.29. Tarski’s parallel axiom implies Euclid 5 in neutral ECG.

Proof. Let L be a line, a a point not on L, and suppose C is parallels to L througha. Let B be another line through a, and points e, b, and t satisfy B(e, b, t) withe on C and t on L, and b not on L or Line (a, t), as in the hypothesis of Euclid

29The cited proof is in a book with two co-authors, but Tarski used this axiom from thebeginning of his work in geometry, and it seems certain that he had this proof before Szmielewand Schwabhauser were involved.

136 MICHAEL BEESON

5. Specifically at is a transversal of the parallel lines C and L, with which theadjacent interior angles of line B make less than two right angles on the sidewhere b is, as witnessed by B(e, b, t). Formally, the variables (L,K, p, r, a, q)of Euclid 5 are matched to (L,C, a, e, b, t) in this application. We must showthat B meets L. The situation is illustrated in Fig. 35. Since b does not lie onLine (a, t), but does lie on Line (e, t), we have e 6= a. Hence Line (e, a) is defined.Define

c = IntersectLineCircle2 (Line (e, a),Circle (a, e)).

Then c is on C and ae = ac and B(e, a, c).Then let b be the point of intersection of segment et with B. Now let d be the

intersection point of Line (b, c) and Line (a, t). By Outer Pasch (Lemma 9.57),applied to triangle eat and segment bc, we have B(a, d, t) and B(b, d, c). Now weare in a position to apply Tarski’s parallel axiom to angle bac and point t, whichis in the interior of angle bac as witnessed by crossbar bc and point d. Accordingto Tarski’s parallel axiom, there exist points x and y on Ray (a, b) and Ray (a, c)respectively such that B(x, t, y). Then x and y are on opposite sides of L, sincepoint t is on L between x and y. But ya does not meet L, since ya lies on C,which is parallel to L, so y is on the same side of L as a, by Lemma 9.83. Thenby the plane separation theorem, x and a are on opposite sides of L. Hencesegment ax meets L. But then B is not parallel to L. That completes the proof.

§11. Comparisons of ECG to other formalizations.

11.1. Comparison of ECG to Hilbert-style theories. By a Hilbert-styletheory, we mean one with variables for lines and whose axioms follow those ofHilbert. For example, the textbook [12] that we used in [3] to compare the firstformulation of ECG with classical logic ues a Hilbert-style theory. The presentversion of ECG does have variables for lines, and for circles too, but its axiomsfollow Tarski more closely than Hilbert. Just as in classical geometry, one maychoose formalisms with more sorts of variables and more axioms, in which casethe formal development looks a bit more like Euclid, or one may choose points-only, and seek for a minimal axiom set, in which case some infrastructure mustbe developed. Here we have taken some steps towards reducing the number ofaxioms, in order to see how this infrastructure is can be constructively devel-oped. We have retained variables of type Line and Circle, but they are easilyeliminated if one so desires. One should then show (directly or indirectly) thatthere are translations in both directions, i.e. the Tarski-style axioms are provablein Hilbert systems, and the Hilbert sytems are interpretable in the Tarski-stylesystems.

11.2. Comparison of ECG to Tarski’s theories. There are several dif-ferences: Tarski’s language is smaller, with only variables for points and therelations of non-strict betweenness and equidistance. He uses non-strict be-tweenness, while ECG uses strict betweenness. His logic is classical insteadof intuitionistic. His theory has no function symbols. His axioms are (conse-quently) not quantifier-free, though they are all in AE (or ∀∃) form. Since hehas no terms for the results of constructions, the question of the continuity ofthose constructions does not arise.


The use of a larger language is a matter of convenience, and of the attemptto make ECG faithful to Euclid. One can eliminate the variables for lines andcircles, introducing new versions of the function symbols that work only withpoints. One can also eliminate the function symbols in favor of quantifiers,and (in view of Markov’s principle for beweenness and stable equality) one cantranslate back and forth between strict and non-strict betweenness. There aretwo remaining innovations in ECG: the use of intuitionistic logic, and the effortto ensure the continuity of the terms for constructions, making use of the conceptsof right and left turns. These are independent innovations, in the sense thateither could be considered without the other. We next demonstrate that one canconsider intutionistic logic without construction terms.

It is some interest, both foundationally and technically, to formulate a geomet-rical theory in Tarski’s language with intuitionistic logic, that is constructivelyacceptable. One cannot just adopt Tarski’s axioms because he has (purposely)not excluded degenerate cases, and some of those cases (for example in his “innerPasch” axiom) make the axiom as formulated constructively invalid.

We modify the axiom of segment construction ∃(T(q, a, x) ∧ ax = bc) to thisversion:

q 6= a → ∃x(¬(¬T(q, a, x) ∧ x 6= a) (Ax. 4I)

This avoids extending a null segment but allows extending a segment by a nullsegment. The case of extending a null segment, i.e. Euclid I.2, is dealt with byconstructing a circle of given radius. But in the circle axiom of Tarski’s theory,one cannot get the intersection point of a line and circle without already havingsome point on the circle (to specify it). Therefore one either has to modify thecircle axiom, or explicitly assume a version of Euclid I.2:∃yxy = uv (Axiom for Euclid I.2)

It is interesting to see how the necessity to assume Euclid I.2 in constructivegeometry, which we saw already in ECG, comes up in Tarski’s theories.

It turns out that the inner Pasch axiom is the only other one that needs amodification. Here is the modified axiom:

B(a, p, c) ∧ B(b, q, c) ∧ p 6= b ∧ q 6= a∧¬B(c, a, q) ∧ ¬B(a, q, c) ∧ ¬(q, c, a) →∃x[B(p, x, b) ∧B(q, x, a)] (Ax. 7I-T)

One can also formulate a similar theory using B instead of T. If one replaces Twith B throughout (including in the modified Ax. 7I-T, which is then called Ax.7I-B), then one has to also add back a few axioms that were present in Tarski’soriginal theories but were derivable using T in degenerate cases. Specifically, onehas to add back the symmetry of betweenness B(a, b, c) → B(c, b, a) and thenull segment axiom ab = cc → a = b and the congruence axiom ab = ba. Wealso take B(a, b, c) → a 6= b and B(a, b, c) → b 6= c as axioms. The identityaxiom for betweenness becomes ¬B(a, b, a) instead of T(a, b, a) → a = b.

To either of these theories we should add stability of atomic formulas. We callthese theories “Intuitionistic Tarski-T” and “Intuitionistic Tarski-B.”

Theorem 11.1. Tarski’s geometry with intuitionistic logic, either with B orwith T, is satisfied in the recursive model. With classical logic, these theories are

138 MICHAEL BEESON

equivalent to Tarski’s theory. The two versions are equivalent when T is definedin terms of B or vice-versa, without using disjunction.

Proof. All the axioms of intuitionistic Tarski-B are evidently provable in ECG.Hence this theory holds in the recursive model. We turn to the equivalence ofthe two versions. First we define B by

B(a, b, c) := T(a, b, c) ∧ a 6= b ∧ b 6= c

Then we have to check that the interpretations of all the axioms of Tarsi-B areprovable in Tarski-T. First we note that the extra axioms added to Tarksi-B that are not in Tarski-T are provable (that was why they are not alreadyincluded). See [26] for details. Since B(a, b, c) implies T(a, b, c), hypotheses be-come stronger; we only need to check that the stronger conclusions are provable.For example, we check the Pasch axiom. Under the hypotheses of Ax 7I(B) onegets from Ax. 7I(T) the existence of x such that T(p, x, b) ∧ T(q, x, a), and onehas to prove B(p, x, b) ∧ B(q, x, a). So one has to rule out the degenerate casesand use Markov’s principle for betweenness. The degenerate case are p = x,x = b, q = x, and x = a. If p = x then a, b, and c are collinear, which is ruledout by hypothesis. If x = b then q = b, contradicting B(c, q, b). If q = x then(since x 6= b), p lies on Line (c, b) and hence p = c, contradicting B(a, p, c). Ifx = a then p = a, contradicting B(a, p, c). The other axioms are very easy andwe omit the verifications. That completes the proof that the two intuitionisticversions of Tarski’s theory are equivalent.

To check that with classical logic, these theories are equivalent to Tarski’stheory, we just need to check that the extra hypotheses imposed in axioms Ax.4-I and Ax. 7-I do not classically restrict the power of the axioms. With regardto the segment extension axiom Ax. 4, we need to prove that a null segment canbe extended; but we added the “Axiom for Euclid 1.2”, which takes care of this.With regard to Ax. 7, we need to check the case when a, c, and q are collinear,which is excluded by hypothesis in the intuitionistic theories, as well as the casesp = b and q = a. The latter two cases are easy: take x = p or x = q, respectively.Suppose a, c, and q are collinear. Then we have B(a, p, c) and B(c, q, b), and allthe points collinear, from which it follows that B(p, c, b) by the inner transitivityof betweenness. So technically, we have to check that this theorem is provablein Tarski’s intuitionistic theory. We do check below that it is provable in ECG,and the same proof works in Tarksi’s intuitionistic theory. That completes theproof of the theorem.

In subsequent sections, we will show how to interpret ECG in intuitionisticTarski-B, and we will also prove that Tarski’s geometry with intuitionistic logic,either with B or with T, proves every AE theorem that is provable with classicallogic in Tarski’s theories, and that these are the same AE theorems provable inECG.

§12. Connections between geometry and Euclidean fields. We nowturn to the characterization of models of ECG as planes over Euclidean fields.Classically, the models of Euclidean geometry (for example Tarski’s theory) areall planes over a Euclidean field, that is, an ordered field in which every positiveelement has a square root. The main point of this section is to prove a similar


theorem for the models of ECG. There are two interesting parts of this work:First, it is not trivial to defined signed arithmetic without case distinctions, i.e.to give a single construction for the sum of two points on Line (0, 1), regardlessof the positions of those points relative to the fixed points 0 and 1. Second, thedifferent forms of the parallel axiom turn out to correspond to different axiomsfor Euclidean fields. In Section 6, we have already discussed the constructive ver-sion(s) of the axioms for Euclidean fields. Recall that there were four versions,which we called “Playfair”, “weakly Euclidean”, “Euclidean”, and “strongly Eu-clidean”, according to what axiom about the existence of reciprocals is assumed.Euclidean means nonzero elements have reciprocals; weakly Euclidean meanspositive elements have reciprocals; Playfair means that elements without recip-rocals are zero; and strongly Euclidean means that nonzero elements are eitherpositive or negative (in addition to weakly Euclidean or Euclidean, which arethen equivalent).

We will show that these different versions of the Euclidean field axioms cor-respond directly to the different versions of the parallel postulate we have con-sidered. Euclid 5 corresponds to positive elements having reciprocals; the strongparallel axiom corresponds to nonzero elements having reciprocals; and Playfair’saxiom corresponds to “elements without reciprocals are zero.”

12.1. Signed addition. In order to establish the connection between ECGand the theory of Euclidean fields, it will be necessary to formalize the definitionsof Perp, Project, and Rotate that are required to define Add and Multiply as inSection 5.

Definition 12.1. Minus (x) = Reflect (x, 0,Line (0, 1)).

Equivalently, Minus (x) = Reflect (x,Line (0, I)).Next we need a way to construct a strict lower bound for given points x and

y on Line (0, 1). We also define the absolute value of x.

Definition 12.2.

Abs (x) = Extend (Minus (1), 0, 0, x)

UpperBound (x, y) = Extend (0, 1, 0,Extend (0, 1, 0,Extend (0, 1, 0, x)))

LowerBound (x, y) = Extend (0,Minus(1), 0,Extend (0, 1, 0,Extend (0, 1, 0, y)))

Lemma 12.3. If p = LowerBound (x, y) and q = UpperBound (x, y), we haveB(p, x, q) and B(p, y, q), and p < x ∧ x < q and p < y ∧ y < q, and B(p, 0, q).

Proof. Since the conclusions are stable, we may argue by cases on the possiblebetweenness relations of 0, x, and y. We omit the details.

There is nothing special about Line (0, 1); we can make the same constructionson any line Line (p, q) with p replacing 0 and q replacing 1. We denote thoseconstructions by LowerBound (a, b, L) and UpperBound (a, b, L). Note that if wechange L(p, q) to L(q, p), then “upper” and “lower” bounds switch roles. Acorollary of the lower bound and upper bound construction is the possibility ofthe “uniform midpoint”; that is a way to construct the perpendicular bisectorof ab that extends to the case a = b. As just stated, that would be impossible,since the perpendicular bisector does not depend continuously on a and b as they

140 MICHAEL BEESON

approach each other. But if we restrict a and b to lie on a fixed line, it can bedone:

Lemma 12.4. There is a construction t(a, b, L) such that for all a and b online L, t(a, b, L) is a point x on L such that ax = xb.

Remark. Then Perp(x, L) is the perpendicular bisector of ab when a 6= b, and isperpendicular to L in any case.

Proof. Let t(a, b, L) = Midpoint (LowerBound (a, b, L),UpperBound (a, b, L)). SinceLowerBound (a, b, L) and UpperBound (a, b, L) are different, t(a, b, L) is definedfor all a, b, on L. That completes the proof.

We are now in a position to define signed addition. We follow the informalLemma 5.5, but now making use of the formal details above. Refer to Fig. 3 andFig. 4 for illustrations.

Definition 12.5. The term Add of ECG is given by collapsing the followingconstruction script into a term:

Add(A,B)

S = UpperBound(A,B)

R = UpperBound(S,0)

H = Perp(Line(0,1),B)

U = Rotate(R,0,I,A)

V = Project(U,H)

D = Project(I,H)

W = Rotate(D,B,R,V)

return W

This script differs from the informal description in Lemma 5.5 only in that Ispecifies the point C of that lemma; and it also gives the details about the con-struction of the point R that were omitted there. In the preceding sections, wehave formally defined and proved the properties of all the constructions men-tioned in the definition of Add; so now we are prepared to verify the propertiesof Add itself.

Lemma 12.6. Add (x, y) obeys the commutative and associative laws and thelaws Add (x, 0) = 0 and Add (x,Minus (x)) = 0, as well as Add (A,Minus (B)) =Add (Minus (A), B) and Minus (Add (A,B)) = Add (Minus (A),Minus (B)).

Proof. First we take up the law Add (A, 0) = 0. We use the notation of the scriptfor Add in Definition 12.5 (which also matches Fig. 3 and Fig. 4, except thatC = I). In this case since B = 0, line H coincides with Line (0, I). Then V = Usince V lies on line H . Rotating V back onto Line (0, 1) produces U again, byLemma 9.102. Hence W = U . That completes the proof of Add (A, 0) = 0. Thelaw Add (0, A) = A can be similarly treated.

Next we prove the law Add (Minus (B), B) = 0 (in which case letA = Minus (B)),and at the same time Add (A,Minus (A)) = 0 (in which case let B = Minus (A).In either case A and B are each other’s reflections in the point 0 and we wantto prove Add (A,B) = 0. The situation is as shown in Fig. 36. First A is ro-


Figure 36. Additive inverse: Addition when OA = OB.A is rotated to U , then projected to V , then rotated to W = O.

b bb

b

b

b

B 0 A

V U

C H

R

tated onto the line H = 0I. Then it is projected onto the line perpendicular toLine (0, 1) at B; that projection is V . Then it is rotated back to Line (0, 1) atpoint W . We have WB = BV = OU = OA. Hence WB = 0A. But 0B = 0Asince B = Minus (A) (or A = Minus (B)). Since A and B are reflections in 0, wehave B(B, 0, A), so 0 is on Ray (B,A). Then by Lemma 9.5 we have B = 0. Thatcompletes the proof of Add (A,Minus(A)) = 0 and Add (Minus (B), B) = 0.

Evidently Minus (0) = 0, since 0 is its own reflection in Line (0, I). Theconstructively difficult part of this proof was getting Add (A,B) to be defined(without a case distinction). Since equality is stable, we may argue for theequations in the lemma by cases, according to the signs of the arguments. Recallthat x is positive (x > 0) if x lies on Ray (0, 1) and x 6= 0, and similarly for“negative”; hence if x is positive, Minus (x) is negative and vice-versa.

The law Add (Minus (x),Minus (y)) = Minus (Add (x, y)) follows from the ob-servation that just as Minus (x) is the reflection in 0 of x, Add (Minus (x),Minus (y))is the reflection in 0 of Add (x, y); in fact the whole diagram for the definition ofaddition is reflected in 0 when A and B are reflected in 0.

Next we take up the commutative law Add (A,B) = Add (B,A). The casewhen either of A or B is zero has already been proved; we now take up the casewhen A and B are both positive, which is illustrated in Fig. 37.

Starting with A, we rotate to U , project to V , rotate to W = B+A. Startingwith B, we rotate to X , project to Y , and rotate to Z = B + A. We mustprove W = Z. As mentioned, we will argue by cases. First assume A and Bare both positive, i.e. they lie on Ray (O, 1). By the properties of Rotate inLemma 9.101, the points Z and W lie on Ray (0, 1), as illustrated in Fig. 37.We have B(0, A,W ), by construction of W , and B(0, B, Z), by construction ofZ. We also have OA = BW and AZ = AY = OX = OB, so AZ = OB. Thussegment OW is composed of OB and BW , and segment OZ is composed ofsegment OA and AZ, and we have OB = AZ and BW = BV = OU = OA, soBW = OA. Then by Lemma 9.3, we have OW = ZO = OZ. Hence W = Z byLemma 9.5, since both W and Z lie on Ray (0, 1). That completes the proof ofcommutativity of addition in case the arguments are positive.

If A is negative and B is positive, then Fig. 38 illustrates the situation. Ais rotated to U , projected to B, and rotated back to F = Add (A,B), labeled

142 MICHAEL BEESON

Figure 37. Commutativity of addition with A and B positive.

b b b

b

b

b

0 A B W = A+B = B +A

U V

C

H

R

b

Xb

Y

Figure 38. Commutativity of addition, one argument nega-tive.

bA

b0

bB

b

U

bA+B

b

V

b

Xb

Y

A+ B in the diagram. B is rotated to X , projected to Y , then rotated back toG. We have to prove F = G. Since A is negative and B is positive, we haveB(A,O,B). We have OB = OX = AY = AG, so AG = OB. Therefore byLemma ?? we have GB = AO, since AB is composed of AO and OB, but also ofAG and GB = AO. We have AO = OU = BV = FB. Hence AO = FB. HenceGB = FB. But both F and G lie on Ray (A,B), since B(A,O,B), according tothe properties of Rotate . Hence by Lemma 9.5, we have F = G as desired. Thatcompletes the proof of commutativity in case A is negative and B is positive.


The other two cases are similar to the two we have treated, and we omit thedetails.

Now we turn to the associative law, Add (Add (A,B), C) = Add (A,Add (B,C)).Again, since the conclusion is an equality and equality is stable, we can argue bycases. Since we have already proved Add (0, x) = Add (x, 0) = x, if any one of A,B, or C is zero we are done. Since Add (Minus (A), B)) = Add (A,Minus (B)) =Minus (Add (A,B)), general associativity can be reduced to the case when A, B,and C are all positive. That case is illustrated in Fig. 39.

Figure 39. Associativity of addition, positive arguments.

b b b

b

b b

C

b

0 A B A+ B G = B + CA+B + Cb

b b

b

U V

b

X Y

Z

b

E

P

The reader should follow the computation of A+B, then (A+B)+C, arrivingat the point labeled A+B+C. Then follow the computation of B+C, and thenA+ (B + C), arriving again at the point labeled A+B + C. We have to provethat you really arrive at the same point. The proof is a diagram-chase producinga chain of conguences, using the properties of projection and rotation. Here arethe details: Let W = Add (A,Add (B,C)) and let W ′ = Add (Add (A,B), C), andF = Add (A,B), with other points as in Fig. 39. We have

OA = OU = V B = BF where F is labeled A+B

OB = OZ = CE = CG

OF = OX = Y C = CW ′

OA = OU = GP = GW

Then OW is composed of OG and GW , and GW = OA. In turn OG is composedof OC and CG, and CG = OB as shown above. So OW is composed of OC,CG, and GW , which are respectively congruent to OC, OB, and OA. Similarly,OW ′ is composed of OC and CW ′. By Lemma 9.6, we have CW = CW ′

(cutting segment OC of both OW and OW ′). Now CW ′ is composed of CGand GW ′, and CW is composed of CG and GW ; applying Lemma 9.6 again wehave GW = GW ′. Then by Lemme 9.5, which applies since B(O,G,W ) andB(O,G,W ′), we have W = W ′. That completes the proof of the associative law,and also the proof of the lemma.

144 MICHAEL BEESON

There are, of course, several other cases according to the signs of A, B, andC; but we omit all the other cases, on the grounds that they are very similar tothis case. That completes the proof of the lemma.

12.2. Signed multiplication. We will now show how to adapt Descartes’sdefinition of multiplication to allow signed arguments. We specify that B inDescartes’s diagram is 0, and A is 1, and the angle at B is a right angle (whichdoes not correspond to Descartes’s diagram, but is convenient). Descartes’sdiagram (Fig. 5) is for multiplying BD by BC, but we want both points to startout on Line (0, 1), so the first step is to obtain C in the diagram by rotatingthe given point from Line (0, 1) to Line (0, I). Fig. 40 illustrates the followingconstruction script:

Multiply(Point a, Point b)

c = Rotate(1,0,I,a)

J = Line(1,c)

K = para(b,J)

e = IntersectLines(K,\Line(0,I))

d = Rotate(I,0,1,E)

return d

Figure 40. A variation of Descartes’s multiplication; here d = Multiply(a, b).

Lb

10

b

a

c

b

b

b

J

Kbe

d

b

Lemma 12.7. The strong parallel axiom implies Multiply (x, y) is defined forall x and y on Line (0, 1).

Remark. The script works also in the problematic cases a = b, a = 0, b =, andeven a = b = 0.

Proof. We work through the script line by line, checking that each point con-structed is provably defined. Suppose given points D and Q to multiply. Since


Rotate is everywhere defined, C in the first line of the script is defined. Since Cis on Ray (0, I), and Line (0, I) meets Line (0, 1) only in 0, C 6= 1, so L in line 2 ofthe script is defined. Since Para is everywhere defined, K in line 3 of the scriptis defined, and is parallel to Line (1, C) through D, if D 6= C, and coincides withLine (1, C) if D = C. Now we wish to prove that K meets Line (0, I).

Assume K fails to meet Line (0, I). Then Q is not zero, since if Q = 0 alsoC = 0 so C lies on K and Line (0, I). Also D is not 1, since if D = 1 thenthe two lines both pass through C. Then K is parallel to Line (1, C) throughD. If it does not meet Line (0, I) then by Lemma ??, Line (1, C) is parallel toLine (0, I), which is false since they both contain C. Hence the assumption thatK fails to meet Line (0, I) has led to a contradiction. That is, we have proved

¬¬ (IntersectLines (K,Line (0, I)) ↓)

Then by the strong parallel axiom, IntersectLines (K,Line (0, I)) ↓). Then E inline 4 of the script is defined, and then N , the script’s result, is defined in line 5,since the points I, 0, and 1 are distinct. That completes the proof of the lemma.

Next we exhibit a construction script for Hilbert’s multiplication method, de-scribed in Section 5.4. It uses the construction Reflect from Lemma 9.105, whichreflects a point e in line L, without a case distinction as to whether e is on Lor not. Note the use of LowerBound to ensure that we are extending a non-nullsegment in the next line.

HilbertMultiply(Point a, Point b)

A = LowerBound(a,b)

B = UpperBound(a,b)

m = Midpoint(A,B)

J = Perp(m,L)

x = Midpoint(I,a)

K = Perp(x,Line(I,a))

e = IntersectLines(J,K)

C = Circle(e,a)

M = Perp(e,Line(0,1))

Q = Perp(e,M)

w = Reflect(I,Q);

return w

Lemma 12.8. The strong parallel axiom implies that HilbertMultiply (a, b) isdefined for all points a and b on Line (0, 1), and returns the result of Hilbert’smultiplication construction; that is, the point w produced by the script lies on thecircle C and is different from I if there is another intersection point of C andLine (0, 1), and is equal to I if not.

Remark. This script, like Multiply , works in the degenerate cases, includinga = b = 0, thanks to the use of LowerBound and UpperBound , and the uniformperpendicular Perp. It is instructive to trace through what happens when a =b = 0.

146 MICHAEL BEESON

Proof. By the properties of LowerBound and UpperBound , we have B(A, a,B)and B(A, b,B), so A 6= B. Hence m is defined, and is the midpoint of AB, andif a 6= b m is the midpoint of ab, and if a = b then m = a. Since Perp is alwaysdefined, J is defined; since I is not on L = Line (0, 1) but a is, x is defined; sincePerp is always defined, K is defined. Then since I does not lie on Line (a, b),by the strong triangle circumscription principle (Lemma 10.20), there is a circleC′ through I, a, and b. Let E be the center of that circle. Then aI is a chordof C′, so its perpendicular bisector K passes through E. If a 6= b then J is theperpendicular bisector of chord ab, so J passes through E. If a = b then L istangent to C′ at x, by the strong triangle circumscription principle. Then J alsopasses through E. But ¬¬(a = b ∨ a 6= b); hence ¬¬on(E, J). By Axiom S3,on(E, J). Hence E lies on J and on K. Then the e defined in the constructionscript is defined and equal to E, the center C′. Hence C′ coincides with C, sincethey have the same center e and both pass through a. Since Perp is alwaysdefined, M and Q are defined; and since Reflect is always defined, so is w. Noww lies on circle C since, if I is not on C and z = IntersectLines (Q,Line (0, I)),then triangle Iez is congruent to triangle wez; and if I is on C then w = I is onC; since ¬¬ (On(I, C)∨¬On(I, C)), we have ¬¬On(w,C), and hence by AxiomS5 we have On(w,C). That completes the proof.

Lemma 12.9. The strong parallel axiom implies that Multiply satisfies thecommutative and associative laws and the law Multiply (1, x) = Multiply (x, 1) =x, and

HilbertMultiply (a, b) = Multiply (a, b).

Proof. The verification of the commutative and associative laws for Multiplyis carried out, for positive arguments, in Hilbert [15], pp. 46–55. (Hilbert ismultiplying segments, and he starts by laying them off to the right of 0 onLine (0, 1), so only multiplication of positive arguments is defined.) Each casewhere the arguments have known signs can be argued similarly; for example ifa < 0 and b < 0, then the picture in Fig. 6 is reflected in the center point e; ifonly one argument is negative, then the picture is reflected in one of the axes,and Hilbert’s arguments for the equivalence and correctness of the two definitionscarry over easily. There are also the cases in which one or more arguments arezero. One can check that Multiply (0, x) and Multiply (x, 0) are 0, and the samefor HilbertMultiply . Hence our conclusion holds in those cases also.

But the double negation of the disjunction of those cases holds, since theycannot all fail. Pushing the double negation inwards, and using the stability ofequality, the desired equations hold without requiring a case distinction.

It is true that Hilbert’s axiom system is different than that of ECG, whichis more closely related to Tarski’s axioms. But Chapter 12 of [26] is primarilydevoted to a very rigorous proof of the theorem of Pappus-Pascal (which Hilbertcalls Pascal’s theorem), which is the key to the correctness of Multiply. Wetherefore do not repeat either proof here. That completes the proof of the lemma.

We now check the validity of Descartes’s construction of square roots. LetSquareRoot (x) be the term defined by the construction script in Section 5.5.Then we have


Lemma 12.10. Playfair’s parallel axiom implies that iff x lies on Ray (0, 1),and z = SquareRoot (x), then

HilbertMultiply (z, z) = x.

Remark. We only need Playfair’s axiom, not the strong parallel axiom. The as-sumption implies SquareRoot (x) is defined, and the fact that HilbertMultiply (z, z)is defined is part of the conclusion.

Proof. We first give a “proof by diagram.” In the case of squaring, Hilbertmultiplication has the circle tangent to the horizontal axis, since a and b coin-cide, as illustrated in Fig. 41. Replacing a by

√x and reflecting the figure in

a diagonal line from upper left to lower right, we obtain the first diagram inFig. 42. Relabeling that diagram, and omitting the bottom half of the circle,we obtain Descartes’s method of calculating the square root of x, shown in thesecond diagram in Fig. 42.

We will now give a rigorous proof. Since what is to be proved is an equation,by the stability of equality we can argue by cases. The first case is when x = 0.Then SquareRoot (x) = 0 and HilbertMultiply(0, 0) = 0, disposing of that case.So now we can assume x 6= 0. Let b and d be constructed by Descartes’s methodfrom input b, where by assumption bp is congruent to 01, a segment whose lengthis taken as unity. The horizontal line at the bottom is a diameter of the circle,whose endpoints are p and s. Erect the perpendicular to this diameter at p andlet cp = bd. Then d 6= c since if d = c then b = p, by the uniqueness of droppedperpendiculars, but b 6= 0 since 0 6= 1 and bp 01. Hence L = Line (c, d) is defined;it meets the circle at d. Let r be the other intersection point of the circle withLine (c, d) (which is defined without a case distinction as to whether the circlehas two or only one intersections with the line). Now triangle dbp is congruent totriangle cpb by SAS, since db = cp and bp = pb and all right angles are congruent,so angle dbp is congruent to angle cpb. Then dp = bc as corresponding sides ofcongruent triangles. Now dcpb is a quadrilateral whose diagonals are equal. ByPlayfair’s axiom and Lemma 10.25, dcpb is a rectangle. Hence by Lemma 10.26,its opposite sides are equal. Hence dc = bp.

The segment that SquareRoot constructs (before the last rotation in theSquareRoot script) is db, which is labeled

√x in the figure. The actual result

of SquareRoot is a point z such that 0z = db. Since db = cp, the result ofHilbertMultiply (z, z) is a point w such that 0w = rc. What we must prove thenis that rc = sb. Point q is defined as the foot of the perpendicular from r tothe diameter. (Point t has not yet been defined.) Define m to be the midpointof rd and e to be the center of the circle. Then em ⊥ L since triangle rme iscongruent to triangle dme, by the definition of triangle congruence (SSS). ThenL ⊥ em by Lemma 9.38, so r is the reflection of d in Line (e,m). Define t =Reflect (c,Line (e,m)). Then quadrilateral trqs is the reflection of quadrilateralcdbp, and hence it is also a rectangle. Hence tr = dc = bp = sq = 01.

Now all the angles of quadrilateral rdbq are right angles. Therefore its oppositesides are equal, by Lemma 10.26. Hence rd = qb. Hence by Lemma 9.3, togetherwith dc = sq, which was proved above, we have rc = sb. That completes theproof of the lemma.

148 MICHAEL BEESON

Definition 12.11. Point HilbertSquare(Point x)

return Multiply(x,x)

Figure 41. Hilbert multiplication in the special case of squar-ing.

ba2

bI

ba

Figure 42. If z = SquareRoot (x), then HilbertMultiply (z, z) =x. The left circle is Fig. 41 reflected in the diagonal line.The right circle is a relabeling that is recognized as Descartes’ssquare root.

b

a

b

1b

a2

b

b

x = a2 1

√x


Figure 43. Rigorous proof that if z = SquareRoot (x), thenHilbertMultiply (z, z) = x.

b

e

bm

bd

b

b

b c

x = a2 1

√x

b

b

pq

rb

s

t

Figure 44. Proof of the distributive law. The shaded trianglesare congruent and the opposite sides of the parallelogram areequal.

b

b+ cb

cb

1b

b

ba(b+ c)

bac

babab

Lemma 12.12. on (x,Line (0, 1) → HilbertSquare (x) ≥ 0.

Proof. Let L = Line (0, 1). Suppose on (x, L). Let C be the circle tangent to Lat x, and passing through I. Then the other intersection point p of C and they-axis J = Perp(0, L) (which might be I itself) is on the same side of L as I,since if not then p 6= I and the y-axis meets C in a third point, contradictingLemma 9.52. By definition z = HilbertSquare (x) = Rotate (I, 0, 1, p). Then byLemma 9.101, z lies on Ray (0, 1). Hence z ≥ 0. That completes the proof.

12.3. The distributive law. I do not know how to prove the distributivelaw directly from the definitions of Add and HilbertMultiply . On the other hand,from Descartes’s definition of Multiply , it is relatively easy. Hilbert [15], p. 55,gives the diagram in Fig. 44.

150 MICHAEL BEESON

Lemma 12.13. Multiply (and hence HilbertMultiply ) satisfy the distributivelaw with Add .

Proof. See Fig. 44. The point on the y-axis labeled a is actually Rotate (1, 0, I, a).labeled ab is the point which, when rotated clockwise ninety degrees, will beMultiply (a, b). Similarly for ac. Assuming b and c and a are all positive, theaddition shown is that produced by Add . Then the shaded triangles are con-gruent, since they are right triangles each with horizontal leg congruent to 0b.Hence the vertical sides of the shaded triangles are congruent (to ab). Hencethe upper left side of the parallelogram is also congruent to ab. Hence the pointlabeled a(b + c) is also ac + ab. Since addition is commutative, that proves thedistributive law in case a, b, and c are all positive. If a < 0 we can use the laws(−a) · b = −(a · b) and (−a) + (−b) = −(a + b), which we have already provedfor HilbertMultiply and Add , to reduce the problem to the case a > 0. Thereare then three more cases to treat, when one or both of b and c is negative. Ifboth are negative, the diagram is just reflected in the y-axis and the argument isessentially unchanged. To treat the case when b is negative, just replace b+ c inthe diagram by c, and c by c−b, and change the labels on the y-axis accordingly.Then the same argument applies. And if any of the three points is zero, the lawsimplifies to an identity using x+0 = x and x·0 = 0. Now comes the point wherewe must consider constructive logic. Since classically these cases are exhaustive,constructively not not one of them holds; hence we have proved

¬¬ a(b + c) = a · b+ a · c

(where the normal notation abbreviates Multiply and Add ). But equality is sta-ble; hence we can drop the double negation. That completes the proof of thedistributive law for Multiply . But Hilbert proved that HilbertMultiply (a, b) =Multiply (a, b), so the distributive law also holds for HilbertMultiply . That com-pletes the proof of the lemma.

12.4. Reciprocals. According to the definition of HilbertMultiply , a recip-rocal of a nonzero point a on L = Line (0, 1) is a point b on L such that thereis a circle C through a, b, and I, and C is tangent to K = Line (0, I) at I. Thereciprocal of x can be constructed by the following script. It uses the scriptOther for the other intersection point of a line and circle, which was definedand proved correct in Theorem 9.108. The script is illustrated in Fig. 45. Thedashed line N is used in the proof, but not in the construction.

Point Reciprocal(Point a)

K = Perp(L,0);

H = Perp(I,K);

M = Line(a,I);

m = Midpoint(a,I);

J = Perp(m,M);

e = IntersectLines(H,J);

C = Circle(e,a);

b = Other(a,L,C);


Figure 45. Construction of the reciprocal of a

L

K

be

bI

b

ab

b

H

M

bm

J

C

N

The following lemma proves the correctness of this script. The hard part is toshow that Reciprocal (x) is defined for x 6= 0.

Lemma 12.14 (Reciprocals). Let Reciprocal be defined as in the constructionscript above. The strong parallel postulate implies that for a 6= 0, Reciprocal (a)is defined, and if b = Reciprocal (a), then HilbertMultiply (a, b) = 1.

Proof. Assume a 6= 0. We will show Reciprocal (x) is defined, by going throughthe script line by line. Since Perp is everywhere defined, K and H are de-fined; since I does not lie on L, a 6= I, so M is defined. Since a 6= I, m isdefined; since Perp is everywhere defined, J is defined. Now we have to provethat lines H and J meet in some point e. That is where we use the strongparallel axiom. Let N = Perp(I,M). Then by Lemma 9.48, N and J are par-allel, since both are perpendicular to M . Both N and J pass through I. Mis not perpendicular to H at I, since if it were, then Ia and I0 would be twoperpendiculars to H at I, so they would coincide, and it follows that a = 0,contrary to hypothesis. Then let x be any point on H other than I, for example,x = IntersectLineCircle1 (H,Circle (I, 0, 1). Then x is not on N , since if it werethen H and N would coincide. Hence we can apply the strong parallel axiom toconclude that H meets J . Then e in the script is defined. Since Circle is alwaysdefined, C is defined. Since a lies on both L and C, b in the penultimate line isdefined. Hence Reciprocal (a) is defined.

By Lemma 9.54, C is tangent to K at I. Then by definition of HilbertMultiply ,we have HilbertMultiply (a, b) = 1, since circle C passes through a and b and meetsK in I, and the “fourth point of intersection” is the point I, since C is tangentto K at I. That completes the proof.

12.5. Interpreting ECG in field theory. For readers who are not expertin logic, we begin with a short discussion of reducing one theory T (for examplea theory of ordered fields) to another theory S (for example ECG or some other

152 MICHAEL BEESON

geometric theory). Classically there are two approaches: we try to characterizethe models of T in terms of models of S (for example, models of a geometry areplanes over a field); or more technically we try to interpret T in S, which meansto give a formal translation of sentences φ in the language of T into sentences

φ in the language of S, such that if T proves φ then S proves φ. The twoapproaches are connected by Godel’s completeness theorem, since if S does not

prove φ then there is a model F of S in which φ fails, and if T does not prove φthere is a model of T where φ fails, and usually a connection between the modelsis established by what amounts to an interpretation. When we use intuitionisticlogic, we do not have Godel’s completeness theorem, so when we speak of amodel-theoretic characterization of a geometric theory, this could be interpretedas (i) meaning just what it says, using a constructive logic at the meta-level, or(ii) as a shorthand for the claim that a formal interpretation exists. In the caseof geometry and field theory, to interpret field theory in geometry we have todefine addition and multiplication and order on the points of a line, by formulasin the language of geometry, and verify the field axioms. That corresponds toshowing how to construct a field within each model of the geometry.

Logicians write T ⊢ φ to mean that theory T proves formula φ. An interpre-

tation is sound if T ⊢ φ implies S ⊢ φ. It is faithful if S ⊢ φ implies T ⊢ φ.Here is a model-theoretic description of the results in this section. Assume F

is a Euclidean field. We will show how to turn F 2 into a model of ECG (or, todescribe the construction more formally, we will show how to interpret geometryin field theory). As usual in the corresponding classical theories, we take thepoints to be elements of F 2, and let lines, circles, arcs, and segments have theirusual analytic definitions; in particular we define circles so that circles of zeroradius are allowed. Hence Circle3 can be interpreted. Markov’s principle in Fallows us to verify that axiom of ECG. The intersection points of circles andlines, and the intersection points of circles and circles, can be defined by thesolution of quadratic equations. Before beginning the more technical part of theproof, we make a general remark about mathematics in EF. The axioms permitone to do ordinary analytic geometry in EF; since we can take square roots ofnonnegative elements, and sums of squares are nonnegative, we can define |x|;dot product and cross product can also be defined; vectors are simply treated aspairs (or, in principle, n-tuples) of field elements. Linear transformations can berepresented as two-by-two matrices, which in turn can be represented in EF bymentioning their entries. When one says “there exists a linear transformation. . .”one means “∃a, b, c, d . . .”. One can prove, for example, that given points (thatis, pairs of field elements) a, b with a 6= b, and points A,B with A 6= B, thereis a linear transformation that takes (a, b, c) to (A,B,C). Namely, translate byA − a, then rotate and dilate to bring b to B; of course for a formal proof onehas to write out the transformation explicitly in terms of the coordinates of thegiven points and verify by algebraic calculation that it works.

We turn now to the technical problem of making the definitions sketched aboveinto a formal interpretation. The idea of the interpretation is that each pointx should be represented by a pair of field elements (x1, x2). Circles will berepresented by their centers, and by one point on the circle, which requires fourfield elements; similarly, lines are represented by two points, which requires four


field elements. To manipulate these pairs and quadruples we make use of lists ofpoints, which we write as (x1, x2, . . . xn); usually n will be 2,4, or 6.

It will be necessary that one variable for a point, circle, or line correspondsin the interpretation to two, three, or four (respectively) variables for field ele-ments. Officially, there is a fixed list of variables p0, p1, . . . of type Point, a listℓ0, ℓ1, . . . of variables of type Line, and a list c0, c1, . . . of variables of type Circle,and in field theory EF there is similarly a fixed list of variables v0, v1, . . . , vn.(Technically when logicians write “x is a variable of type Point”, they mean thatx is one of the pn; thus “x” is a metavariable ranging over the true variables.)We associate to each variable of ECG, two, four, or three variables of EF in aunique way. Specifically, pn is associated to v9n, v9n+1, and ℓn is associated tov9n+k for k = 2, 3, 4, 5, and cn is associated to v9n+k for k = 6, 7, 8. If x is avariable of type Point, we denote the variables of EF associated to it by x1 andx2; if x has type Line we use x1, x2, x3, and x4, and if x has type Circle, we usex1, x2, and x3. Here the subscript notation is used in a special, technical way.Usually x1 would just be another metavariable ranging over official variables.Now, the subscript is used to associate a variable of EF to a variable of ECG.This will mean that, for the duration of this proof, we will abstain from usingsubscripts in the usual way, to avoid confusion. For example, if the variable xhas type Point, and is really u5, then x1 is v50 and x2 is v51. As a typographicalconvenience, if ℓ is a list of terms, we will write ℓn to mean the n-th member ofthe list. For example, when x is a variable of type Point, we will define belowx = (x1, x2); then x1 = x1, so there is no ambiguity with the use of subscriptson variables.

When working in or with the theory EF or related theories, we use somecommon abbreviations: xy means x · y for field elements. If x and y are lists ofthe same length, x·y means the dot product defined using the multiplication signof EF. x2 means x ·x. If t is a list of two terms (a 2-vector) then |t| =

√

t21 + t22.

Hence, if t is a term of type Point, |t| means√

(t1)2 + (t2)

2. If t is a term of

EF+ then |t| means√t2, which means

√t · t. If t and s are two terms of type

Point, t + s is the list (t1 + s1, t2 + s2); binary minus x − y means x + (−y);t− s means (t1 − s1, t2− s2). Cross product x× y is definable in EF on lists oflength 2 as x1y2 − x2y1 (so the result is a scalar). If t is a list (t1, . . . , tn), thent ↓ means the conjunction of the tk ↓ for k = 1, . . . , n.

We want to interpret ECG in the field theory EF defined in Definition 8.2,and vice-versa; but there is a technical detail concerning the treatment of theconstants α, β, and γ that must be dealt with without delay. In geometrythese constants stand for “unspecified” noncollinear points. When interpretinggeometry in field theory, we intend to interpret α as (0, 0) and β as (1, 0). It mayseem natural to pick a third specific point, such as (0, 1), for the interpretation ofγ. The problem with this is that it introduces “extra” or “accidental” theorems,since any specific point definable in EF will be definable in geometry too. If wewant our interpretations to be faithful (the interpretation of φ is provable if andonly if φ is provable), then we cannot pick a specific point as the interpretationof γ. The way around this difficulty is to augment EF with two new constants.At the same time we will solve another problem, namely that ECG does not

154 MICHAEL BEESON

specify which point on a circle is given by pointOnCircle , while we must pick aparticular point in the interpretation in EF. We can pick one arbitrarily, such asalways picking the “northernmost” point, and the interpretation will sound, butthen it will not be faithful, as when we go from ECG to EF and then back, wemay not get back the point we started with. We fix this by introducing anotherarbitrary constant to serve as the x-coordinate of pointOnCircle (Circle (0, 1)).

Definition 12.15. The theory EF+ is EF with three additional constants γ1,γ2, and γ3, and the axioms 0 6= γ2 and γ2

3 ≤ 1. “Weak EF+” is EF+ with AxiomEF1 replaced by Axiom EF7, positive elements have reciprocals. “Playfair EF+”is EF+ with Axiom EF1 replaced by Axioms EF9 and EF10.

Using EF+ instead of EF does not introduce any new theorems:

Lemma 12.16. EF+ is conservative over EF. That is, every theorem of EF+

that does not contain the constants γ1 and γ2 is provable in EF.

Proof. Let φ be any formula and let π be a formal proof of φ. Replace γ1 andγ2 throughout π by 0 and 1, and γ3 by 0. The axiom γ2 6= 0 becomes 0 6= 1,which is Axiom EF0. The axiom γ2

3 ≤ 1 is an abbreviation for ¬P (γ3 · γ3 − 1);this becomes ¬P (0 · 0 − 1), which is provably equivalent to ¬P (−1), which isprovable in EF. Hence the result of the substitution is still a formal proof. If φdoes not contain γ1 and γ2, then the last line of the proof after substitution isstill φ. That completes the proof.

In the interpretation we are about to define, points will be represented as pairsof field elements; we treat these pairs as lists of length two, (a, b). Lines will berepresented as lists of four field elements, two for each of two points on the line.We write these lists as (a, b, c, d). Circles similarly will be represented as lists ofthree field elements, two for the coordinates of the center, and one for the radius.

Definition 12.17. The definition of t for terms t of ECG will now be given.

x is (x1, x2) for x a variable of type Point

L is (L1, L2, L3, L4) for L a variable of type Line

C is (C1, C2, |C3|) for C a variable of type Circle

α is (0, 0) a list with two elements

β is (1, 0)

γ is (γ0, γ1)

pointOn1 (L) is (L1, L

2)

pointOn2 (L) is (L3, L

4)

Line (t, s) is Append(t, s)

In the last line, Append is list concatenation, and multiplication of a list bya field element means to multiply each member of the list by the element. Thepoint of using Append and Nonzero is to ensure that Line (a, b) is undefinedwhen a = b, or more generally when a − b is not invertible, when working withweaker reciprocal axioms than EF1. With these definitions, the direction of aline is recoverable from the order of the two associated points.


A circle is represented by its center and radius:

Circle (a, b, c) is (a1, a2, |b − c|)

center (C) is (C1 , C

2 )

The interpretation of pointOnCircle (C) is a point on C such that the radiusvector from the center to that point makes an angle with the horizontal whosecosine is γ3 (the third new constant introduced in EF+). This is formally ex-pressed by

pointOnCircle (C) is (C1 + γ3 · |C

3 |, C2 +

√

1 − γ23 · |C

3 |)

Since EF+ includes the axiom γ23 ≤ 1, we have pointOnCircle (C) ↓ provable in

EF+.Next we want to define the interpretation of IntersectLines (L,K). The idea

is that from L and K we can get two point on each line, then compute thecoordinates of the point of intersection, and the formulas giving those two coor-dinates are the members of the list IntersectLines (L,K). Note that for this towork, there must be a single formula, i.e. no definitions by cases are allowed, e.g.according as the lines are vertical or not. This does work out, as the formula forthe intersection turns out to have a denominator that is non zero exactly whenthe two lines are neither parallel nor coincident. To work this out, suppose thata = (L

1, L2), b = (L

3, L4), c = (K

1 ,K2 ), and d = (K

3 ,K4 ). Then solve for the

intersection point x of the line through a and b with the line through b and c.One way to do that is to note that on (x,Line (a, b)) means (x−a)× (x− b) = 0,so we have that equation and (x− c)× (x− d) = 0. Writing those equations outby components and using Cramer’s rule to solve them we find

D =

∣

∣

∣

∣

a2 − b2 b1 − a1

c2 − d2 a1 − c1

∣

∣

∣

∣

x1 =1

D

∣

∣

∣

∣

a2b1 − a1b2 b1 − a1

c2d1 − c1d2 d1 − c1

∣

∣

∣

∣

x2 =1

D

∣

∣

∣

∣

a2 − b2 a2b1 − a1b2c2 − d2 c2d1 − c1d2

∣

∣

∣

∣

The denominator D of the solution is zero just when its rows are linearly depen-dent; but the rows are the vectors determining the lines K and L respectively, sothe determinant is nonzero just when L and K are lines in different directions;hence, the terms of EF that gives the intersection points are defined exactlywhen lines L and K intersect. Hence, using the axiom EF1, lines that are nei-ther parallel nor coincident will meet, verifying the strong parallel axiom; if infield theory we only know positive elements have reciprocals then we must beable to determine the sign of D; and in Playfair field theory, we must show thatD is invertible before two lines can be proved to intersect.

Some of the algebra needed to define IntersectLineCircle1 (L,C) andIntersectLineCircle2 correctly has been done in Section 6.2, but we still needto pay attention to whether one can get a single term of EF that interpretsIntersectLineCircle1 (L,C).

156 MICHAEL BEESON

To define IntersectLineCircle1 (L,C) and IntersectLineCircle2 (L,C), let a =(L

1, L2) and b = (L

3, L4), and let e = (C

1 , C2 ) be the center of C and r = C

3

be the radius of C. Then the equations for the intersection points x are

(x− a) × (x− b) = 0 (where × is cross product)

(x− e)2 = r2 (squaring a vector means dot product)

We need to prove that these equations can be solved, and the solutions given bytwo terms defined without a case distinction.

We look for the intersection points x in the form x = a + λ(b − a) (so theywill automatically lie on Line (a, b), and then the second equation becomes aquadratic equation for λ:

(λ(b − a) + a− e)2 = r2

λ2(b − a)2 + 2(a− e) · (b− a)λ+ (a− e)2 − r2 = 0

This equation can be solved by the quadratic formula:

λ = − (a− e) · (b− a) ±√

((a− e) · (b − a))2 − (b− a)2((a− e)2 − r2)

(b − a)2(1)

and the denominator is zero just when b = a; hence just when the interpretationof Line (a, b) is undefined. We define the interpretation ofIntersectLineCircle2 (L,C) to be the term given by taking the positive sign forthe square root and IntersectLineCircle2 (L,C) to be the term given by takingthe negative sign. Here we understand that in the above expressions for λ, wewill also make the following substitutions:

e := (C1 , C

2 )

c := (C3 , C

4 )

a := (L1, L

2)

b := (L3, L

4)

It remains to check that if p and q are the two points that interpretIntersectLineCircle1 (L,C) and IntersectLineCircle2 (L,C), respectively, thenthey occur on L in the same order as (a, b). We will come to that point after wehave finished defining the interpretation.

Our next task is to define the interpretations of IntersectCircles1 andIntersectCircles2 . Much of the required work has already been done in Sec-tion 6.2, where we saw that the equation for the intersection points is qua-dratic, so it can be solved using the quadratic formula (which is both express-ible and provable in EF); that gives us two terms for the two intersectionpoints, according as the plus or the minus sign is used in the quadratic formula.But do these two terms correspond exactly to IntersectLineCircle1 (C,K) andIntersectLineCircle2 (C,K)? And which solution of the equations serves as theinterpretation of which term? The latter question is easy to answer for the circlesC and K considered in Section 6.2, which have their centers at (0, 0) and (1, 0)respectively. In that case IntersectCircles1 (C,K) is the one with y-coordinate≤ 0, i.e. where we take the negative square root to solve the last equation inthat section, and IntersectCircles2 (C,K) is the other solution, where we take thepositive square root. To correctly define the interpretations of IntersectCircles1


and IntersectCircles2 , we cannot assume the circles have their centers at (0, 0)and (1, 0). Let λ = |b − a|, where a and b are the centers of C and K. Letr = (1/λ) times the radius of C, and s = (1/λ) times the radius of b. Let e be aunit vector (b − a)/|b − a|, which is defined since a 6= b. Let f be a unit vectororthogonal to e such that the cross product e× f = 1. Then solve the equationsin Section 6.2 as is done there; if (x, y) solves those equations then a+λxe+λyfis a point on both circles. The (term defining the) solution with y ≥ 0 is theinterpretation of IntersectCircles2 (C,K), and the (term defining the) solutionwith y ≤ 0 is the interpretation of IntersectCircles2 (C,K).

Since we have proved that IntersectCirclesSame and IntersectCirclesOpp areredundant, we do not need to interpret them; or putting the matter another way,their interpretations can be defined using their definitions in terms of the rest ofECG.

Now we have completed the definition of a term t of EF for every term tof ECG. Next we define a formula φ of EF for each formula φ of ECG. Ifa and b are terms of type Point, then a = b means a0 = b0 ∧ a1 = b1, anda 6= b means ¬(a = b). Similarly, if t is a list of terms of EF, then t ↓ is theconjunction of the tk ↓ for tk in the list t. The following predicate will be usedin defining the interpretation of lines.

Definition 12.18.

L(x) := ¬ (x1 = x3 ∧ x2 = x4)

For terms t of type Point or Circle, we define

(t ↓) is t ↓and for terms L of type Line, we define

(t ↓) is t ↓ ∧L(t)

(a = b) is a = b

B(a, b, c) is a 6= b ∧ b 6= c ∧ |c− a| = |b− a| + |c− b|E(a, b, c, d) is (a− b)2 = (c− d)2

on(p, L) is (p − (L1, L

2)) × ((L

3, L4) − (L

1, L2)) = 0

On(p, C) is (p − (C1 , C

2 ))2 = ((C

3 , C4 ) − (C

1 , C2 ))2

We define for x a variable of type Point or Circle,

(∀xA) is ∀xA

(∃xA) is ∃xA

where when we quantify over a list of variables, as in ∃x, we mean to quan-tify over each of the variables separately. The interpretation of quantificationover variables of type Line has a twist. We need to ensure that Line (a, a) isundefined (as that is Axiom CA4); but as it stands, there is nothing to prevent(a0, a1, a0, a1) from being L for a line L. We could try to rule that out by defin-ing Nonzero (x) = |x|/|x|, which in EF would be defined and equal to 1 whenx 6= 0, and then just multiply each element in the list defining Line (a, b) by

158 MICHAEL BEESON

Nonzero (a− b). That works in EF but it causes trouble when we work in fieldtheories with weaker reciprocal axioms, as Nonzero (x) is only defined when x isinvertible, which may not be the same as x 6= 0. The solution to this problemis to simply interpret variables of type Line to range over lists (a1, a2, b1, b2) inwhich we do not have a1 = b1 ∧ a2 = b2. To accomplish this we recall fromDefinition 12.18 that

L(x) := ¬ (x1 = x3 ∧ x2 = x4)

the condition just mentioned. Then we define for L a variable of type Line,

(∀LA) is ∀L(L(L) → A

(∃LA) is ∃L(L(L) ∧A)

We write φ[x := t] for the result of substituting t for x in φ. If x is a listof variables and t is a list of terms, we use the same notation for simultaneoussubstitution.

Lemma 12.19 (Substitution commutes with the interpretation). For each for-mula A of ECG, with free variables x = x1, . . . , xn, EF+proves A(x)[x : t]is equivalent to A[x := t].

Proof. By induction on the complexity of the formula A. We have definedthe interpretation so that the lemma is true for atomic formulas. For example,B(p, q, r) is B(x, y, z)[(x, y, z) := (p, q, r)] for terms p, q, and r, and similarlyfor the other atomic sentences. Since both the interpretation and substitutioncommute with the propositional connectives, the induction steps correspondingto propositional operations are immediate. The two quantifiers are treated thesame way; we write out the proof for ∀. In the following calculation, each lineis provably equivalent to the next in EF. Here x is a variable of type Point orCircle and x is a list of two or three variables as explained above.

(∀xA(x, y))[y := t]

∀x (A(x, y))[y := t] by definition of the interpretation

∀x (A(x, y)[y := t]) by the induction hypothesis

(∀xA(x, y))[y = t] by definition of substitution

If x is a variable of type Line, then we have to account for the use of L. Wehave

(∀xA(x, y))[y := t]

∀x (L(x) → (A(x, y))[y := t]) by definition of the interpretation

∀x (L(x)[y := t] → (A(x, y)[y := t]) since y is not in the list x

∀x (L(x) → (A(x, y)[y := t]) by definition of the interpretation

∀x (L(x) → (A(x, y)[y := t]) by the induction hypothesis

(∀xA(x, y))[y = t] by definition of substitution


The following lemma will be needed below, and it also may help the readerunderstand the purpose of L:


Lemma 12.20. Let L be a variable of type Line. Then ECG proves L(L).

Proof

L(L) ↔ ¬(L1 = L3 ∧ L2 = L4)

↔ ¬(L1 = L

3 ∧ L2 = L

3)

Now, for the k such that L is ℓk (the k-th variable of type Line), we haveL

j = v9k+j , for j = 1 to 4, by the definition of L. Then, by the definition ofv9k+j (see Definition 12.47), we have

L1 =X(pointOn1 (L))

L2 = Y (pointOn1 (L))

L3 =X(pointOn2 (L))

L4 = Y (pointOn2 (L))

Putting this result into the previous calculation we have

L(L) ↔ ¬(X(pointOn1 (L)) = X(pointOn2 (L))

Y (pointOn1 (L)) = Y (pointOn2 (L)))

But ECG proves y = z↔X(y) = X(z) ∧ Y (y) = Y (z). Hence

L(L) ↔ pointOn1 (L) 6= pointOn2 (L)

But pointOn1 (L) 6= pointOn2 (L) is provable from Axioms CA1 and CA4, as

pointed out just after the statement of axiom CA7 on page 47. Hence L(L) isprovable in ECG. That completes the proof of the lemma.

Theorem 12.21 (Interpretation of ECG in EF). (i) Suppose ECG proves φ.Then EF+proves φ. That is, ECG is interpreted by the theory of Euclideanfields (where nonzero elements have reciprocals).

(ii) If the strong parallel axiom is replaced by Euclid 5, then EF1 can be re-placed by axiom that positive elements have reciprocals (EF7). That is, Euclid 5is interpreted by the theory of weakly Euclidean fields.

(iii) If the strong parallel axiom is replaced by Playfair’s axiom, then the axiomthat nonzero elements have reciprocals (EF1) can be replaced by axioms EF9 andEF10 (elements without reciprocals are zero, and elements greater than a positiveinvertible are invertible). That is, Playfair geometry is interpreted by the theoryof Playfair fields.

Remark. The issues about multiplicative inverse come up only in verifying thatIntersectLines (L,K) is defined when it needs to be. There are only two axiomsrequiring the intersection points of lines to be defined; those are Pasch’s axiomand the parallel postulates. We need Axioms EF9 and EF10 to verify Pasch’saxiom, and EF9 is needed only for Pasch’s axiom.

Proof. By induction on the lengths of proofs in ECG. We first check the axiomsof the logic of partial terms. There are the axioms c ↓ for c = α, β, γ. Since α

is (0, 0), α ↓ becomes 0 ↓ ∧0 ↓, which is provable in EF by that same axiom of

160 MICHAEL BEESON

LPT. Now β is (1, 0), so β ↓ becomes 1 ↓ ∧0 ↓, which is provable. The axiomγ ↓ becomes γ1 ↓ ∧γ2 ↓, which are axioms of EF+ since EF+ uses the logic ofpartia terms. Similarly, the interpretation of the axiom x ↓ is x ↓; since x is alist of variables, this is a conjunction of formulas of the form xk ↓, each of whichis provable by this same axiom of LPT.

We turn to the strictness axioms φ(p) → p ↓ for atomic φ. Using abbrevia-tions introduced above, this notation includes the case where p is a list of severalvariables. Arguing in EF, suppose φ(p). By Lemma 12.19, that is equivalentto φ(x)[x := p]. So what has to be proved is φ(t) → t ↓. In other wordswe must show that A[x := t] → t ↓ whenever A is φ for some φ. That leadsus to ask, for what A we have A[x := t] → t ↓. That property holds for atomicA that contain x, and holds for conjunctions if it holds for at least one conjunctthat actually contains x, as one easily checks. (Of course it does not hold ingeneral, as the example A(x) := ¬x ↓ shows, since we certainly do not have¬t ↓ → t ↓.) One can check by referring to the definitions of B, E, on , andOn above that in each case, the interpretation of an atomic formula is either anequation containing all the variables, or a conjunction, one of whose conjuncts isan equation containing all the variables. That completes the verification of thestrictness axioms

We consider the axiom ∀x(t ↓ ∧ A(x) → A(t); more formally we should writeA[x := t] instead of A(t). Consider the case when x is a variable of type Point.The interpretation of the formula is

∀x(t1 ↓ ∧ t2 ↓ ∧ A(x) → A(t)).

Argue in EF as follows: suppose t1 ↓ ∧ t2 ↓ ∧ A(x). Then by this same axiom,we have (A(x))[x := t]. Then by Lemma 12.19 we have A[x := t], which isthe desired conclusion.

Now consider the axiom t ↓ ∧ A[x := t] → ∃xA. The interpretation of this ist ↓ ∧ A[x := t] → ∃xA. Argue in EF as follows: Suppose t ↓ ∧ A[x := t].By Lemma 12.19 we have A[x := t]. Since t ↓, we can use this same axiom toconclude ∃xA, which is the desired conclusion. That completes the verificationthat the logical axioms of LPT are correctly interpreted.

We now turn to the nonlogical axioms of ECG. First we take up the stabilityaxioms S0 to S5. Each of these has the form ¬¬φ → φ, where φ is atomic,with main symbol =, E, on , On , or B. The interpretation preserves logicaloperations, so we have to check that the interpretations of these are stable in EF.For t = s, E(a, b, c, d), on (p, L) and On (p, C), the interpretations are polynomialequations in t, s, a, etc. The interpretation of B(a, b, c) is a conjunction ofpolynomial equations and inequations. All these are stable. That completes theverification of the stability axioms.

Next we turn to the incidence axioms. Consider Axiom I1, on (a,Line (a, b)).The interpretation of the term Line (a, b) is the list (a1, a

2, b

1, b

2). So the inter-

pretation of on (a,Line (a, b)) is

(a − (a1, a2)) × ((b1, b

2) − (a1, a

2)) = 0

Since a = (a1, a2), the first factor in the cross product is zero; since polynomial

identities can be verified in EF, this is provable in EF. Axiom I2, which is


on (b,Line (a, b)) is treated similarly–this time the second factor in the crossproduct is zero.

Axiom I3 is on (p, L)∧ on (q, L)∧on (r,Line (p, q)) → on (r, L). For notationalconvenience let a = (L

1, L2) and b = L

3, L4, so that a and b are two distinct

points on L (since L(L)). Then, assuming the of the hypothesis of AxiomI3, we have p 6= q (since Line (p, q) ↓), and

0 = (q − a) × (q − b) since on (q, L)

0 = (p − a) × (p − b) since on (p, L)

0 = (r − p) × (r − q) since on (r,Line (p, q))

and we must prove (r − a) × (r − b) = 0. The above equations are invariantunder translation, and cross product is invariant under rotations about the origin.Therefore we can assume without loss of generality that a = (0, 0) and b = (b1, 0).Then we have (dropping the superscript for simplicity)

0 = (q − a) × (q − b)

= q × (q − b)

= q1q2 − q2(q1 − b1)

= q2b1

But b1 6= 0 since a 6= b. Hence q2 = 0, i.e q lies on the x-axis. Similarly fromthe second equation in the hypothesis we obtain p2 = 0. Now we have from thethird equation,

0 = (r − p) × (r − q)

= (r1 − p1)r2 − r2(r1 − q1)

= (q1 − p1)r2

But q1 6= p1 since p 6= q and both p and q lie on the x-axis. Hence r2 = 0; thatis r also lies on the x-axis. Now we calculate

(r − a) × (r − b) = r × (r − b)

= r1(r2 − b2) − r2(r1 − b1)

= 0 since r2 = 0, b2 = 0, and r2 = 0

That completes the verification of Axiom I3.Axiom I4 is p = IntersectLines (L,K) → on (p, L) ∧ on (p,K). We have

sketched the proof that it is correctly interpreted when we explained how tocalculate IntersectLines (L,K); the complete calculation will not be given here.

Axiom I5 is IntersectLines (L,K) ∼= IntersectLines (K,L). The interpretationsof the two sides are the solution of the same two equations, given in the oppositeorder; so they agree.

Axiom I6 says that if p = IntersectLineCircle1 (L,C), then p is on both L andC; and Axiom I7 says the same for IntersectLineCircle2 . Since the interpre-tations of IntersectLineCircle1 and IntersectLineCircle2 are defined by solvingthe the equations for L and C, verifying this axiom amounts to proving thequadratic formula in EF, which is straightforward.

Axiom I8 and Axiom I9 similarly treat the intersection points of two circles;again the interpretations of the terms give solutions of equations that say the

162 MICHAEL BEESON

points are on the circles, so by the quadratic formula in EF, the interpretationsof these axioms can be verified.

Axiom I10 and Axiom I11 have been shown to be redundant in Theorem 9.96.Therefore we do not have to verify them.

Axiom I12 says that if L and K do not coincide (because p is on both ofthem) then their intersection point is defined. As remarked in the definition ofIntersectLines (L,K), the term giving the solution has a denominator that isnonzero just when L and K do not coincide and are not parallel. Since termswith zero denominator are undefined in EF, since if a/0 ↓, then 0 · a/0 would beboth 1 and 0, but 1 6= 0 by Axiom EF0.

Axiom I13 says that if IntersectLines (L,K) ↓, then the only intersection pointof L and K is the one given by that term. As just remarked, if the interpretationof that term is defined, then the lines (specified by) L and K do not coincideand are not parallel, so there is only that solution.

Axiom I14 through I17 say that if lines and circles intersect, or circles andcircles intersect, then the terms giving the intersection points are defined. Underthe interpretation, that means that the equations for the intersection pointshave solutions given by the terms we have defined to be the interpretations ofIntersectLineCircle1 (L,C), etc. Again, this is just verifying in EF that thesolution terms do indeed solve the equations.

Axiom I17 and Axiom I18 specify that if the terms for the intersection points ofcircles are defined, then the circles are not concentric. In Theorem 6.1 we deriveda formula for the intersection points, but we began by assuming the circleswere not concentric and making a linear transformation to bring the centersto (0, 0) and (1, 0). The formula for that transformation, which will be part ofthe interpretation of IntersectCircles1 (C,K) and IntersectCircles2 (C,K), willhave a denominator that is zero when the circles are concentric. That will resultin IntersectCircles1 (C,K) and IntersectCircles2 (C,K) being undefined when(C

1 , C2 ) = (K

1 ,K2 ). That completes the verification of Axioms I17 and I18.

Axioms I19 and I20 concern IntersectCirclesSame , which has been shown re-dundant; so we do not have to verify these axioms.

Axiom I21 says that if IntersectLines (L,K) ↓, then K and L do not coincide.We have already remarked that the denominator will be zero if K and L

coincide, which verifies Axiom I21.We now turn to the constructor and accessor axioms. The verifications of these

are just a matter of checking that the bookkeeping is correct; lines and circles arerepresented by lists of four field elements in a consistent way. Consider AxiomCA1,

Line (pointOn1 (L), pointOn2 (L)) = L.

Here L is a list of four elements (L1, L

2, L

3, L

4), of which the first two represent

one point on L, and the second two represent another point. The interpreta-tion of pointOn1 (L) will be pointOn1 (L) = (L

1, L2), and the interpretation

of pointOn2 (L) will be pointOn2 (L) = (L3, L

4). Then the interpretation of

the left hand side Line (pointOn1 (L), pointOn2 (L)) concatenates those two lists,producing L. That verifies Axiom CA1.


Axiom CA2 says that if a 6= b then pointOn1 (Line (a, b)) = a. If a = (a1, a2)

and b = (b1, b2) then Line (a, b) = (a1, a

2, b

1, b

2) and

pointOn1 (Line (a, b)) = (a1, a2) = a.

The hypothesis a 6= b tells us that L(Line (a, b)). Axiom CA3 is verified simi-larly, with pointOn2 .

Axiom CA4 says that Line (a, a) is not defined. According to the definition of(t↓), the interpretation of (Line (a, a) ↓) is (equivalent to)

L(Line (a, a)) ∧ a ↓ .But Line (a, a) is the list (a1, a

2, a

1, a

2), which does not satisfy L (see Definition

12.18). That completes the verification of Axiom CA4.Axiom CA5 is center (Circle3 (a, b, c)) = a. We have

Circle3 (a, b, c) = (a1, a2, d, e)

where the terms d and e do not matter here. Then

center (Circle3 (a, b, c)) = (a1, a2) = a.

That completes the verification of Axiom CA5.Axiom CA6 is Circle (center (C), pointOnCircle (C)) = C. By definition,

C = (C1 , C

2 ,

√

(C3 )2)

pointOnCircle (C) = (C1 + γ3 · |C

3 |, C2 +

√

1 − γ23 · |C

3 |)center (C) = (C

1 , C2 )

Circle (center (C), pointOnCircle (C)) is an abbreviation for

Circle3 (center (C), center (C), pointOnCircle (C)).

So its interpretation is (C1 , C

2 , d), where

d =

√

(C1 + γ3 · |C

3 | − C1 )2 + (C

2 +√

1 − γ23 |C

3 | − C2 )2

=

√

γ23 |C

3 |2 + (√

1 − γ23 |C

3 |)2

= |C3 |

=√

(C3 )2 officially, since |x| is just an abbreviation

Hence Circle (center (C), pointOnCircle (C)) = C as desired. That completesthe verification of Axiom CA6.

Consider Axiom CA7, which says Circle3 (A,B,C) ↓. Note that the term forCircle3 includes a square root (to calculate the radius); but the argument ofthe square root is a sum of squares, so it can be proved non-negative in EF. Todeal with circles of zero radius, we do need to prove closure of the non-negativeelements (those whose additive inverses are not positive) under sum and product,but that is easy.

Now we come to the betweenness axioms. We have defined the interpretationof B(a, b, c) to be an algebraic formula expressing that the distance from a to c

is the sum of the distances from a to b and from b to c, and that a 6= b and

164 MICHAEL BEESON

b 6= c. (Without these last two conditions we would get non-strict between-ness.) These two conditions make Axioms B1-ii and B1-iii automatically valid,and Axiom B1-iv (symmetry of B) reduces to the commutativity of addition.

Axiom B1-i requires us to check that this condition on the distances can only besatisfied when b is on the line connecting a and c. We can check that algebraicallydirectly, or we can first derive the triangle inequality in EF and use that. Thedetails are of approximately equal length, so we do the direct verification; forsimplicity we drop the superscripts, writing a instead of a. We will prove thatfor a, b, and c distinct pairs of field elements (thought of as 2-vectors),

|a− c| ≤ |a− b| + |b− c|with equality only when b is on the line connecting a and c. It suffices by a linearchange of variables to prove it when a = 0. Then it becomes |c| ≤ |b| + |b − c|.We have the following reversible calculation (in which b2 means the dot productof vector b with itself, etc.):

|c| ≤ |b| + |b− c|c2 ≤ b2 + 2b · (b− c) + |b− c|2

= b2 + 2b2 − 2b · c+ c2 + b2 − 2b · c+ c2

= c2 + 4b2 − 4b · c+ c2

= c2 + (2b− c)2

0 ≤ (2b− c)2

This calculation can be checked in EF from the bottom up. That verifies AxiomB1-i.

Axiom B2 says that the point calculated by the Midpoint (a, b) script, wheninterpreted, lies between a and b. Let p be that midpoint; then to verify AxiomB1 we need to check that

|a − p| + |b − p| = |a − b|.That follows if p is indeed the midpoint of segment ab. To verify this wemake the calculation one would make in ordinary analytic geometry, and checkthat we used only the axioms of EF to do it. In particular, the square roots oneneeds to solve for the intersection points exist, because what is under them isnonnegative. We omit the details.

We turn to Axiom B3, which says that if a, b, and c are distinct points ona line, and two of the three possible betweenness relations among them fail,then the third one cannot fail, i.e. it not not holds. Since equality is stablein EF we might as well say that the third one holds. Suppose then (droppingthe superscripts for convenience) we have ¬B(a, b, c) and ¬B(b, c, a). Then wemust show that B(c, a, b). That means that we have

|c− a| 6= |b− a| + |c− b|

|b− a| 6= |c− b| + |a− c|and we must show

|c− b| = |c− a| + |a− b|


Since one can prove in EF that linear transformations preserve distance, we canmake a linear transformation that brings the line to the x-axis, and point a tozero. Then we can assume that b, and c are scalars rather than vectors. Sinceequality is stable we can argue by cases on the possible order relations of b andc. Our hypotheses are now |c| 6= |b|+ |c−b| and |b| 6= |c−b|+ |c|, and our desiredconclusion is |c− b| = |c| + |b|. That will be true if b and c have opposite signs,i.e. not both are nonnegative and not both are nonpositive. But the hypothesestell us that b and c both do not have opposite signs to the same expression; hencethey do have opposite signs to each other. The other clauses in Axiom B3 aretreated similarly. That completes the verification of Axiom B3.

Before going on to the next axiom, we note that the interpretation of B(a, b, c),which is |a − c| = |a − b| + |b − c|, implies in EF that a, b, and c are collinear,in the usual algebraic sense of satisfying a linear equation, which can be writtenwithout a case distinction (as to to whether the line is vertical or not) by sayingthat the cross product of (a− b) × (b− c) is zero. If we know that the directionof the line is not vertical then the usual form of a linear equation applies. Thenone can prove in EF that the interpretation of B(a, b, c) is equivalent to theconjunction of the two coordinate inequalties (for i = 1, 2)

¬¬((ai < bi ∧ bi < ci ) ∨ (ci < bi ∧ bi < ai )).

Axiom B4 is Explicit Inner Pasch. The hypothesis is

B(a, p, c) ∧ B(b, q, c) ∧ p 6= b ∧ q 6= a ∧ ¬ on (c, Line(a, q)

Then set

z := IntersectLines (Line (p, b),Line (a, q)).(2)

We must show that z is defined, i.e. that using the axioms of EF (or therestricted axioms for reciprocals) we can calculate the coordinates of z. Wemay simplify the algebra by first applying a linear transformation to bring thepoints to some standard position, since linear transformations preserve B (eventhough they may not preserve distance). By a translation, we can bring a to(0, 0); then by the transformation with the matrix

(

p1 p2

p2 −p1

)

we can bring p to the point (p21 + p2

2, 0) on the positive x-axis. Note that nodivision was required so far. Then c lies on the x-axis.

We continue to use only the axioms of Playfair field theory, i.e., we cannot justdivide by any nonzero element, but only by one known to be invertible. Sincep is the intersection point of Line (a, c) and Line (b, p), and these lines do notcoincide since b does not lie on Line (a, c), the cross product (b− p) × (a− c) isinvertible. Since p2 = 0 and a = (0, 0) and c2 = 0 we have (b−p)×(a−c) = b2c1,so b2c1 is invertible. Generally if xy is invertible then x and y are invertible,since if (xy)z = 1 then x(yz) = 1 and y(xz) = 1. Hence b2 is invertible. Thenthe linear transformation with matrix

(

b2 −b1 + p1/b20 1

)

166 MICHAEL BEESON

is defined; it fixes the x-axis and brings the point b directly over point p, asillustrated in Fig. 46.

Figure 46. Verification of Axiom B4 (Explicit Inner Pasch)

b

a = (0, 0)b

pb

c

bb

b q

z

Now recall the formula for for

z = IntersectLines (Line (p, b),Line (a, q)).

Its denominator in this case is

D = (p− b) × (a− q)

= b2 · q1 since p2 = a2 = 0 and b1 = p1

The desired intersection point z will be defined if and only D is invertible. Sinceb2 is invertible and D = b2 · q1, it suffices to show that q1 is invertible.

Note that (b− p) × (a− p) is invertible, since it is the denominator of

IntersectLines (Line (b, p),Line (a, p))

and this is term is equal to p, so it is defined. But (b− p)× (a− p) = −b2p1, andb2 is invertible. Therefore p1 is a quotient of invertibles, hence it is invertible.But q1 > b1 since q is between b and c and c1 > p1 = b1. Hence q1 > p1 = |p1|.By Axiom EF10, anything greater than a positive invertible is invertible, so q1is invertible. That completes the proof that z is defined.

It remains to verify the conclusion of the axiom, namely B(p, z, b)∧B(q, z, a).Since a = (0, 0) the collinearity of a, z, and q is expressed by z × q = 0.That is, z2q

1 − z1q

2 = 0. But z1 = 1, so q2 = z2q

1 . Then (dropping the super-

script for readability), we have by the above remarks that the interpretation ofB(q, z, a) is q1 > 1. (That can also be derived by a six-line calculation directlyfrom the definition.) The interpretation of the hypothesis B(a, p, c) tells us that


c1 > 1, since a1 = 0 and b1 = 1; and then since b1 = 1 and c1 > 1, the hypoth-esis B(b, q, c) tells us 1 < q1 < c1. Hence the interpretation of the conclusionB(p, z, q) is verified. The interpretation of the conclusion B(p, z, b) is equivalent,by the above remarks, to 0 < |z2| < |b2|. We have |z2| < |q2| since B(a, z, q) hasbeen verified, and |q2| < |b2| since B(b, q, c) is a hypothesis. Hence |z1| < |b2| asrequired to verify B(p, z, b). That completes the verification of Axiom B4.

We turn now to the handedness axioms. Note that Right (a, b, c) implies a 6= b,as does Left (a, b, c), since it is defined using the intersection points of circles withcenters a and b, and Axiom I17 and A18 guarantee that these intersection pointsare not defined for concentric circles; this observation will used below withoutrepetition. Although Right and Left are not official predicate symbols of ECG,it will still be convenient to give them interpretations.

The concept “ABC is a left turn” can be defined as usual in computer graphics,by the cross product. Specifically the cross product is (c1 − b1)(a2 − b2) − (c2 −b2)(a1−b1), where a = (a1, a2) and b = (b1, b2). Note that no division is required.The cross product is defined even when a = b = c, while Right (a, b, c) should beundefined when a = b. We therefore define Right (a, b, c) to mean that a 6= b andthe cross product (a−b)×(c−b) is non-negative. We have already discussed thisidea when giving the interpretation of IntersectCircles1 and IntersectCircles2 .With this definition of Right and Left , we claim that the interpretation ofDefinition 8.8 is provable. (That definition would be taken as an axiom, if Rightand Left were taken as an official part of the theory.) Arguing in EF (i.e., usinganalytic geometry), let C be a circle with center a, and K be a circle with centerb 6= a. Let

p = IntersectCircles1 (C,K)

q = IntersectCircles2 (C,K)

We must show, using the axioms of EF, that bap is a right turn (as definedusing cross product) and baq is a left turn. To do this we recall the defini-tion of IntersectCircles1 (C,K). That definition says that with λ = |b − a|,and e and f a pair of orthogonal unit vectors with e = (b − a)/|b − a| ande × f = 1, we can put p and q in the form a + λxeλyf , and the solutionwith y ≥ 0 is IntersectCircles2 (C,K), that is q; and the one with y ≤ 0 isIntersectCircles1 (C,K), that is q. Thus

q = a+ λxe+ λyf with y ≥ 0

p = a+ λxe+ λyf with y ≤ 0.

Now to verify the interpretation of Definition 8.8, we must check that abp isa right turn, that is, (a − b) × (p − b) ≥ 0; and also that abq is a left turn,that is, (a − b) × (q − b) ≤ 0. If we make a linear transformation of positivedeterminant, then the signs of cross products are preserved; so we can make alinear transformation with positive determinant that brings a to (0, 0) and b to(1, 0); that makes e = (1, 0), f = (0, 1), and λ = 1. Then we have q = (x, y)

168 MICHAEL BEESON

with y ≥ 0 and p = (x, y) with y ≤ 0. We calculate

(a− b) × (p− b) = ((0, 0) − (1, 0)) × ((x, y) − (1, 0))

= (−1, 0) × (x− 1, y)

= −y≥ 0 since y ≤ 0

Hence abp is a right turn. Similarly (a−b)×(q−b) = −y ≤ 0 since for q, y ≥ 0; soabq is a left turn. That completes the verification that Definition 8.8 is correctlyinterpreted; or to put it another way, that the definitions of IntersectCircles1

and IntersectCircles2 work as intended.Having verified Definition 8.8, we may use that result in verifying the handed-

ness axioms; in other words, we can proceed as if Right and Left were symbolsof the language. By definition, the interpretation of on (x,Line (a, b)) is just theusual analytic-geometry condition for x to lie on the line connecting a and b(formulated in terms of cross product, but equivalent to the formulation withlinear equations if we know the line is not vertical). Linear transformations withpositive determinant preserve incidence, betweenness, and the signs of cross prod-ucts, as well as collinearity; and the handedness axioms H1-H12 mention onlythose predicates (i.e., there are no occurrences of the equidistance relation E).Hence, in verifying them, we may first apply a linear transformation of positivedeterminant to bring the points into a more convenient position. For example,consider Axiom H1. Applying a linear transformation, we may assume a andb lie on the X-axis, indeed we may assume a = (0, 0) and b = (1, 0); and thefurther hypotheses are that p and q do not lie on the x-axis, and abp and abq areboth right turns. We must show that no point on the x-axis is between p and q.Let p = (x, y). We have (a − b) × (p− b) = −y as above, so since abp is a rightturn, we have y ≤ 0. Similarly if q = (u, v), we have (a− b) × (q − b) = −v, sov ≤ 0. If z is a point between p and q, then we must have z2 < 0, so z does notlie on the x-axis. That verifies Axiom H1.

Axiom H2 is similarly, but with Left instead of Right , so we have y ≥ 0 and v ≥0, and again z cannot lie on the x-axis. Axiom H3 has Right (a, b, p)∧Left (a, b, q),so when we make the same calculation, we find y ≤ 0 and v ≥ 0. Hence the linebetween p and q does cross the x-axis, so the intersection of the x-axis (whichis the interpretation of Line (a, b)) meets the interpretation of Line (p, q). Thatverifies Axiom H3.

Turning to Axiom H4, again we apply a linear transformation of positivedeterminant to bring a to (0, 0) and b to (1, 0). As before, the hyptothesis thatRight (a, b, p) is equivalent to p2 ≤ 0, and the desired conclusion Right (a, b, q)is equivalent to q2 ≤ 0. By hypothesis there is a point on the x-axis betweenp and y; hence y2 > 0 and p2 < 0. From the interpretation of B(q, z, y) ∧on (z,Line (a, b)) we have z on the x-axis and since y2 > 0, we have q2 < 0;hence Right (a, b, q), which is the desired conclusion. That verifies Axiom H4.Axiom H5 is similar with Left instead of Right ; the signs are reversed, so wefind Left (a, b, q) in the end.

Turning to Axiom H6, again we apply the same linear transformation; sothe hypothesis Right (a, b, p) tells us that p2 ≤ 0; the hypothesis B(p, q, x) ∧


on (x,Line (a, b)) tells us that q2 > 0, which we have seen above implies Left (a, b, q).That verifies Axiom H6. H7 is similar with Right and Left interchanged.

Axioms H8 through H11 concern turns in a triangle. By means of a lineartransformation with positive determinant, we may assume that a = (0, 0), b =(1, 0), and c lies on the y-axis; but we do not know anything about the sign of c2(it may even be zero). In this situation, as we have proved above, Right (a, b, c)is equivalent to c2 ≤ 0, and Left (a, b, c) to c2 ≥ 0. Let us verify Axiom H8. Thehypothesis tells us c2 ≤ 0 and b 6= c. To check that Right (b, c, a), we computethe cross product

(b− c) × (a− c) = (1 − c1,−c2) × (−c1,−c2)= (1 − c1)(−c2) − (−c2)(−c1)= −c2 + c1c2 − c2c2

= −c2≥ 0 since c2 ≤ 0 by hypothesis

Hence Right (b, c, a) as required. That verifies Axiom H8. Axioms H9, H10, andH11 are verified similarly.

Axiom H12 is Left (α, β, γ). We have specified α = (0, 0), β = (1, 0), andγ = (γ1, γ2). We compute

((α − β) × (γ − β)) = ((0, 0) − (1, 0)) × ((γ1, γ2) − (1, 0))

= (−1, 0)× (γ1 − 1, γ2)

= −γ2

≤ 0 since γ2 > 0 is an axiom

Hence Left (α, β, γ) as desired. That completes the verification Axiom H12, whichis the last handedness axiom.

The lower dimension axioms D1 through D4 are immediate, since we havechosen α, β to lie on the x-axis and be distinct, γ does not lie on the x-axissince γ2 > 0.

Next we verify the upper dimension Axiom D5, which says that if three dis-tinct circles have a common chord, their centers are collinear. Let three circleswith distinct centers p, q, and r have the common chord ab. Make a linear trans-formation that brings the chord to the y axis with its midpoint at the origin; wemay also assume the endpoints are a = (0, 1) and b = (0,−1). Points equidis-tant from a and b must lie on the x-axis, and hence the centers of the circles arecollinear. Note that linear transformations take vectors with zero cross productinto vectors with zero cross product, so collinearity in the cross product senseused to define on is preserved. That completes the verification of Axiom D5.

We now turn to the congruence axioms. Axiom C1 says that we can lay offsegment cd along segment ba, by taking the point

q = IntersectLineCircle2 (Line (a, b),Circle3 (a, c, d)).

The possibility c = d is allowed. We have to prove that the conclusions aq = cdand ¬B(x, a, b) are correctly interpreted. Of course the interpretation of aq = cdis immediate, since q lies on the circle of radius |c − d| about a. We have to

170 MICHAEL BEESON

prove (dropping the superscript ) that

¬(x 6= a ∧ a 6= b ∧ |x− b| = |x− a| + |b − a|)Suppose x 6= a and a 6= b and |x − b| = |x − a| + |b − a|; we must derive acontradiction. By definition of IntersectLineCircle2 , we know that for z not onLine (a, b), zab is a right turn exactly when zpq is a right turn, where

p = IntersectLineCircle1 (Line (a, b),Circle3 (a, c, d)).

That is, the signs of (z − a)× (z − b) and (z − p) × (z − q) are the same. Againwe make a linear transformation that brings a to (0, 0) and b to (1, 0). Withi = (0, 1) we have Right (−i, a, b). Hence Right (−i, p, q). The two intersectionpoints of the circle with Line (a, b) are (±|c − d|, 0); therefore p = (−|c − d|, 0)and q = (|c − d|, 0). Therefore q1 ≥ 0. But B(x, a, b) means x2 < 0, as iseasily checked. Hence ¬B(q, a, b) as desired. That completes the verification ofAxiom C1.

Axioms C2, C3, and C4 mention only the equidistance relation, and theirinterpretations are immediate consequences of the definition of |x; for exampleconsider C4, which says ab = ba. Its interpretation is |a − b| = |b − a|, whichis immediate from the definition and the ring laws. We say no more aboutthese three axioms. Consider Axiom C5, which defines incidence for circles. Theinterpretation of On (x,Circle3 (a, b, c)) is |x−a−| = |b− c, which is exactlythe interpretation of ax = bc. That verifies Axiom C5.

Axiom C6 is Tarski’s 5-segment axiom. Here we have four points a, b, c, d inthe configuration illustrated in Fig. 8, and four more points A, B, C, D, in asimilar configuration, with the segments shown as solid lines pairwise congruent,i.e. ab = AB, ad = AD, bd = BD, and bc = BC. The conclusion is thatdc = DC. To verify this, we make a linear transformation T that takes (a, b)onto (A,B). Since ab = AB, this can be done by taking T as the compositionof a translation and a rotation, so T has determinant 1 and preserves distances.Since T preserves distances, we have |A−Tc| = |a− c| = |A−C| and |B−Tc| =|b − c| = |B − C. Therefore Tc lies on the circle of radius AC about A and thecircle of radius BC about B. Since a, b, and c are collinear, and T is linear, A,B, and C are collinear; two non-concentric circles with an intersection point onthe line joining the centers have only one intersection point (as is easily proved inanalytic geometry, i.e. in EF); therefore Tc = C. Similarly, Td lies on the circleof radius AD about A (since AD = ad) and on the circle of radius BD aboutB (since BD = bd). Let D′ be the reflection of D in the line joining A and B.Then ¬¬(Td = D ∨ Td = D′). Since we are trying to prove an equality, namely|D−C| = |d− c|, and equality is stable, we can argue by cases. If Td = D thenwe are finished, as we have |D −C||Td−C| = |d− c|. If Td = D′ then we have|D − C = |D′ − C since C lies on the axis of the reflection that takes D to D′,and |D−C| = |D′ −C| = |Td−C| = |d− c|. That completes the verification ofAxiom C6.

For readers who are not trained in formal logic, we offer the following remarks.The “proof” we have just given that Axiom C6 is verified, like the verificationsof the other axioms above, should be regarded as a high-level description of analgebraic calculation explicitly representing the components of the points and the


transformation T . In turn that algebraic calculation would technically need to beformalized in EF, producing a long list of formulas constituting a formal proof.If we wrote out explicitly either the calculation or the formal proof, it would bedifficult for a human to follow, so we prefer to give the high-level description.

Now we turn to the line-circle continuity axioms. In Definition 8.11, wehave defined the concept“inside” in ECG. That definition makes it evidentthat the interpretation of “x is inside Circle (a, b)” is |x − a| ≤ |a − b.Please refer to the definition of the interpretation, where we defined termsIntersectLineCircle1 and IntersectLineCircle2 of EF; see especially equation(1). We proved that if line L meets C, then IntersectLineCircle1 (L, C) andIntersectLineCircle2 (L, C) are defined and lie on line L and circle C. Butthat is just what the interpretation of Axiom Cont 1 says; so we have alreadyverified Axiom Cont 1, at least using the strong parallel axiom. Now we discussthe verification of that axiom using only Playfair field theory. The question iswhether the denominator in (1) is invertible. It is (a− b) · (a− b) where the linethat should intersect the circle is Line (a, b). We only know that a 6= b, not that|a − b|2 is invertible. This problem is easily remedied. By Axiom EF10, it willsuffice if we replace a and b by two points a′ and b′ such that a′b′ > 01, i.e. thedistance between a′ and b′ is more than 1. That can be done, for example, bytaking a′ = a and

b′ = IntersectLineCircle2 (Line (a, b),Circle3 (b, 0, 1)).

Then B(a, b, b′) and bb′ = 01. Since Line (a, b′) coincides with Line (a, b), it isenough to verify that IntersectLineCircle1 (Line (a, b′), C) is defined, since byLemma 9.90 we have

IntersectLineCircle1 (Line (a, b′), C) = IntersectLineCircle1 (Line (a, b), C).

But |(b′) − a| is invertible, by Axiom EF10, since it is > 1 and 1 is invert-ible. IntersectLineCircle2 is treated similarly. That completes the verificationof Axiom Cont 1.

Axiom Cont 2 says that if L has a point strictly inside C, then the two in-tersection points are distinct. To prove this, suppose that the two intersectionpoints are not distinct. Refer to the definition of IntersectLineCircle1 andIntersectLineCircle2 ; then the expression under the square root in the solutionfor λ is zero:

((a− e) · (b− a))2 − (b − a)2((a− e)2 − (e− c)2) = 0(3)

In this formula, e is the center of the circle, a and b are the points determiningline L, and c is a point on C (given by the third and fourth coordinates of theinterpretation of the circle). Technically, all these letters should have superscript, since we are working in EF, but we omit the superscripts. We have to show,using analytic geometry, that this formula implies that line L is tangent to circleC. One can do this in a natural and direct way using the expression for a dotproduct in terms of the cosine of the included angle; but trigonometry is notavailable in EF, at least not directly, so that proof is not legal here. It is alsonot necessary. In EF, the quadratic formula tells us that there are at mosttwo points of intersection; and since we have already verified Cont 1, both theterms giving points of intersection are defined, so there is at least one. Then

172 MICHAEL BEESON

all we have to do is verify in EF that if point (t, s) on L lies inside C, then thetwo intersection points of L and C are distinct. Linear transformations preservecircles and collinearity, so we may assume the center of the circle is at the originand the line is horizontal. The equation of C then is x2 + y2 = r2 for some r,and the equation of L is y = c for some c. Since (t, s) is on L, we have s − c.Since (t, s) is inside C we have t2 + c2 < r2. The intersection points of the line

with C are the solutions of x2 + c2 = r2, i.e. x = ±√r2 − c2. If these are equal

then r2 = c2. Then from t2 + c2 < r2 we have t2 < 0; but squares are positive,so this is a contradiction. Hence the two intersection points are not equal, whichwas what we had to prove. That completes the verification of Axiom Cont 2.

Axiom Cont 3 says that IntersectLineCircle1 (L,C) and IntersectLineCircle2 (L,C)depend extensionally on the circle C, i.e. if two circles C and K have the samepoints on them, then

IntersectLineCircle1 (L,C) = IntersectLineCircle1 (L,K)

and

IntersectLineCircle1 (L,C) = IntersectLineCircle1 (L,K)

Suppose the interpretation of “C and K have the same points on them” holds.Then C and K have the same points on them, i.e. the same coordinate pairssatisfy the equation of C and the equation of K. The equation of C is

(x− C1 )2 + (y − C

2 )2 = (C3 − C

1 )2 + (C4 − C

2 )2

according to the definition of On . But according to the definition of C, wewill have C = K, since the centers will be the same, and the third and fourthcoordinates are determined by the center and radius (they specify a point onthe circle due “north” of the center). Hence the equations of C and K are thesame. That completes the verification of Axiom Cont 3.

We turn to Axioms Cont 4 and Cont 5, which say that the order of the two in-tersection points on line L = Line (a, b) is the same as the order of a and b, in thesense that, if p = IntersectLineCircle1 (L,C) and q = IntersectLineCircle2 (L,C),then for each point z not on L, zpq is a right (or left) turn if zab is a right (orleft, respectively) turn. Suppose that z is not on L, and zab is a right turn, i.e.Right (a, z, b). We must show that zpq is a right turn. Since Right is stable (it isdefined by an equality), we can argue by cases. If p = q then zpq is automaticallya right (and a left) turn, so there is nothing to prove. Therefore we can assumep 6= q. Hence C is not a degenerate circle. By the handedness axioms, which wehave already verified in EF, we can change z to any convenient point on the sameside of L as the original z. A convenient point is one of the two intersections ofthe diameter M of C that is perpendicular to L. Also by the handedness axiomswe can change a and b to any other points a′ and b′ on the line in the same orderas the original a and b, in the sense that the points are moved one at a timealong L without crossing. Let us use “west” and “east” to refer to directions onL, with a originally west of b. Then we can slide a west of C, and b east of C,and then slide a and b back towards the center of e until they intersect C, thusdefining a′ and b′; and we have Right (z, a′, b′) by the handedness axioms. Nowa′ and b′ coincide, in some order, with p and q. But looking at the definition ofIntersectLineCircle1 (L,C), we find that p = a− λ(b− a) and q = a+ λ(b− a),


where λ ≥ 0 because it is given by a square root. Hence p is the intersectionpoint to the “west” and q is the intersection point closest to the “east”. Hence a′

coincides with p and b′ coincides with q. Hence Right (z, p, q) as desired. Thatcompletes the verification of Axioms Cont 4. Axiom Cont 5 is verified in thesame way, replacing Right by Left .

Next we turn to the circle-circle continuity axioms. Axiom Cont 6, illus-trated in Fig. 12, says that if circle K has a point p inside circle C and a pointq outside circle C, then both intersection points IntersectCircles1 (C,K) andIntersectCircles2 (C,K) are defined. (An incidence axiom that we have alreadyverified says that they lie on both C and K.) We have already given a detailedverification of this axiom in Theorem 6.1.

Axiom Cont 7 concerns IntersectCirclesSame , and does not need to be verifiedin view of Theorem 9.96.

Axiom Cont 8 specifies that the intersection points of circles depend exten-sionally on the circles. That is evident from our definition of On .

Axiom Cont 9 (illustrated in Fig. 13) says that if a and b are the centers ofC and K, and p = pointOnCircle (C) and q = pointOnCircle (K), and t is non-

strictly between a and b with bt = ap, then pt = ab. Let us write γ4 =√

1 − γ23 ,

and r = C3 , to simplify the formulas. Then by definition,

p = (a1 + γ3 · r, a2 + γ4 · r)q = (b1 + γ3 · r, b2 + γ4 · r)

The hypothesis of Axiom Cont 9 mentions point t, non-strictly between b and q,such that bt = ap. Then we have

t = b + λ(q − b)

where λ is chosen to make |t−b| = |p−a|; that requires λ = |p−a|/|q−b|. Thedenominator is not zero, since K is a non-degenerate circle, by the hypothesisB(u, b, q). Now we calculate

|t − b| = λ|q − b|= |p − a|

as required. We also have to verify the second part of the axiom, which saysthat segment pt does not meet segment ab. That is true in the interpretationsince pt is parallel to ab. That completes the verification of Axiom Cont 9.

We now take up the strong parallel axiom. Let L be a line, and let K be aline such that IntersectLines (L,K) is not defined; that is, K cannot fail both tobe parallel to L and to be coincident with L. Let p be on K and let a be a pointnot on K, and let M = Line (p, a). We have to show that IntersectLines (L,M)is defined, i.e. M meets L.

Let J = Perp(a, L), so J is perpendicular to L; let m be the intersectionpoint of J and L. By applying a linear transformation, we may assume thatL = Line (0, 1) and m = 0. Now J does not coincide with K, since it containspoint a, which is not on K. Then if J does not meet K it is parallel to K,and since then m is on J but not on K, K is parallel to L. Then J and L aretwo parallels to K through m, contradicting even Playfair’s axiom; hence J doesmeet K. Call the point of intersection z. Then z 6= a since a is not on K. We

174 MICHAEL BEESON

have z = (0, z2) since z is on J ; similarly a = (0, a2). Now we can write theequation of line M and solve for its meeting point with the x-axis.

0 = y = a1 +a1 − z2a1 − p1

x

x =(p1 − a1) · a2a2 − z2

and since a 6= z, the denominator is not zero, and hence using the strong recip-rocal axiom, it is invertible, so this equation does provably define x in EF. Thatcompletes the verification of the strong parallel postulate in EF.

Next we check that Euclid 5 can be verified using only P (x) → 1/x ↓ . Thealgebra for this is the same as above; we must use the extra hypotheses of Euclid5 to control the sign of the denominator. The extra hypotheses are that K doesnot meet L, and there are points r on K and q on L with B(r, a, q). Thenwe have r2 = q2 and a2 < r2 , and hence a2 < q2 , controlling the sign of thedenominator as required. That completes the verification of Euclid 5 from theaxiom P (x) → 1/x ↓ .

Finally we check that Playfair’s axiom can be verified using only weak Eu-clidean field theory, i.e. Axioms EF9 and EF10 instead of the strong reciprocalaxiom EF1. Suppose Playfair’s axiom. Then with the hypotheses as for thestrong parallel axiom, but also assuming K and L are parallel (rather than co-incident). Let a be any point on M . We have to verify that M and L cannotfail to meet. That is, x in the above equation cannot fail to be defined. Supposeit does fail to be defined; then the denominator of the expression for x is notinvertible. Hence, by Axiom EF9, it is zero. Then a2 = q2, so a lies on K. Sincea was any point on M , we have proved M and K coincide, contradicting thehypothesis.

That completes the proof of the theorem.

12.6. Area in Euclid, Hilbert, and ECG. In the time of Euclid, area wasnot considered as a real-valued function. In fact, “area” is never mentioned inEuclid; nor is “distance” mentioned. When Euclid states theorems that wouldnow be considered to be about area, he speaks of “equality” between (specific)polygons. Hilbert [15], Chapter IV, has a chapter on Theory of Plane Area inwhich he proceeds as follows. He first makes Euclid’s notion of “equality” of poly-gons precise: two polygons are equidecomposable if they can be decomposed intoa finite number of triangles that are congruent in pairs. Two polygons (or unionsof polygons) A and B are equicomplementable if one can find a finite number ofpairs of equidecomposable polygons (Pi, Qi) and polygons A′, B′, such that A′ iscomposed of A and the Pi and B′ is composed of B and the Qi. Euclid’s “equal”in the latter part of Book I should be read as “equicomplementable”, accordingto Hilbert. He gives an argument for reading it as “equicomplementable” ratherthan “equidcomposable”.

Of course neither of these notions is a first-order (geometric) notion, becauseof the mention of “finite number”; but to formalize Euclid Book I (including itsculmination, the Pythagorean theorem) a fixed (and relatively small) number oftriangles in the decompositions is required, and we do not have a theorem with“equal polygons” in the hypothesis, so Book I remains first-order. We take it


then that Euclid’s I.47, the Pythagorean theorem, can be proved in ECG. EuclidI.47 says that the square on the side of the hypotenuse is equicomplementable(with, say, less than 100 triangles, to stay first-order) to the two squares on thesides.

To define area as a function from polygons to points on Line (0, 1), we makeuse of the fact that we already know how to introduce coordinates (X(a), Y (a))geometrically. We can then introduce the usual apparatus of vector algebra, butusing geometric definitions for the underlying arithmetic.

Definition 12.22.

Point VectorSubtract(Point a, Point b)

u = Add(X(a),Minus(X(b)))

v = Add(Y(a),Minus(Y(b)))

return MakePoint(u,v)

Point CrossProduct(a,b)

p = Multiply(X(a),Y(b))

q = Multiply(X(b),Y(a))

return Add(p,Minus(q))

Point SignedArea(Point a, Point b, Point c)

u = VectorSubtract(c,b)

v = VectorSubtract(a,b)

w = CrossProduct(u,v)

1/2 = Divide(1,Add(1,1))

return Multiply(1/2, w)

Point Area(Point a, Point b, Point c)

return SquareRoot(HilbertSquare(SignedArea(a,b,c)))

Then the area of a triangle can be defined in the way that is now customaryin computer graphics:

Lemma 12.23. Let y lie on segment xz. Then

SignedArea (0, z, x) = SignedArea (0, y, x) + SignedArea (z, y, 0)

Proof. This is a routine computation in vector algebra, but it has to be doneusing Add and HilbertMultiply and CrossProduct . Nevertheless for readabil-ity we write it in the ordinary notation; so for instance X(y) becomes y1 andCrossProduct (x, y) becomes x× y. Since y lies on segment xz we can write, for

176 MICHAEL BEESON

some t, y = tz + (1 − t)x. Then

y × x = y1x2 − y2x1

= (tz1 + (1 − t)x1)x2 − (tz2 + (1 − t)x2)x1

= t(z1x2 − x1x2 − z2x1 + x2x1)

= t(z1x2 − z2x1)

z × y = z1(tz2 + (1 − t)x2) − z2(tz1 + (1 − t)x1)

= t(z1z2 − z1x2 − z2z1 + z2x1) + z1x2 − z2x1

= t(−z1x2 + z2x1) + z1x2 − z2x1

y × x+ z × y = z1x2 − z2x1

= z × x

That completes the proof.

Lemma 12.24. SignedArea (a, b, c) = SignedArea (b, c, a) and

SignedArea (a, b, c) = Minus (SignedArea (b, a, c).

Proof. These are also proved by routine computation. For readability we writeb− a instead of VectorSubtract (b, a), and so on:

(b− a) × (a− c) = (b× a) + (a× c) − (b× c)

(c− b) × (b− a) = (c× b) − (c× a) + (b× a)

and the two right sides are equal. That completes the proof.We now introduce several more standard concepts in ECG:

Definition 12.25.

Norm(Point a)

xsq = HilbertSquare(X(a))

ysq = HilbertSquare(Y(b))

return SquareRoot(Add(xsq,ysq))

VectorAdd(Point a, Point b)

u = Add(X(a),X(b))

v = Add(Y(a),Y(b))


Distance(Point a, Point b)

u = VectorSubtract(a,b)

return Norm(u)

Radius(Circle C)

e = center(C)

b = pointOnCircle(C)

return Distance(e,b)

ScalarMultiply(Point c, Point u)


x = HilbertMultiply(c,X(u))

y = HilbertMultiply(c,Y(u))

return MakePoint(x,y)

We develop in ECG the theory of two-by-two matrices and matrix multipli-cation Av, where A is a matrix whose entries are points on Line (0, 1) and v isany point (thought of as a 2-vector (X(v), Y (v))).

Point MatrixMultiply( Point a, Point b, Point c, Point d, Point v)

u = Add(HilbertMultiply(a,X(v)),HilbertMultiply(b,Y(v)))

v = Add(HilbertMultiply(c,X(v)),HilbertMultiply(d,Y(v)))


Point Determinant(Point a, Point b, Point c, Point d)

return Add(Multiply(a,d),Minus(Multiply(b,c)))

We can define linear transformations (using six points); we define translations,rotations, and reflections as usual: translations are maps u 7→ VectorAdd (u, b),and rotations about the origin are maps with matrices

(

a b−b a

)

where a2 + b2 = 1

where for readability we have written −b instead of Minus b and a2 instead ofHilbertSquare (a), etc.

Lemma 12.26. Translations preserve both congruence and distance. That is, ifA = VectorAdd (a, t) and B = VectorAdd (b, t) then AB = ab and Distance (A,B) =Distance (a, b).

Proof. First we show translations preserve distance.

A = VectorAdd (a, t)

B = VectorAdd (b, t)

VectorSubtract (A,B) = VectorSubtract (a, b)

Hence Distance (A,B) = Distance (a, b) by definition of Distance .Now we show translations preserve congruence. Let a and t lie on Line (0, 1)

and let p = Add (a, t); then ap = 0t, by Lemma 9.6, since segment 0p is composedon the one hand of 0a and ap and on the other of 0t and tp, but tp = 0a bythe definition of Add (a, b), according to which a is first rotated to the y-axis,then projected onto the vertical line through t, then rotated back to the x-axisto construct p. Those two rotations have the same radius 0a; hence tp = 0a.

Now consider a vertical translation

A = VectorAdd (a,MakePoint (0, t))

B = VectorAdd (b,MakePoint (0, t)).

Then quadrilateral abBA has four right angles (for t 6= 0 and a 6= b) and henceis a rectangle; hence ab = AB. But any translation is the composition of a

178 MICHAEL BEESON

horizontal and a vertical translation; hence if ab lies on Line (0, 1) and AB isthe image of ab under any translation, then AB = ab. Thus any two horizontalsegments that can be transformed into each other by a translation are congruent.Since rotating through ninety degrees by Rotate also preserves congruence, thesame is true of vertical segments, i.e. any translation preserves congruence ofvertical segments. Note, rotation in the sense of Rotate is not (so far proved tobe) the same as rotation by a matrix of the form specified as a rotation.

But now consider an arbitrary segment ab, transformed into A and B by theformulas in the lemma. Then let

a′ = VectorAdd (a,MakePoint (X(t), 0)

A′ = VectorAdd (A,MakePoint (X(t), 0).

Then aa′ = AA′ since aa′ is a horizontal segment and AA′ is its translation byt. Similarly a′b is a vertical segment, so a′b = A′B. Then right triangle aa′b iscongruent to right triangle AA′B; hence their corresponding sides ab and ABare congruent. That completes the proof of the lemma.

Lemma 12.27. Rotations preserve congruence and distance. That is, if T is amatrix of the form specified above as a rotation, and A = Ta and B = Tb, thenDistance (A,B) = Distance (a, b) and ab = AB.

Proof. First we show rotations preserve distance. By a routine computation, oneproves that if A is a matrix and v is a point then

Norm (Av) = HilbertMultiply (Determinant (A),Norm (v))).

(But the routine computation takes place in ECG, using the geometric defini-tions of arithmetic.) Hence matrices of determinant one preserve distance. Theseinclude rotations (defined as above), as one easily computes that the determinantof a rotation is 1.

Next we prove that rotations preserve congruence. Let A be a rotation. Thenfor some u and v with Norm (u, v) = 1, we have

A =

(

u vMinus (v) u

)

.

Let p = MakePoint (u, v). (See Fig. 47.)Then we claim

A(MakePoint (x, 0)) = Rotate (1, 0, p, x).

This is proved by a calculation:

A(MakePoint (x, 0)) =

(

u Minus (v)v u

) (

x0

)

=

(

HilbertMultiply (u, x)HilbertMultiply (v, x)

)

= MakePoint (HilbertMultiply (u, x),HilbertMultiply (v, x))

Hence the point q = A(MakePoint (x, 0)) lies on Line (0, p), since it satisfies theequation of that line (as shown by Hilbert, using similar triangles, see [15], p. 57),and

X(q) = HilbertMultiply (u, x)


Figure 47. Rotations preserve congruence. Point q is definedby matrix multiplication, but it is the same as the point com-puted by Rotate (1, 0, p, x).

b

0b

b

X(p)

p

b

xb

X(q)

b

q

u

v

Y (q) = HilbertMultiply (v, x).

On the other hand Rotate (1, 0, p, x) is the point s on Ray (0, p) with 0s = 0x, atleast for x ≥ 0. We are trying to prove s = q. We know now that they lie on thesame ray and that they have the same distance x from 0. That is tantalizinglyclose, but still some argument is required to show s = q. It will suffice to provethat there is only one point on Ray (0, p) at a given distance from 0. Any point onthat ray has the formw = MakePoint (HilbertMultiply (c, u),HilbertMultiply (c, v))for some c > 0, so its distance from 0 is c, since MakePoint (u, v) has norm 1.Then c = Divide (X(w), u) if u 6= 0 and c = Divide (Y (w), v) if u = 0. In eithercase c is determined by w so there cannot be two points on the ray at the samedistance from 0. Since equality is stable, we are allowed to argue by cases here.That completes the proof of the lemma.

Lemma 12.28. If ab = AB then Distance (a, b) = Distance (A,B). Converselyif Distance (a, b) = Distance (A,B) then ab = AB.

Proof. By a translation and rotation we can bring a to the origin and b to lieon Ray (0, 1). By the previous lemmas, ab is congruent to this new segmentand distance is preserved. Without loss of generality, then, we may assumea = A = 0 and both b and B lie on Ray (0, 1). Now assume ab = AB. Then byLemma 9.5, b = B. Hence Distance (a, b) = Distance (a,B) = Distance (A,B).

180 MICHAEL BEESON

Conversely, assume Distance (a, b) = Distance (A,B). But Distance (a, b) =X(b), and Distance (A,B) = X(B). Since b and B lie on Ray (0, 1), we haveY (b) = Y (B) = 0. Hence b = MakePoint (X(b), 0) = B. Hence 0b = 0B; but0 = a = A so ab = AB. That completes the proof.

Lemma 12.29. Let triangle abc be congruent to triangle ABC. Then Area (a, b, c) =Area (A,B,C).

Proof. By a routine computation, one proves that if A is a matrix and v is apoint then

Norm (Av) = HilbertMultiply (Determinant (A),Norm (v))).

(But the routine computation takes place in ECG, using the geometric defini-tions of arithmetic.) Hence matrices of determinant one preserve area. Theseinclude rotations (defined as above) and reflections. We can bring a and A to 0by translations, and then by a rotation we can bring b and B to lie on Ray (0, 1).By the previous lemmas, these translations and rotations preserve both congru-ence and distance. Hence, without loss of generality we can assume a = A = 0and b = B, with b lying on Ray (0, 1). By the uniqueness of triangle construction(Lemma 9.68), then not not C = c or C is the reflection of c in Line (0, 1). In thelatter case, we can apply a reflection (which also has determinant one) to makec = C. Therefore not not Area (a, b, c) = Area (A,B,C); but by the stability ofequality, we can drop the double negation, so Area (a, b, c) = Area (A,B,C) asclaimed.

Hilbert [15], defines the area of a triangle more traditionally, using the formula“half the base times the altitude”, but our way is perhaps easier, as it reduceseverything to mechanical computations with the cross product. Once the areaof a triangle is defined, however, we follow Hilbert:

Lemma 12.30 (Hilbert’s theorem 49). Let point o lie outside triangle abc. Then

SignedArea (a, b, c) = SignedArea (o, a, b)+SignedArea (o, b, c)+SignedArea (o, c, a).

Proof. Again, what is to be proved is an equality, so we may argue by cases asto the position of o, as Hilbert does. Hilbert’s argument depends only on theprevious two lemmas, so it goes through without change.

After this theorem, one can define the area of a positively oriented simplepolygon as the sum of the areas of triangles in a triangular decomposition of thatpolygon, and prove that it is independent of the decomposition. The culminationof Hilbert’s theory of area is his Theorem 51 (see [15], p. 69), which says Twoequicomplementable polygons have the same area, and two polygons with the samearea are equicomplementable. Since “equicomplementable” is not a first-orderconcept, Hilbert’s Theorem 51 is not a first-order theorem as it stands (even ifwe restrict the number of sides of the polygon to 4 or some other fixed number).The first half could be taken as a first-order schema, one theorem for each fixednumber of triangles allowed for the decompositions. The second half raises thequestion whether a fixed number of triangles suffices. Hilbert’s proof showsthat the required number of triangles is bounded by a small constant times thenumber of sides of the polygon; hence, for a fixed number of sides of the polygon,this is a first-order theorem. Since the conclusion is an equality, and equality is


stable, we are free to prove the theorem by contradiction, so we do not have toworry about whether Hilbert’s proof is constructive or not.

From Hilbert’s Theorem 51 and Euclid I.47, we can draw as a corollary thefollowing version of the Pythagorean theorem:

Lemma 12.31. If p is on Circle(0, r), then

Add (HilbertSquare (X(p)),HilbertSquare (Y (p))) = HilbertSquare (r).

Proof. Consider the triangle with vertices 0, MakePoint(X(p), 0), and p. Thesquare on the hypotenuse has area HilbertSquare (r), as one can compute usingthe cross product and a decomposition of the square into two triangles. Thesquare on the horizontal side has area HilbertSquare (X(p)). The square onthe vertical side has area HilbertSquare (Y (p)). By Euclid I.47, the two smallersquares are equicomplementable to the large square. The lemma then followsfrom Hilbert’s Theorem 51. That completes the proof.

Actually, Lemma 12.31 has a much simpler proof: Since p and the point b =MakePoint (r, 0) both lie on Circle (0, r), then by Axiom C5, segment 0p is con-gruent to segment 0b. Hence, by Lemma 12.28, Distance (0, p) = Distance (0, b).But

Distance (0, p) = Add (HilbertSquare (X(p)),HilbertSquare (Y (p))),

and Distance (0, b) = HilbertSquare (r). Hence Distance (0, p) = Distance (0, b),proving Lemma 12.31 the easy way.

We can then use Hilbert’s Theorem 51 to prove Euclid’s version of the Pythagoreantheorem, but without the specific decompositions given by Euclid’s proof to es-tablish that the two small squares and the large square are equicomplementable.Nevertheless, the fact that proof is constructive shows that some decompositioncan be found. It would be possible to trace through the proof and extract theEuclid-style proof in ECG of the Pythagorean theorem that is implicit in theshort proof of Lemma 12.31 plus Hilbert’s Theorem 51.

12.7. Handedness, cross product, and linear transformations. In thissection we show that Right (a, b, c) is related to the cross product (a− b) × (c−b) as one would expect. This turns out to be somewhat more difficult thanexpected. In this section we also formalize some other lemmas in vector algebra;we need these results because in the course of studying the faithfulness of theinterpretation of ECG in field theory, we need to formalize in ECG a numberof algebraic arguments.

We start with the theory of similar triangles.

Definition 12.32. Triangle abc is similar to triangle ABC if the threecorresponding angles are pairwise congruent, i.e. angles abc, bca, and cab arecongruent to ABC, BCA, and CAB respectively.

One easily checks that similarity is an equivalence relation.

Definition 12.33. The proportion a : b = c : d means HilbertMultiply (a, d) =HilbertMultiply (b, c). The proportion ab : AB = bc : BC means

Distance (a, b) : Distance (A,B) = Distance (b, c) : Distance (B,C).

182 MICHAEL BEESON

Lemma 12.34. Triangles abc and ABC are similar if and only if correspondingsides are proportional. A : a = B : b and B : b = B : c and A : a = C : c.

Proof. See [15], pp. 55-57. Since the conclusions are stable, we do not have tocheck line by line whether the proofs are constructive. The proof goes back toDescartes’s definition of Multiply , so as we have stated the lemma, we also needto appeal to the equivalence of Multiply and HilbertMultiply .

Lemma 12.35. . Let ABC and abc be right triangles with right angles at b andB and legs ab and AB parallel to Line (0, 1) and legs bc and BC perpendicularto Line (0, 1), and suppose BC : bc = AB : ab. Then triangle ABC is similar totriangle abc.

Remark. Of course the lemma is true without the restriction that the legs beparallel to the coordinate axes; but to prove that we would naturally use thePythagorean theorem in the form |A−C|2 = |A−B|2 + |C −B|2, and to provethat we need a chain of lemmas reaching back to this lemma; so at this point weprove only the special case where the legs are parallel to the axes. In that specialcase, the Pythagorean theorem is immediate from the definitions of Norm andDistance .

Proof. Using normal notation (to abbreviate geometric arithmetic) let λ = (B−A)1/(b− a)1. Then by hypothesis also λ = (C −B)2/(c− a)2, and since the legsare parallel to the axes we have C1 = B1 and A2 = B2, and c1 = b1 and a2 = b2.Then

|C −A|2 = (C −A)21 + (C −A)22

= (A−B)21 + (C −B)22

= λ2(b− a)21 + λ2(b− c)22

= λ2((b− a)21 + (b − c)22)

= λ2((c− a)21 + (a− c)22

= λ2|a− c|2

and taking the square root of both sides we have |C − A| = λ|c − a|. It followsthat the third pair of sides is also proportional, and then by Lemma 12.34, thetriangles are similar. That completes the proof of the lemma.

Next we show that the points on a line are exactly the points satisfying theequation of the line, in either of two senses:

Lemma 12.36. The following three conditions are all equivalent:(i) Point p is on Line (a, b)(ii) for some λ we have p = VectorAdd (a,ScalarMultiply (λ,VectorSubtract (b, a))).(iii) CrossProduct (VectorSubtract (p, a),VectorSubtract (b, a)) = 0.

Proof. We first show that (ii) implies (i). By Axiom S3, ¬¬ on (p, L) →on (p, L), so we can argue by cases. Suppose p is given by the formula in (ii).

Case 1, X(a) = X(b) (i.e. the line is vertical). Then X(p) = X(a), from theformula for p. Then Line (b, a) is perpendicular to Line (0, 1) at X(a), by Lemma


9.81. Then p is on Line (a, b) because, by definition of X(p), the perpendicu-lar from p to Line (0, 1) meets Line (0, 1) at the same point as Line (a, b) does,namely at X(a), so by Lemma9.81 that perpendicular coincides with Line (b, a).

Case 2, Y (p) = Y (a) (i.e. the line is horizontal). This is analogous to Case 1,with Y and X interchanged.

Case 3, p = a. Then there is nothing to prove as p is already on Line (a, b).Case 4, X(b) 6= X(a) and Y (b) 6= Y (a) and p 6= a.

Let c = MakePoint (X(b), Y (a)). Then a, b, and c are not collinear, sincec lies on a horizontal line through a, so if they were collinear, then b wouldalso lie on that line, so Y (b) would be equal to Y (a), contradiction. Let d =MakePoint (X(p), Y (a)). Then (since p 6= a) similarly a, d, and p are notcollinear. We wish to show triangles acb and adp are similar. We have

VectorSubtract (p, a) = HilbertMultiply (λ,VectorSubtract (b, a)).

Taking the norm of both sides we have

Distance (a, p) = HilbertMultiply (λ,Distance (b, a)).

Taking X instead of norm, we have

Distance (a, d) = HilbertMultiply (λ(Distance (a, c)))

and taking Y instead of norm, we have

Distance (p, d) = HilbertMultiply (λ,Distance (b, c)).

Therefore the corresponding sides of triangles acb and adb are proportional.Hence by Lemma 12.34, the triangles are similar. Hence angle pad is congruentto angle bac. Hence p lies on Line (a, b), as one can prove from the definition ofangle congruence. That completes the proof that (ii) implies (i).

Now we prove (i) implies (iii). Again we can argue by cases, since equalityis stable. For readability we use normal notation for components and crossproducts, but it only abbreviates geometric arithmetic. Let points d and c be asdefined above. Then We have

(p− a) × (b − a) = (p− a)1 · (b− a)2 − (p− a)2(b− a)1

In the cases of a vertical or horizontal Line (a, b), the right side is zero. Hencewe may assume the line is not horizontal or vertical. Then by Definition 12.33,the cross product is zero if and only

pd : bc = ad : ac

Since triangles pda and bca are right triangles with legs parallel to the coordinateaxes, by Lemma ?? we conclude that the cross product is zero if and only iftriangle pda is similar to triangle bca. But triangle pda is similar to triangle bcaif and only if point p lies on Line (a, b). Hence the cross product is zero if andonly if p lies on Line (a, b). That completes the proof that (i) implies (iii).

Now we prove (iii) implies (ii). Suppose (p− a)× (b− a) = 0. Then as shownabove

0 = (p− a)1 · (b − a)2 − (p− a)2(b − a)1

Note that we are not allowed to argue by cases for a conclusion that startswith ∃λ. We must define λ without a case split, which is slightly tricky since

184 MICHAEL BEESON

we must avoid a zero denominator that might arise in some cases. But we doknow a 6= b, so the denominator should be Distance (b, a), which is equal toNorm (VectorSubtract (b, a)). At the same time we have to take care about thesign of λ, which could be positive or negative. The correct value of λ is thenfound by taking the component of p− a in the direction b − a, and dividing byDistance (b, a):

λ :=(p− a) · (b − a)

HilbertSquare (Distance (b, a))

The numerator abbreviates a DotProduct term. Now we claim that with thisvalue of λ, (ii) holds. (Once we have specified λ, what has to be proved is anequality, so now we could argue by cases, but it turns out not to be necessary todo so.) In normal notation, what has to be proved is p− a = λ(b − a). Puttingin the value of λ, what has to be proved (in normal notation) is

p− a =(p− a) · (b− a)

|b− a|2

This can be obtained by setting x = p− a and y = b− a in the formula

x =x · y|y|2 y for y 6= 0

or equivalently

|y|2x = (x · y)y

which is easily verified by a componentwise calculation:

|y|2x = (y21 + y2

2)(x1, x2)

= (x1(y21 + y2

2), x2(y21 + y2

2)

= (x1y1 + x2y2)(y1, y2)

= (x · y)(y1, y2)


Lemma 12.37. The sign of CrossProduct (a, b) is invariant under linear trans-formations with positive determinant. In fact if A is a linear transformation then

CrossProduct (A(x), A(y)) = HilbertMultiply (Determinant (A),CrossProduct (x, y)).

Proof. A straightforward computation, but it has to be performed using geomet-ric arithmetic inside ECG. Nevertheless for readability we write it in normal


notation, writing for example x1 instead of X(x1) and x2 instead of Y (x).

A(x) =

(

a bc d

) (

x1

x2

)

=

(

ax1 + bx2

cx1 + dx2

)

A(y) =

(

ay1 + by2cy1 + dy2

)

A(x) ×A(y) =

(

ax1 + bx2

cx1 + dx2

)

×(

ay1 + by2cy1 + dy2

)

= (ax1 + bx2)(cy1 + dy2) − (cx1 + dx2)(ay1 + by2)

= bcx2y1 − adx2y1 + adx1y2 − bcx1y2

= (ad− bc)(x1y2 − x2y1)

= Determinant (A)(x × y)


The following lemma formalizes in ECG a well-known test for parallelism orcoincidence. Note how the use of linear transformations simplifies its proof.

Lemma 12.38. L = Line (a, b) and K = Line (c, d) are weakly parallel (thatmeans IntersectLines (L,K) is not defined) if and only if

CrossProduct (VectorSubtract (b, a),VectorSubtract (d, c)) = 0.

Proof. We note that if all four points are transformed by the same linear trans-formation, since lines are carried to lines, the transformed lines intersect if andonly if the original lines do, provided the determinant of the transformation isnonzero. By Lemma 12.37, the cross product is nonzero after such a transfor-mation if and only if it is nonzero before. Hence we may assume that a = 0 andb lies on Ray (0, 1). Then the cross product in the lemma is (in normal notation)b1(d2 − c2). Since b1 6= 0, the cross product vanishes if and only if d2 = c2;but that is exactly the condition IntersectLines (K,Line (0, 1)) to be undefined.That completes the proof of the lemma.

Lemma 12.39. Right (a, b, c) is invariant under rotations about b.

Proof. To study this issue we make use of “local coordinates” for the first time.We treat Line (b, a) as we formerly treated Line (0, 1), and temporarily consider0 as another name for b and 1 as another name for the point on Ray (b, a) suchthat αβ = 01. Let J = Perp(0,Line (b, a)) be the new y-axis, and let X andY be projections on Line (b, a) and on J . In order to define addition in thesenew coordinates we need to determine an orientation of J ; that is, we need todetermine a point on J to temporarily play the role of I. That should be a pointat distance 1 from b on the same side of Line (b, a) as c. It is not quite trivial toconstruct such a point, but the following script does the job:

Point NewI(Point a, Point b, Point c)

r = Y(c)

C = Circle(b,alpha,beta)

186 MICHAEL BEESON

L = Line(b,a)

K = Perp(a,L)

s = IntersectLineCircle1(K,C)

K = Circle(r,s)

I = IntersectCirclesSame(C,K,c)

return I

Then I and c are on the same side of Line (b, a) = Line (0, 1), so by Axiom H4,Right (1, 0, I) if and only if Right (a, b, c).

Now consider a rotation T , whose matrix in these coordinates is

T =

(

u Minus (v)v u

)

Consider first the case v > 0, i.e., a non-trivial rotation counterclockwise by lessthan 180 degrees. Let A = T (a) and C = Tc(c). We have to prove Right (a, b, c) ifand only if Right (A, b, C). Suppose Right (a, b, c). We will prove Right (A, b, C).We need to construct an auxiliary point d. We need d to be in the upper halfplane (the side of Line (0, 1) containing c) and also to be on the same side ofLine (b, A) as C. See Fig. 48 for an illustration.

Figure 48. Right (a, b, c) is invariant under rotation. abcchanges to abd, then to Abd, then to AbC.

b

bb

a

bc

bA

bC

b

q

bd

Since we are trying to prove Right (A, b, C), and Right is defined by an equality,we may legally argue by cases. Case 1 is when C lies in the lower half plane.Then segment AC meets the x-axis in a point q. We have A 6= q since A lies inthe upper half plane. Let d be the midpoint of Aq. Then d is in the upper halfplane. We claim d is on the same side of Line (b, A) as C. To prove that, supposedC meets Line (b, A). Then Line (b, A) and Line (b, C) coincide, so A × C = 0;but we have proved that rotations preserve cross products, so A×C = a×c 6= 0.Hence dc does not meet Line (b, A). Then by Lemma 9.83, d and C are on thesame side of Line (b, A). That completes the construction of the desired point din Case 1.


Case 2, C lies in the closed upper half plane. Then let d be the midpoint ofAC. Then d lies in the upper half plane, since its y-coordinate is the average ofa positive and a non-negative number. As in Case 1, d lies on the same side ofLine (b, A) as C.

Thus in either case we have the desired point d. Now we can argue as follows:

Right (a, b, c) by assumption

↔ Right (a, b, d)

by Axiom H4, since d and c are on the same side of Line (b, a)

↔ Right (A, b, d)

by Axiom H4, since a and A are on the same side of Line (b, d)

↔ Right (A, b, C)

by Axiom H4, since d and C are on the same side of Line (b, A)

Thus in either case we reach the conclusion Right (A, b, C). Since not not (Case 1or Case 2), we have ¬¬Right (A, b, C), and hence by the stability of equality andthe definition of Right as an equality, we have Right (A, b, C). Since we assumedRight (a, b, c), we have proved without an assumption that

Right (a, b, c) → Right (A, b, C).

Similarly, appealing to Axiom H5, we can prove Left (a, b, c) → Left (A, b, C).Hence ¬Left (A, b, C) → ¬Left (a, b, c). We wish to prove

Right (a, b, c) ↔ Right (A, b, C).

Since Right is stable, we are allowed to prove this by cases. If a, b and c arecollinear, then also A, b and C are collinear, so both sides are automaticallytrue. Otherwise Right (a, b, c) is equivalent to ¬Left (a, b, c) and we are done.That completes the proof in case v > 0, i.e., rotation T is a rotation by less than180 degrees.

To finish the proof, we decompose any rotation into a product of two rotations,each of which is by less than 180 degrees. Then we can apply the above argumentto each rotation separately. We need to prove formally that any rotation can beso decomposed. Given rotation

T =

(

u −vv u

)

with u2 + v2 = 1 and v < 0 we define

V =

√

1 − u

2

U = −√

1 + u

2(These formulas are based on the usual half-angle formulas for sine and cosine,but of course no trigonometry is involved here. How we thought of these formulasis not relevant for the proof.) Now define

S =

(

U −VV U

)

188 MICHAEL BEESON

Then routine algebraic computation (done using geometric arithmetic, here in-dicated with “normal” notation), shows that S2 = T . That completes the proofof the lemma.

Lemma 12.40. Right (a, b, c) is invariant under translations along Line (a, b).That is, if

A(x) = VectorAdd (x,HilbertMultiply (t,VectorSubtract (a, b)))

then Right (A(a), A(b), A(c))↔Right (a, b, c).

Proof. For any t, the point A(x) lies on Line (a, b) if x lies on Line (a, b), becauseit satisfies the equation of that line. We claim that Right (a, b, c) is equivalent toRight (a,A(b), A(c). By the stability of equality, we can argue by cases on thesign of t. If t = 0 there is nothing to prove. If t > 0 then A(a) lies on Ray (b, a),so by Lemma 9.89, Right (a, b, c) is equivalent to Right (A(a), b, c). Then A(b) lieson Ray (A(a), b), so a second application of Lemma 9.89 makes Right (A(a), b, c)equivalent to Right (A(a), A(b), c). Combining the two we have Right (a, b, c)equivalent to Right (A(a), A(b), c). Now we claim A(c) is on the same side ofLine (a, b) as c. By Lemma 9.83 it suffices to show that line segment cA(c) doesnot meet Line (a, b). For that it suffices to show that Line (c, A(c)) is parallel toLine (b, a). That follows from Lemma 12.38, since one easily computes that

CrossProduct (VectorSubtract (A(c), c),VectorSubtract (b, a)) = 0.

Hence A(c) is on the same side of Line (a, b) as c. Hence by Axiom H4,Right (A(a), A(b), A(c)) is equivalent to Right (A(a), A(b), c), which we have shownis equivalent to Right (a, b, c). That completes the case t > 0. The case t < 0 istreated similarly. That completes the proof of the lemma.

The following lemma is the computer-graphic definition of Right , but now“internalized” using geometric arithmetic:

Lemma 12.41. For a 6= b, we have Right (a, b, c) if and only if

CrossProduct (VectorSubtract (a, b),VectorSubtract (c, b)) ≥ 0.

Proof Since both Right and ≥ are stable, we can argue by cases. If a, b, andc are collinear then Right (a, b, c), and the cross product is zero, so the claimedresult is valid in that case. Therefore we can assume that a, b, and c are notcollinear. We also argue by cases according as b = 0 or b 6= 0. Assume b 6= 0.By Lemma 12.39, we can rotate abc about b until a lies on Line (0, b). Then,by Lemma 12.40, we can translate abc along Line (0, b) until b = 0. By thecited lemmas, Right (a, b, c) still has the same value as originally. Moreover, byLemma 12.37, the sign of the cross product in the lemma is also not changed.Hence, we can assume b = 0. Now, by Lemma 12.39 and Lemma 12.37, wecan rotate again, bringing a onto Ray (0, 1). Finally we can compute. We knowRight (1, 0, I) by Lemma 9.104. Hence Right (a, b, c) is equivalent to Y (c) > 0,by Axiom H4, since Y (c) > 0 if and only if c is on the same side of Line (0, 1)as I. Since b = 0, the cross product in the lemma is CrossProduct (a, c), whichby definition, and since Y (a) = 0, is HilbertMultiply (a, Y (c)). Since a lies onRay (0, 1), we have a > 0, so the cross product has the same sign as Y (c). HenceRight (a, b, c) and CrossProduct (a, c) > 0 are equivalent. Since we have assumed


a, b, and c are not collinear, the cross product is not zero, so we can replace >by ≥. That completes the proof.

Corollary 12.42. Right (a, b, c) is invariant under arbitrary linear transfor-mations with positive determinant.

Proof. By Lemma 12.41, Right (a, b, c) is equivalent to a (a − b) × (c − b) ≥0 (written in normal instead of geometric notation). By Lemma 12.37, thelatter is invariant under linear transformations with positive determinant. HenceRight (a, b, c) is also invariant. That completes the proof.

Lemma 12.43 (Triangle inequality). Norm (a+b) ≤ Norm (a)+Norm (b), withequality if and only if 0, a, and b are collinear, if and only if a× b = 0.

Proof. For readability we write ‖x‖2 instead of HilbertSquare (Norm (x)), a + binstead of Add (a, b), etc.

‖a+ b‖2 = ‖a‖2 + ‖b‖2 + 2DotProduct (a, b)

≤ ‖a‖2 + ‖b‖2 + 2‖DotProduct (a, b)‖≤ ‖a‖2 + ‖b‖2 + 2‖a‖‖b‖≤ (‖a‖ + ‖b‖)2

Taking SquareRoot of both sides yields the inequality. We have equality onlyif DotProduct (a, b) = ‖a‖‖b‖. A short computation shows this is equivalent toa× b = 0. We write a1 and a2 for X(a) and Y (a), etc.:

(DotProduct (a, b))2 = (a1b1 + a2b2)2

= a21b

21 + a2

2b22 + 2a1b1a2b2

‖a‖‖b‖ = (a21 + a2

2)(b21 + b22)

= a21b

21 + a2

2b22 + a1

2b21 + a2

1b22

Subtracting, we find equality in the lemma if and only if

2a1a2b1b2 = a22b

21 + a2

1b22

0 = (a2b1 − a1b2)2

0 = a× b

It only remains to show that a × b = 0 if and only if 0, a, and b are collinear.But b is on the line containing 0 and a if and only if b2a1 = a2b1 (that is, bsatisfies the equation of the line, which is x2a1 = a2x1). But that is exactly thecondition a× b = 0. That completes the proof of the lemma.

Lemma 12.44. B(a, b, c)↔Add (Distance (a, b),Distance (b, c)) = Distance (a, c).

Proof. Recall Distance (a, b) = Norm (VectorSubtract (b, a)). We have (in moreconventional notation)

c− a = (c− b) + (b − a)

‖c− a‖ = ‖(c− b) + (b− a)‖≤ ‖c− b‖ + ‖b− a‖ by the triangle inequality

190 MICHAEL BEESON

When the conventional notation is replaced by the geometrically defined conceptshere abbreviated, we have the statement of the lemma. That completes the proof.

Lemma 12.45. If center (Q) = center (K) and Radius (Q) = Radius (K) thenQ = K.

Proof. Suppose we can prove pointOnCircle (Q) = pointOnCircle (K). Then wehave

Q = Circle (center (Q), pointOnCircle (Q)) by Axiom CA6

= Circle (center (K), pointOnCircle (K))

=K by Axiom CA6 again

Therefore it suffices to prove pointOnCircle (Q) = pointOnCircle (K). The ideais that Axiom Cont 9 forces pointOnCircle (K) to lie in the same direction fromcenter (K), for any circle K. Consider the following construction script:

f(Circle K)

C = Circle(0,1)

p = pointOnCircle(C)

b = CircleCenter(K)

r = Radius(K)

s = MakePoint(HilbertMultiply(X(p),r),HilbertMultiply(Y(p),r))

z = VectorAdd(b,s)

return z;

Then we claim that f(K) = pointOnCircle (K). To prove this we will useAxiom Cont 9. Refer to Fig. 13. In the statement of Axiom Cont 9, take Cto be Circle (0, 1), and let p = pointOnCircle (C) and q = pointOnCircle (K).Let b = center (K). Lay off 0p on Ray (b, q), to produce point t on Ray (b, q)with bt = 0p. Then according to Axiom Cont 9, we have pt = 0b. Then 0ptbis a parallelogram; then b and t are respectively the images of 0 and p underthe translation by b. But translations take lines to lines; so Line (0, p) goesto Line (b, t) under translation by b. Indeed, translations take rays to rays, soRay (0, p) goes to Ray (b, t). The point s in the script for f lies on Ray (0, p) (sinceit satisfies the equation of Line (0, p) and r ≥ 0), so its image z = VectorAdd (b, s)under that translation lies on Ray (b, t). Since translations preserve distance, andDistance (0, s) = r, we have Distance (b, z) = r. By Lemma 12.28, bz = bq. Henceby Axiom C5, On (z,K). But Line (b, t) intersects K in at most two points, byLemma 9.51, and by Lemma 9.12, only one of those two points lies on Ray (b, t).Hence z = q as claimed. That completes the proof of the lemma.

The following lemma shows that Axiom Cont 9 really does specify whatpointOnCircle (C) is, relative to the arbitrary choice of pointOnCircle (Circle (0, 1)).

Lemma 12.46. Let b be the center of circle K, and let p = pointOnCircle (Circle (0, 1)).Then

pointOnCircle (K) = VectorAdd (b,ScalarMultiply (Radius (K), p))).

Proof. Refer to Fig. 13. Let q = pointOnCircle (K). Since what has to be provedis an equality, by the stability of equality we can argue by cases. First consider


the case when b = 0 and Radius (K) ≥ 1. Then let point t lie on Ray (b, q) withbt = 1; that point t is inside K because the radius of K is at least 1, so T(b, t, q)as in the hypothesis of Axiom Cont 9 (where T is non-strict betweenness). Thenby Axiom Cont 9, p and q are collinear; so X(p) is to X(q) as Y (p) is to Y (q).That is,

HilbertMultiply (X(q), Y (p)) = HilbertMultiply (X(p), Y (q)).

Since e = 0, we have

q′ := VectorAdd (b,ScalarMultiply (Radius (K), p))

= ScalarMultiply (Radius (K), p))

We are trying to show that q = q′. We will first show the sum of the squaresof the coordinates of both sides are the same. For readability we write this innormal notation, rather than in geometric arithmetic notation, using q1 = X(q)and q2 = Y (q). Here is the computation:

|q′|2 = HilbertSquare (Norm (ScalarMultiply (Radius (K), p)))

=

(

p1

√

q21 + q22

)2

+

(

p2

√

q21 + q22

)2

=

(

√

q21 + q22

)2

(p21 + p2

2)

= (q21 + q22)(p21 + p2

2)

= (q21 + q22) since p lies on the unit circle

At the last step we have used Lemma 12.31. The previous steps are valid becausewe have checked that the laws of EF work in geometric arithmetic. But the lastexpression is HilbertSquare (Radius (K)). We have now shown that |q′ = |q|; andq′ is a scalar multiple of p, it lies on Line (0, p); But q lies on Line (0, p) too.According to Axiom Cont 9, pt = ab; but in this case ab is a null segment, sop = t. Then B(b, t, q) means B(0, p, q). Since Radius (K) ≥ 0, q′ is a positivescalar multiple of p, so it lies on the same side of 0 as p does; but that is the sameside as q. Hence q′ and q are two points on Ray (0, q) with the same distancefrom 0. By Lemma 12.28, 0q = 0q′. Then by Lemma 9.5, q = q′ as desired.That completes the case when e = 0 and Radius (K) ≥ 1. The case when e = 0and Radius (K) < 1 is handled similarly, changing the roles of C = Circle (0, 1)and K.

We now take up the case when b 6= 0 and Radius (K) ≥ 1. Then we can find tas in the hypothesis of Axiom Cont 9 (note that the case of radius 1 is coveredbecause Axiom Cont 9 only requires t to be non-strictly between b and q). ThenAxiom Cont 9 tells us that ptba is a parallelogram. Note that the part about ptnot meeting ab is essential here. Hence t = VectorAdd (b, p). Then since tb = 0p,by Lemma 12.28 we have Distance (t, b) = 1. Since q lies on Ray (b, t) we have (innormal notation for readability, but it abbreviates geometric vector arithmetic)q − b = r(t − b), where r = |q − b| since |t − b| = 1. But since t = p and b = 0,we have q − b = rp. But that is (translated into normal notation) what had tobe proved. That completes the proof in the case b 6= 0 and Radius (K) > 1.

192 MICHAEL BEESON

Now consider the case when Radius (K) < 1. Then we interchange the rolesof C = Circle (0, 1) and K in Axiom Cont 9, and a similar calculation proves thetheorem in that case.

These cases are classically exhaustive; and in each case we have reached thedesired conclusion; hence we have proved not not the desired conclusion. Butsince the conclusion is an equality, and equality is stable, we then finally reachthe conclusion itself. That completes the proof of the lemma.

12.8. Interpreting field theory in ECG. We now define an interpretationfrom field theory to the language of ECG. That involves defining multiplication,addition, and square root in geometry. The definition of multiplication requiresconstructing a circle through three points, for which we use the strong parallelaxiom, so we do not obtain interpretations from the theory of weak Euclideanfields or Playfair fields into the corresponding geometries. We return to this issuein a subsequent section; in this section, we concentrate on interpreting EF+ inECG.

We begin by associating a term t of ECG to each term t of EF+. Of course wewant x, when x is a variable, to be a variable of ECG. One may be tempted tojust say x is x, since after all one can consider that the list of variables v1, v2, . . .is the same for any first-order theory; but we want to define our interpretation sothat it will be the inverse of the interpretation φ defined in the previous section.That interpretation takes the k-th point variable pk of ECG onto (p9k, p9k+1),etc. We therefore define x in a more complicated way. The reasons for arrangingthe definition in this way will become apparent in Lemma 12.59.

Recall that X(b) is a term giving the intersection point of Perp(b,Line (0, 1))with Line (0, 1); in other words the “x-coordinate of b”. Similarly Y (b) givesthe intersection point of Perp(b,K) with K = Perp(0,Line (0, 1)), that is, the“y-coordinate of b”. Here is the complete definition of x for variables x of EF+.

Definition 12.47.

v9k :=X(pk)

v9k+1 := Y (pk)

v9k+2 :=X(pointOn1 (ℓk))

v9k+3 := Y (pointOn1 (ℓk))

v9k+4 :=X(pointOn2 (ℓk))

v9k+5 := Y (pointOn2 (ℓk))

v9k+6 :=X(center (ck))

v9k+7 := Y (center (ck))

v9k+8 := Radius (ck)

Thus when x is a variable of EF+, x is generally not a variable, but a term ofECG. That term contains a variable, which we call x∗. Specifically

Definition 12.48. Let vk be the variables of EF+, and pk, ℓk, and ck thePoint, Line, and Circle variables of ECG. To each variable x of EF+ we asso-ciate a variable x∗ of ECG as follows:


If x is v9k+j then x∗ is pk if j = 0, 1, or ℓk if j = 2, 3, 4, 5, or ck if j =6, 7, 8. If x is a list of variables then x∗ is the list of u∗ for u in list x. A“group of variables” is either v9k, v9k+1, or v9k+2, v9k+3, v9k+4, v9k+5, orv9k+6, v9k+7, v9k+8.

Then x and y are in the same group if and only if x∗ is the same variable as y∗.The point of the definition is that the groups will be used to represent coordinatesspecifying points, lines, and circles, respectively. It takes two coordinates for apoint, four for a line, and three for a circle (center and radius).

Here is the definition of t for any term t of EF+, or any list of two, three orfour terms:

Definition 12.49.

0 is 0 (another name for α)

1 is 1 (another name for β)

γ1 is X(γ) where X is as defined just above

γ2 is Y (γ) where Y is as defined just above

γ3 is X(pointOnCircle (Circle (0, 1)))

x is as defined above, when x is a variable

t+ s is Add (t+ s)

t · s is HilbertMultiply (t, s)√t is SquareRoot (t)

1/t is Reciprocal (t)

(a, b) := MakePoint (a, b)

(a, b, c, d) := Line (MakePoint (a, b),MakePoint (c, d))

(a, b, c) := Circle (MakePoint (a, b),

MakePoint (Add (a,HilbertMultiply (γ3, c)),

Add (b,HilbertMultiply (SquareRoot (Add (1,

Minus (HilbertSquare (γ3)))), c))))

In the last line, c is the radius of the circle, so to get a point on the circle,we take a point one radius away from the center, at an angle whose cosine isγ3. Since the term given there is a bit long, we give an equivalent constructionscript (but with bars omitted):

f(Point a, Point b, Point c)

e = MakePoint(a,b) // the center of C

x = Add(a,HilbertMultiply(gamma3,c))

// x-coordinate of pointOnCircle(C)

s = Sqrt(Add(1,Minus(HilbertSquare(gamma3)))) // sine of the angle

y = Add(b,HilbertMultiply(s,c)) // y-coordinate of pointOnCircle(C)

p = MakePoint(x,y) // pointOnCircle(C)

return Circle(e,p)

194 MICHAEL BEESON

We next want to define φ for every formula φ of EF+. The definition of ∀xφ(x)is somewhat complicated. There is only one type of variable in EF, for fieldelements; but there are three types of variables in ECG, and the correspondencewe have already set up uses the variables of EF in different ways, according totheir index (subscript) in the official list of variables. Recall that of the ninevariables starting with v9k, the first two represent (under the interpretation)coordinates of a point pk, the next four represent coordinates of two points ofthe line ℓk, and the last three represent the center and radius of a circle ck. Thefollowing definition is designed to make φ 7→ φ the inverse of that interpretation.

Definition 12.50.

P (t) is ¬B(t, 0, 1) ∧ t 6= 0)

t = s is t = s

t ↓ is t ↓φ ∧ ψ is φ ∧ ψφ ∨ ψ is φ ∨ ψ

φ → ψ is φ → ψ

¬φ is ¬ φ∀xφ is ∀x∗ φ∃xφ is ∃x∗ φ

The strange thing about this definition is that different variables x can cor-respond to the same x∗. For example, v9k and 9k+1 both correspond to pk.Hence if a formula has the form ∀v9k∀v9k+1φ, its interpretation is going to gettwo quantifiers over pk. That does make sense: when we introduce an arbitrarypoint to represent an arbitrary field element, the other coordinate of the pointis not yet used. These double, triple, or quadruple quantifiers can of course beerased (up to provable equivalence), so we have

∀v9k∀v9k+1φ ↔ ∀pkφ

∀v9k+2∀v9k+3φ∀v9k+4φ∀v9k+5φ ↔ ∀ℓkφ∀v9k+6∀v9k+7∀v9k+8φ ↔ ∀ckφ

∃v9k∃v9k+1φ ↔ ∃pkφ

∃v9k+2∃v9k+3φ∃v9k+4φ∃v9k+5φ ↔ ∃ℓkφ∃v9k+6∃v9k+7∃v9k+8φ ↔ ∃ckφ

It might seem natural to take this as the definition, but then one has to worryabout defining the interpretation when not all the variables in the same group arequantified; in essence the definition given above says, when some of the variablesin a group are not quantified over, the right hand side is still the same as itwould be if they all were quantified over.


Example. Suppose we start with the formula φ equal to ∃x(on (x, L)). Let ussuppose L is actually the variable ℓ3 and x is the variable p2. We compute

φ ↔ (∃ℓ3(on (p2, ℓ3))

↔ ∃v9·3+2, v9·3+3, v9·3+4, v9·3+5(on(p2, ℓ3)

↔ ∃v29, v30, v31, v32((p2 − ((ℓ3)1, (ℓ3)2)) × (p2 − ((ℓ3)3, (ℓ

3)4)) = 0)

(Recall that the interpretation uses a cross product to interpret on .) Nowp2 = (v18, v19), and ℓ3 = (v29, v30, v31, v32). Then

φ ↔ ∃v29, v30, v31, v32(((v18, v19) − (v29, v30)) × ((v18, v19) − (v31, v32)) = 0)

Now we interpret back into ECG. We want to compute φ and get back theoriginal φ. The sequence of quantifiers ∃v29, v30, v31, v32 will be converted backto ∃ℓ3, so we get

φ ↔ ∃ℓ3ψwhere ψ is the equation

((v18, v19) − (v29, v30)) × ((v18, v19) − (v31, v32)) = 0.

Let us take this piece by piece. We have

(v18, v19) = MakePoint (v18, v19)

= MakePoint (X(p2), Y (p2))

= p2 by Lemma 10.10

Similarly

(v29, v30) = MakePoint (v29, v30)

= MakePoint (X(pointOn1 (ℓ2)), Y (pointOn1 (ℓ2)))

= pointOn1 (ℓ2) by Lemma 10.10

Thus the variables come out right in ECG. That was our point for now; toexplain why the quantifiers are handled as they are in Definition 12.50. Butsince we started this example, we remark on how it will finish: the subtractionand cross product will be converted to arithmetic between points using Add ,Minus , and HilbertMultiply . What we get back is thus not the original φ, buta formula in geometry expressing that the cross product used to interpret on iszero (not a formula in EF, a formula in ECG); and to prove that it is equivalentto the original φ, we will need to give geometric proofs of some formulas ofalgebra. That will be done below; for now we are just motivating the definitionof φ. That completes the example.

Next we have to investigate how substitution commutes with this interpre-tation. First we note that if x is a variable, generally x is a term of the formX(MakePoint (a, b)) or Y (MakePoint (a, b)), where a and b are variables of ECG;hence by Lemma 10.10, x is provably equal to a variable; moreover, that variableis a subterm of x, which could be specified by a long definition by cases. Wetherefore define A[x := s] to mean the result of substituting s for the variableequal to x in A. In other words, we can substitute for x even though technicallyx is not a variable.

196 MICHAEL BEESON

Lemma 12.51. For terms r of EF+, we have r[x := t] = r[x := t].

Proof. If x and y are different variables of EF, then r and x may contain morethan one variable (each), but these variables will not clash: suppose x is v9J andy is v9K+1, with 0 ≤ j < 9 and 0 ≤ k < 9. Then the variables of x are p9J andp9J+1, and the variables of y are p9K and p9K+1, so they do not clash. Similarlyfor every other possible combination of indices j and k where x is v9J+j and yis v9K+k; it would take a long time to write out all the combinations, but thedefinition has been concocted to make this true.

We proceed by induction on the complexity of r. If r is a constant or a variableother than x, the left side is just r, and the right side is r too, since r does notcontain x. If r is the variable x, then r[x := t] = t and r[x := t] is also t.That takes care of the base case of the induction. Since substitution commuteswith function symbols, there is nothing to prove for the induction step. Thatcompletes the proof.

Example. We give an example to illustrate the previous lemma. Suppose r isv18 and x is also v18 and t is v20. Then

r = v18 = X(p2)

x = r = X(p2)

t = v20 = X(pointOn1 (ℓ2))

r[x := t]i = v20

r[x := t] = v20 = X(pointOn1 (ℓ2))

On the other hand

r[x := t] =X(p2)[X(p2) : X(pointOn1 (ℓ2))]

=X(pointOn1 (ℓ2))

= r[x := t] as calculated above

The example is meant to illustrate why r does not contain x: namely, all thevariables mentioned in x, where x is vk, have subscripts between 9k and 9k+ 8,so they do not clash with variables arising from y where y is another variable ofEF.

Lemma 12.52. Let φ be a formula of EF. Then ECG proves φ[x := t] is equiv-alent to φ[x := t].

Proof. By induction on the complexity of φ. When φ is atomic, say r = s, wehave

φ[x := t] ↔ r = s[x := t]

↔ r[x := t] = s[x := t]

↔ r[x := t] = s[x := t]

↔ r[x := t] = s[x := t] by Lemma 12.51

↔ (r = s)[x := t]

↔ φ[x := t]


as required.When φ is atomic, of the form r ↓, we have

φ[x := t] ↔ r ↓ [x := t]

↔ r[x := t] ↓↔ r[x := t] ↓ by Lemma 12.51

↔ r ↓ [x := t]

↔ φ[x := t]

as required.When φ is atomic, of the form P (r), we have

φ[x := t] ↔ P (r)[x := t]

↔ P (r[x := t])

↔ 0 < r[x := t]

↔ 0 < r[x := t]

↔ (0 < r)[x := t]

↔ P (r)[x := t]

↔ φ[x := t]

as required.For the induction step, we consider the case of φ ∧ ψ; the other propositional

connectives are treated in the same way.

(φ ∧ ψ)[x := t] ↔ φ[x := t] ∧ ψ[x := t]

↔ φ[x := t] ∧ ψ[x := t]

↔ φ[x := t] ∧ ψ[x := t] by the induction hypothesis

↔ (φ ∧ ψ)[x := t]

↔ φ ∧ ψ[x := t]

as required.Now consider the case of a formula that begins with a quantifier. Refer to

Definition 12.50 for the definition of φ in this case; it is complicated. In particularwe do not have

∀yφ ↔ ∀yφsince in general y is not even a variable; it is a term like X(pk). Instead, we have

∀yφ ↔ ∀y∗φ,where y∗ is given in Definition 12.48. Then

(∀yφ)[x := t] ↔ ∀y(φ[x := t])

↔ ∀y∗φ[x := t]

↔ ∀y∗φ[x := t] by the induction hypothesis

↔ ∀yφ[x := t]

as required.

198 MICHAEL BEESON

The case of an existential quantifier is treated exactly the same, just changing∀ to ∃. That completes the proof of the lemma.

Theorem 12.53. Euclidean field theory is interpretable in ECG. That is, ifEF+ proves φ then ECG proves φ.

Proof. We first show that on (t,Line (0, 1)) is provable for every term t of EF.This is proved by induction on the complexity of the term t. When t is 0 or 1, tis α or β, and these both lie on Line (0, 1), since α = 0 and β = 1. When t is γor a variable, t is produced by the coordinate functions X and Y , whose valuesby definition lie on Line (0, 1). For all other terms, t is either produced directlyby X and Y , or is produced by Add , HilbertMultiply , SquareRoot , or Reciprocal ,all of which are defined so that if their arguments lie on Line (0, 1), then so dotheir values. That completes the proof that t lies on Line (0, 1) for all terms t.

In case t is a list of two or three terms, then t is (provably) a point or circle,respectively, according to the definition of t, since Circle3 is everywhere defined,and MakePoint is everywhere defined since we are assuming the strong parallelaxiom. In case t is a list of four terms such that L(t), then t is a line, becausethe two points specified by the first two and last two members of the list will bedistinct, by definition of L.

We have already checked that the interpretations of the commutative ringaxioms are provable, using the strong parallel axiom of ECG. Technically, weought to exhibit formal proofs in ECG of the (interpretations of) the ring axioms,and it may well be possible to produce such proofs using a theorem-prover orproof-checker, but we rely on the reader to be convinced that such proofs existbased on an examination of the detailed constructions given above.

We turn to the verification of the (interpretations of the) axioms involvingP (x). Let L = Line (0, 1). We note that by Lemma 9.91, P (x) is equivalent (forx on L) to Right (x, 0, I), where as above I is a point on the line K perpendicularto L at 0 such that Right (1, 0, I) and 0I = 01. Recall also that we definedx < y in Section 9.3 to mean Left (x, y, I), so 0 < x means Left (0, x, I), which isequivalent to Right (x, 0, I), and hence to P (x).

We first check axiom EF2, which says that the sum and product of positiveelements are positive. First we check the sum. Suppose 0 < x and 0 < y.Since 0 < x we have Right (x, 0, I). To construct Add (x, y), we first rotate xcounterclockwise 90 degrees to a point u. Specifically, we first construct R suchthat B(0, x, R) and B(0, y, R), and set u = Rotate (R, 0, I, x). Then u is onRay (0, I) by the properties of Rotate . Then let H be perpendicular to L at y;we project u onto the point v, the foot of the perpendicular from u to H . Thenuv and L are both perpendicular to K, so by Lemma 9.48, uv does not meet L.Then u and v are on the same side of L, by Lemma 9.83. Let D be the projectionof I onto H ; then we claim v lies on Ray (0, D). Once we prove that, we arefinished, since then x + y lies on Ray (y,R) by the properties of Rotate , and itfollows that x+ y > y > 0. Suppose then that v does not lie on Ray (0, D); thenv and D are on opposite sides of L, since y is between them. ID does not meetL, since by Lemma 9.48 and the fact that ID and L are both perpendicular toK. Hence by Lemma 9.83, D and I are on the same side of L. Since v and D areon opposite sides of L, by Lemma 9.58, v and I are on opposite sides of L. Since


I and u are on the same side of L, u and v are on opposite sides of L. But uvand L are both perpendicular to K, and they do not coincide since x > 0. Henceby Lemma 9.48 they are parallel, so uv cannot meet L, which is a contradiction,since u and v are on opposite sides of L. That shows that v cannot fail to lie onRay (0, D). But since lying on a ray is defined by a negation, v lies on Ray (0, D).That completes the verification that Add(x, y) is positive if x and y are positive.

Next we check that HilbertMultiply (a, b) is positive if a and b are positive.Refer to Fig. 6. Let C be the circle passing through a, b, and I. We mustshow that if a and b lie on Ray (0, 1) and are not zero, then the fourth pointof intersection z lies on Ray (0, I) and is not 0. Since lying on a ray is definedby a negation, we can prove this by cases. The first case is z = I, i.e., a andb are reciprocals and C is tangent to K at I. Then z = I lies on Ray (0, I).The second case is z = 0; then there are three points of C on L, contradictingLemma 9.52. Now suppose z 6= 0 and z 6= I. If z and I are on opposite sides ofL, then L meets chord zI of circle C. By Lemma 9.76, L meets C in two pointson opposite sides of K, which coincides with Line (z, I). But it meets C in a andb, which are on the same side of K, and by Lemma 9.52, those are the only twointersection points. This is a contradiction. Hence z and I are not on oppositesides of L; that is, ¬B(z, 0, I). Hence z lies on Ray (0, I), which was what we hadto show. That completes the verification that the product of positive elementsis positive.

Next we check axiom EF3, which is x + y = 0 → ¬ (P (x) ∧ P (y)). LetL = Line (0, 1). The interpretation of this axiom is

on(x, L) ∧ on(y, L) → (Add (x, y) = 0 → ¬ (P (x) ∧ P (y))

Assume x and y are on L and Add (x, y) = 0. Then y = Minus (x), as can beproved as follows using the laws of ring theory that we have already checked inLemma 12.9:

Add (x, y) = 0

Add (Minus(x), Add(x, y)) = Add (Minus(x), 0)

Add (Add (Minus(x), x), y) = Minus (x)

Add (0, y) = Minus (x)

y = Minus (x)

Thus we have to prove ¬ P (x)∧ P (Minus (x)). Suppose P (x)∧ P (Minus (x)); wemust derive a contradiction. Then ¬B(x, 0, 1) and x 6= 0 and ¬B(Minus (x), 0, 1).Then by definition of SameSide , x and Minus (x) are on the same side of 0, i.e. onthe same side of the line perpendicular to L at 0. But by definition of Minus (x),if x 6= 0 then B(x, 0,Minus (x)), so x and Minus (x) are on opposite sides of 0.Then by the plane separation theorem (Lemma 9.58), x and x are on oppositesides of L; that is, for some y we have B(x, y, x), contradicting Axiom B1-ii.That completes the verification of axiom EF3.

Next we check EF4, which is x+ y = 0∧¬P (x)∧¬P (y) → x = 0. As above,Add(x, y) = 0 implies y = Minus (x), so what has to be proved is that, for x onL, ¬ (x > 0) and ¬ (Minus (x) >)0 implies x = 0. Since equality is stable, wecan prove x = 0 by contradiction, so assume x 6= 0 and ¬ (x > 0) and ¬ (y > 0).

200 MICHAEL BEESON

Then with y = Minus (x) we have B(y, 0, x) and B(x, 0, 1) and B(y, 0, 1). SinceB(x, 0, 1) and B(x, 0, y), 1 and y are on the same side of 0. Hence by Lemma 9.58,¬B(1, 0, y). But this contradicts B(y, 0, 1). That completes the verification ofaxiom EF4.

Now we turn to axiom EF5, which is x+ y = 0∧¬P (y) → ∃z(z · z = x). Theinterpretation is that, for x, y on L, if Add (x, y) = 0∧¬P (y) then for some z onL, HilbertMultiply (z, z) = x. Suppose x and y are on L and Add (x, y) = 0; theny = Minus (x). Suppose ¬P (y); then ¬y > 0, that is, Minus (x) ≤ 0; that is x lieson Ray (0, 1). Let SquareRoot be defined by the construction script in Section5.5. Let z = SquareRoot (x). Then by Lemma 12.10, HilbertMultiply (z, z) = x.That completes the verification of axiom EF5.

There are axioms about the special constants γ2 and γ3. The axiom γ2 ≥ 0,or officially ¬P (−γ2), becomes 0 ≤ γ2, which is 0 ≤ Y (γ); according to thedefinition of ≤ in ECG, this means Right (Y (γ), 0, I). By Lemma 10.9 this isprovable. The axiom γ2

3 ≤ 1 becomes HilbertSquare (γ3) ≤ 1. By definition,γ3 = X(pointOnCircle (C)), where C = Circle (0, 1). Let p = pointOnCircle (C);then we have to prove HilbertSquare (X(p)) ≤ 1. By Lemma 12.31, we have

Add (HilbertSquare (X(p)),HilbertSquare (Y (p))) =

Add (HilbertSquare (X(1)),HilbertSquare (Y (1))).

But X(1) = 1 and Y (1) = 0, so the right-hand side is equal to Add (1, 0),which is 1, because we have already checked the commutative ring axioms. ByLemma 12.12, HilbertSquare (Y (p)) ≥ 0. Adding HilbertSquare (X(p)) to bothsides, which is legitimate since a ≤ b is equivalent to Add (b,Minus (a)) ≥ 0, andwe have already verified the field laws, we obtain

Add (HilbertSquare (X(p),HilbertSquare (Y (p)) ≥ HilbertSquare (X(p)).

But the left side is 1, so we have 1 ≥ HilbertSquare (X(p)) as desired.Next we verify Markov’s principle ¬¬P (x) → P (x). The interpretation of

this is

on(x,Line (0, 1)) → (¬¬(¬B(x, 0, 1) ∧ x 6= 0) → (¬B(x, 0, 1) ∧ x 6= 0)).

The double negation can be pushed inside the conjunction and dropped, makingthe statement equivalent to one of the form A → (B → B), which is provable.

We now turn to the verification of Axiom EF7′, which is

x 6= 0 → (x · (1/x) = 1),

i.e. nonzero elements have reciprocals. The interpretation of this is that ifa is a point on L different from 0, then there is a point b on L such thatHilbertMultiply (a, b) = 1. We take b to be the point Reciprocal (a), whereReciprocal is defined by the construction script given just before Lemma 12.14.Then Lemma 12.14 gives the desired conclusion.

We have now verified the interpretation of every non-logical axiom of EF+.Let us check the logical axioms of LPT. The first one is ∀xA ∧ t ↓ → A[x := t].


We have

∀xA ∧ t ↓ → A[x := t]

↔ ∀xA ∧ t ↓ → A[x := t]

↔ ∀xA ∧ t ↓ → A[x := t] by Lemma 12.52

Now in the formula ∀xA, A might begin with other universal quantifiers overvariables in the same group as x, where “same group” is defined just after Defi-nition 12.47. Throwing these variables into a list y we can write ∀xA in the form∀x∀y B. Then Definition 12.50 tells us

∀x, y B ↔ ∀x∗BThen we can continue the argument:

∀xA ∧ t ↓ → A[x := t]

↔ ∀x, y B ∧ t ↓ → A[x := t]

↔ ∀x∗B ∧ t ↓ → A[x := t]

↔ ∀x∗B ∧ t ↓ → ∀y B[x := t]

↔ ∀x∗B ∧ t ↓ → (∀y∗ B)[x := t]

But y∗ and x∗ are the same variable (which was why we did not write y∗ onthe left); so the last formula is another instance of the axiom schema whoseintepretation we are verifying. That completes the verification of that axiom.

The axiom A[x := t] ∧ t ↓ → ∃xA is treated similarly; the only differencesare that we have ∃ instead of ∀, and the quantifier appears on the right insteadof the left of the implication sign.

That completes the base case of an induction on the length of proofs inEF+(except for any logical axioms there may be). In addition to checking thelogical axioms, for the induction step we have to check that the rules of inferencepreserve the interpretation. For example if modus ponens is a rule of inference,we have to check that if φ and φ → ψ are provable, then so is ψ. That is evidentsince φ → ψ is just φ → ψ. Actually, we did not specify the exact rules ofinference of any of our theories, having been content to mention “multi-sortedintuitionistic predicate calculus with the logic of partial terms.” But since thedefinition of φ commutes with the propositional connectives, and since it al-most commutes with quantifiers (except for changing variables x to x⋆, it doesnot matter whether we choose, for example, a Gentzen-style formulation or aHilbert-style formulation; in either case the induction step is just commutingthe interpretation with the connectives or quantifiers to get another instance ofthe same inference rule or logical axiom.


12.9. Faithfulness of the interpretations. We have now proved that theinterpretation φ from EF to ECG is sound, and the interpretation φ from ECGto EF+ is sound; that is, if φ is provable then its interpretation is provable. Aninterpretation is called faithful if the interpretation of a formula is provable ifand only if the formula is provable. In this section, we prove that both the

202 MICHAEL BEESON

interpretations we have introduced are faithful. The reason is simple: they areinverses.

Before embarking on the details, we comment on the significance of this result.It shows that geometry is completely equivalent to algebra: each is reduced tothe other, and nothing is added or left out by the reductions. With classicaltheories, the issue of faithfulness is handled by model theory and an appeal toGodel’s completeness theorem. With intuitionistic theories, we do not have thatoption. Besides, even for classical theories, it is of interest to have an algorithmfor translating, say, an algebraic proof into an axiomatic geometric proof. Forexample, there are various computer programs that “prove” geometry theoremsby translating them to algebra, and then using computer algebra techniques.Using the work in this section, it would be possible to write computer programsto convert those algebraic “proofs” (which are really calculations) first to formalproofs in field theory, and then to formal proofs in ECG (or ECG plus clas-sical logic). That possibility, however, is not the main reason for studying thefaithfulness of the interpretations. It is simply to check that these things workout as expected; any failure would be a surprise and would be either a “bug” inthe formal systems or the definitions of the interpretations, or would representa new and scientifically interesting phenomenon. The latter is illustrated by thefact that the interpretations are not faithful for Euclid 5 or Playfair, but onlyfor the strong parallel axiom.

The essence of what is required to prove the interpretations faithful is that wehave to formalize analytic geometry in ECG. We had to define arithmetic justto define the interpretation from field theory to geometry, but now we will haveto show, for example, that the points on a line are exactly those satisfying theequation of the line. Euclid proved the Pythagorean theorem, but in a differentsense of “area” than we now use; we will need to derive the modern form, whichuses arithmetic to calculate the areas.

We now begin the technical preparations. We start with some lemmas thathelp us compute t for various terms t.

Lemma 12.54. Perp(p, L) = (q1, q2, r1, r2), where (q1, q2) and (r1, r2) are thecoordinates of the two intersection points of circles C and K, whose centers aregiven by the intersection points of L and the circle centered at p with radiusr = |a− p| + |a− b| where a = (L

1, L2) and b = (L

3, L4); the circles C and K

pass through each other’s centers.

Proof. Perp(p, L) is the line defined by the two intersection points of two cir-cles centered on L and passing through p. Since the interpretation preservesincidence the lemma follows. A detailed proof would require computing the

interpretation of each line of the script defining Perp; we omit those details.That completes the proof.

Lemma 12.55. The interpretation of being perpendicular is that the interpretedline vectors have dot product zero. Formally,

(L ⊥ K) = ((L1 − L

3), (L2 − L

4)) · ((K1 −K

3 ), (K2 −K

4 )) = 0

Proof. By definition of perpendicular, L and K are perpendicular if there existpoints a, m, and b on L and c on K such that m is also on K and triangle amc is


congruent to bmc. Under the interpretation we have |a−m| = |b−m|, etc.The proof then reduces to an exercise in the computation of the dot product,which we omit. That completes the proof.

Lemma 12.56. EF+ proves

a 6= b ∧ c 6= b ∧ on (d,Line (b, c))

→ Rotate (a, b, c, d) = A

(

d1 − b1d2 − b2

)

+

(

b1b2

)

where A is the matrix of the rotation; specifically

A =

(

A11 A12

A21 A22

)

where

A11 = A22 =(a − b) · (c − d)

|a − b||c − d|

A12 = −A21 =(a − b) × (c − d)

|a − b||c − d|Remark. We cannot mention cos θ and sin θ in EF+, but intuitively, the entriesof the rotation matrix involve those quantities, which are here computed in termsof dot product and cross product. Note that no case distinctions are needed.

Proof. The denominators are nonzero by the hypotheses a 6= b and c 6= b; so atleast the matrix A is defined. We claim that the determinant of the matrix A is1. If we define

cos θ =(a − b) · (c − d)

|a − b||c − d|

sin θ =(a − b) × (c − d)

|a − b||c − d|then the claim boils down to sin2 θ + cos2 θ = 1. Since we are working in EF+,this argument is only motivation, not proof. The actual proof consists in writingthe identity out in terms of the components of a, b, c, and d and simplifying.What has to be verified is an instance of the following proposition: if u and vare unit vectors, then (u · v)2 + (u × v)2 = 1. Writing this out by componentswe have

(u · v)2 + (u× v)2

= (u1v1 + u2v2)2 + (u1v2 − u2v1)

2

= (u21v

21 + 2u1v1u2v2 + u2

2v22) + (u2

1v22 − 2u1v2u2v1 + u2

2v21)

= u21v

21 + u2

2v22 + u2

1v22 + u2

2v21

= (u21 + u2

2)(v21 + v2

2)

= 1 since u and v are unit vectors

Hence the determinant of A is indeed 1.According to Lemma 9.101, the value x of Rotate (a, b, c, d) is a point on

Line (b, a) such that bx = bd and Right (d, b, x) if and only if Right (c, b, a).There is only one such point x, so if we show the point defined in the lemma

204 MICHAEL BEESON

and asserted to be Rotate (a, b, c, d) has those two properties, then indeed it isRotate (a, b, d, c).

Let us write Ax for the result of multiplying matrix A by the column vectorwith entries x1 and x2. Then the point defined in the lemma is z = A(d−b)+b.We have

|z − b| = |(A(d − b) + b) − b|= |A(d − b)|= |d − b| since A has determinant 1

Since A is a linear transformation, it takes d to a point on the line determinedby b and a. There are only two points on that line at the distance |d − b|from b. One of them is z and the other is on the other side of b, so just oneof them satisfies the second condition of the lemma. Hence if we show that zsatisfies that condition, we will be finished.

Right (c, b, a) is equivalent to (c − b) × (a − b) > 0, as we verified inthe proof of Theorem 12.21; and that condition is equivalent to A21 > 0. NowRight (d, b, z) is equivalent to (d − b)× (z− b) > 0, so we calculate that crossproduct. We have z = A(d − b) + b. For simplicity we set u = d − b; thenz − b = Au. Then

(d − b) × (z − b) = u×Au

= u1(Au)2 − u2(Au)1

= u1(A21u1 +A22u2) − u2(A11u1 +A12u2)

= A21u21 −A12u

22 + u1u2(A22 −A11)

= A21u21 −A12u

22 since A11 = A22

= 2A21(u21 + u2

2) since A21 = −A12

≥ 0 if and only if A21 > 0

Hence Right (c, b, a) if and only if Right (d, b, z). Hence z is the correct point,i.e. z = Rotate (a, b, c, d) as claimed. That completes the proof of the lemma.

Lemma 12.57. EF+ proves the following:

I = (0, 1)

MakePoint (x, y) = (x, y)

Proof. We follow the script given in Definition 9.103 to define I. For conveniencewe repeat that script here; you may wish to refer to Fig. 31, which we do notrepeat.

I()

L = Line(0,1)

J = Perp(0,L)

K = Perp(gamma,J)

p = IntersectLines(J,K)

return IntersectLineCircle2(Line(0,p),Circle(0,1))


We have

L = (0, 0, 1, 0)

We now calculate J according to Lemma 12.54. The circles C and K mentionedin that lemma have radius 1, and are centered at (−1, 0) and (1, 0). In order topass through each other’s centers they have radius 2. Therefore they meet they-axis at (0,±

√3). Hence

J = (0,−√

3, 0,+√

3)

The order of the two points represented in J is determined by the script forPerp, although it is not stated explicitly in Lemma 12.54 and does not matter.Now we calculate K. Recall γ = (γ1, γ2), and γ2 > 0 is an axiom of EF+.This time, the radius of the circle used to find the centers of the new C and Kis

r =

√

(γ2 +√

3)2 + γ21 + 2

√3

so the coordinates of the centers are(

0, γ2 ±√

r2 − γ21

)

and the intersection points of the two circles are (±c, γ2) for a certain c givenby a solving a quadratic equation. The value c is thrown away anyway when wecalculate p in line 4 of the script, and we find p = (0, γ2). Since γ2 > 0 thenin the last line of the script the return value is (0, 1). That is the value of I.That proves the first claim of the lemma. (It may seem like a lot of work foran obvious result, but the point is that proving these things formally shows thatthe machinery is set up correctly.)

Consider MakePoint (a, b); recall from Definition 10.8 that MakePoint is de-fined by the following script:

MakePoint(Point x, Point y)

L = Line(0,1)

J = Perp(0,L)

U = Perp(x,L)

z = Rotate(1,0,I,y)

V = Perp(z,J)

return IntersectLines(U,V)

We will go through the script line by line and calculate the interpretationof each line. In line 1 we have L = (0, 0, 1, 0). In line 2 as above we have

J = (0,−√

3, 0,√

3). In line 3 we have U = (x,−c, x, c) where again c could beexplicitly calculated but does not matter. According to Lemma 9.101 we havez = (−y2, y1). According to Lemma 9.56, J ⊥ L and U ⊥ L. Let us abusethe dot-product notation by writing L · K for the dot product mentioned inLemma 12.55. Then by that lemma we have J ·L = 0 and U ·L = 0. HenceU and J are in the same direction. Since z lies on V by Lemma 9.56, all pointson V have the same second coordinate as z, which is y1 , and since U contains

206 MICHAEL BEESON

x, all points on U have the same first coordinate as x, namely x1. Hence theintersection point has coordinates (x1, y

1). That completes the proof.

The next lemma says that that EF proves the correctness of the algorithmsHilbertMultiply , Add , and SquareRoot .

Lemma 12.58. EF+ proves the following, assuming a2 = b2 = 0:

HilbertMultiply (a, b) = (a1 · b1, 0)

Add (a, b) = (a1 + b1, 0)

Minus (a) = (−a1, 0)

Reciprocal (a) = (1/a, 0)

¬P (−x1) → SquareRoot (x) · SquareRoot (x) = (x1, 0)

Proof. First we consider the term Reflect (a, L). That is a term in EF thatgives the reflection of a in line L. In particular

Reflect (a,Line (0, I)) = (−a1, a2).We start with the theorem about Minus .

Minus (a) = Minus (MakePoint (X(a), Y (a))

= Reflect (MakePoint (X(a), Y (a)),Line (0, I))

Now apply to both sides.

Minus (a) = Reflect (a,Line (0, I))

Minus (a) = Reflect (MakePoint (X(a), Y (a)),Line (0, I))

= Reflect (MakePoint (a0, a1), (0, 0, 0, 1))

= Reflect ((a0, a1), (0, 0, 0, 1))

= (−a0, a1) as discussed above

= −a

as claimed.HilbertMultiply (a, b) is a term of EF that gives the fourth intersection point

of circle C with the y-axis, where the first three intersection points are (a, 0),(b, 0), and (0, 1). We have to prove in EF+ that that fourth intersection pointis (0, a · b). The way the term HilbertMultiply is defined, the formula will firstconstruct the perpendicular bisector of the line segment from (a, 0) to (0, 1), andthe line x = (a + b)/2, and then find the center of the circle at the intersectionpoint of those lines, which will be

e = ((a · b + 1)/2, (a + b)/2).

The radius of the circle is |e − 1|. Then the fourth intersection point is (0, y)where y solves

|e− (0, 1)|2 = |(0, y) − e|2.That is,

y −(

ab+ 1

2

)2

=

(

1 − ab+

2

)2


It is an easy piece of algebra to check that the solutions are just y = 1 andy = ab; that algebra can be done in EF, since only the ring laws are involved.(Denominators of 2 can be cleared).

Similarly, Reciprocal (a) is the second point on the x-axis lying on the circlethat is tangent to the y-axis at (0, 1), and passes through (a1, 0). That circle’scenter is at (a+ (1/a))/2, 1)

The term Add involves Rotate , so we first check that Rotate really performs arotation. We only need rotations by plus or minus ninety degrees to define Add ,using angles with one side on the x-axis. Rotate is defined by bisecting an angleand dropping two perpendiculars, so it has to be checked that the constructionwe gave for bisecting an angle, at least, a ninety-degree angle with its sides onthe x and y axis, is analytically correct. That algorithm involved constructingthe points at distance one from the vertex (that would be (±1, 0) and (0,±I)),and then drawing the circles with those centers passing through the other point,and then connecting their intersection points, which will be in the four cases athand will be points on the lines y = ±x, so the angle bisectors will be those lines;then when we drop perpendiculars from points on the x-axis or y-axis to thoselines, and then again to the target axis, the effect will be the same as a rotation.For example, Rotate (1, 0, I, x) will be (−x2, x1). To describe this analyticallywe can represent rotations as two-by-two matrices and show that

Rotate (1, 0, I, x) =

(

0 −11 0

) (

x1x2

)

.

Rotate (I, 0, 1, x) =

(

0 1−1 0

) (

x1x2

)

.

More generally to rotate clockwise about the vertex (b1, 0), the result is(

0 1−1 0

) (

x1 − b1x2

)

+

(

b10

)

.

Now, the script (or term) for Add (a, b) calls for a rotation of (a0, 0) by thefirst matrix, followed by a projection onto the vertical line at (b1, 0), followed bya rotation by the second matrix. In order to verify the correctness of Add , theresult should be (a1 + b1, 0). Let us check it in EF. We start with (a1, 0) (wherea1 is not assumed to have any particular sign). Rotating by the first matrix, weobtain (0, a1). Projecting we have (b1, a

1). Using the equation given above for

the second rotation, we have(

0 1−1 0

) (

b1 − b1a1

) (

b10

) (

b10

)

=

(

a1 + b10

)

as desired. That completes the correctness proof of Add .Next we turn to the correctness of SquareRoot . According to the algorithm,

we find SquareRoot (x) = (y1 , 0) by first constructing the circle with center at(x1+1)/2, 0 and passing through 0, and then computing the length of the verticalchord over (x1, 0). So the radius of the circle is r = (x + 1)/2, and if y is the

208 MICHAEL BEESON

height of the chord we have

(

x − x + 1

2

)2

+ (y)2 =

(

x + 1

2

)2

which is easily solved to derive (y)2 = x.That completes the proof of the lemma.

Lemma 12.59. For each term t of EF+, or list of two or four terms,

EF+ proves t ∼= t.

When t is a list of three terms t = (a, b, c), we have

EF+ proves t = (a, b, |c|).

Remark. Lists of three terms represent circles given by center and radius. Wehave arranged it so a negative value r in the third position is converted by r toa positive radius |r|. So when we take r we get back |r| instead of the originalr.

Proof. We proceed by induction on the complexity of t.First consider the basis case when t is a constant. If t is a constant, it is 0,

1, γ1, or γ2. If t is 0, then t = α and t = α = 0. If t is 1, then t = β andt = β = 1. If t = γ1, then t = X(γ1), and

t =X(γ)

Here X(γ) is by definition a term of ECG giving the foot of the perpendicularfrom γ to Line (0, 1). Therefore it (provably in ECG) satisfies the geometricformula A(γ,X(γ)) expressing that it is the foot of that perpendicular, whereA(p, q) is on(q,Line (0, 1))∧qp ⊥ Line (0, 1). By the soundness of the interpre-tation, X(γ) is the field element q satisfying A(γ, q). Since γ = (γ1, γ2),that field element q is γ1. Hence γ1

= γ1 as claimed. Similarly when t is γ2,but using Y instead of X .

Now consider the basis case when t is a variable x. Then x occurs as the j-thvariable vj for some j, and x was defined according to the value of j mod 9.There are 9 cases to consider.

If x is v9k then x = X(pk), where pk is the k-th Point variable of ECG.Then x = (X(pk)) = (pk)0. By definition, pk

is the list (v9k, v9k+1), so(pk)0 = v9k = x as required.

If x is v9k+1 then x = Y (pk), where pk is the k-th Point variable of ECG.Then x = (Y (pk)) = (pk)1. By definition, pk

is the list (v9k, v9k+1), so(pk)1 = v9k+1 = x as required.

If x is v9k+2 then x = X(pointOn1 (ℓk)), where ℓk is the k-th Line variable.Then x = (ℓk)1. Now (ℓk) is (v9k+2, v9k+3, v9k+4, v9k+5), as one can verify bychecking the definition on page 153. Hence (ℓk)1 = v9k+2 = x as required. Sim-ilarly when x = v9k+3, v9k+4, v9k+5, using Y instead of x and/or pointOn2 (ℓk)instead of pointOn1 (ℓk).


If x is v9k+6, then x = X(center (ck)), where ck is the k-th Circle variable.Then

x = (ck)1

= (v9k+6, v9k+7)1

= v9k+6

= x

as required. Similarly if x = v9k+7, with Y instead of x and (Ck)2 instead of(Ck)1.

If x is v9k+8 then

x =X(SquareRoot (HilbertSquare (Y (pointOnCircle (ck)),

Minus (Y (center (ck))))))) by definition of x, Definition 12.47

Applying to both sides we have

x =X(SquareRoot (HilbertSquare (Y (pointOnCircle (ck)),

Minus (Y (center (ck)))))))

By Lemma 12.58, we can convert the geometrized arithmetic of SquareRoot andHilbertSquare to field operations in EF:

x =√

(Y (pointOnCircle (ck)) − Y (center (ck)))2

=√

(pointOnCircle (ck)2 − center (ck)2)2

By definition

pointOnCircle (ck) = (ck)1, (ck)2 + (ck)3

= (v9k+6, v9k+7 + |v9k+8) by Definition 12.17

center (ck) = (v9k+6, v9k+7)

center (ck)2 = v9k+7

so we have

x =√

((v9k+7 + |v9k+8) − v9k+7)2

=√

v29k+7

=√x2

= |x|= x

which is always defined, so x = x as claimed. (That was the reason for using |x|instead of just x in the definition of v9k+6.) That completes the base case of theinduction.

210 MICHAEL BEESON

Consider the case when t is r·s for terms r and s. Then t = HilbertMultiply (r, s).When we apply to both sides we get

(t) = HilbertMultiply (r, s)

= r1 · s1 by Lemma 12.58∼= r · s by the induction hypothesis

= t

Similarly the cases when t = r + s or −r are treated using Add or Minus inplace of HilbertMultiply , and again appealing to Lemma 12.58. Suppose t is

√r.

Then t = SquareRoot (r). Applying to both sides we have

(t) = (√r)

= SquareRoot (r)

=√r by Lemma 12.58

∼=√r by the induction hypothesis

= t

Now consider the case when t is 1/r. Then t = Reciprocal (r), so

t = Reciprocal (r)

∼= 1/r by Lemma 12.58∼= 1/r by the induction hypothesis∼= t

Next consider the case when t is a list of two terms, t = (a, b). Then bydefinition t = MakePoint (a, b), so

t = MakePoint (a, b)

= (a, b) by Lemma 12.57

= (a, b) by the induction hypothesis

= t as desired

Next consider the case when t is a list of four terms, t = (a, b, c, d). The “is”of definition can be replaced formally by ∼=. (We cannot use = since perhapsa = b.) Then by definition

t is Line (MakePoint (a, b),MakePoint (c, d)).

The “is” of definition can be replaced formally by ∼=. (We cannot use = sinceperhaps a = b.)


t ∼= Line (MakePoint (a, b),MakePoint (c, d))

= (X(MakePoint (a, b)), Y (MakePoint (a, b))),∼= X(MakePoint (c, d)), Y (MakePoint (c, d)))∼= (a, b, c, d)∼= (a, b, c, d) by the induction hypothesis∼= t as desired


Finally consider the case when t is a list of three terms, t = (a, b, c). Then bydefinition

t = Circle (MakePoint (a, b),MakePoint (a,Add (b, c))


t = Circle (MakePoint (a, b),MakePoint (a,Add (b, c))

= (X(MakePoint (a, b)), Y (MakePoint (a, b)),

|MakePoint (a, b) − MakePoint (a,Add (b, c))|)= (a, b, |(a, b) − (a,Add (b, c))|)= (a, b, |(a, b) − (a,Add (b, c))|) by the induction hypothesis

= (a, b, |(a, b) − (a, b + c)|) by Lemma 12.58

= (a, b, |(a, b) − (a, b+ c)) by the induction hypothesis

= (a, b, |(0, c)|)= (a, b, |c|)


Lemma 12.60. For each term t of ECG, ECG proves t = t.Moreover the following are also provable:

(i) If p is a term of type Point, then

t = MakePoint (t1, t2)

X(t) = t1

Y (t) = t2

(ii) If C is a term of type Circle, then

X(center (C)) = C1

Y (center (C)) = C2

Radius (C) = C3

(iii) if L is a term of type Line, then

X(pointOn1 (L)) = L1

Y (pointOn1 (L)) = L2

X(pointOn2 (L)) = L3

Y (pointOn2 (L)) = L4

L = Line (MakePoint (L1, L

2),MakePoint (L

3, L4))

Proof. We proceed by induction on the complexity of the term. The basis caseis when t is a variable or constant.

Consider the case of a Point variable t = pk. Then t is (v9k, v9k+1), so t1 isv9k, and t1 is X(pk), that is, X(t). t2 is v9k+1, and t2 is Y (pk), that is, Y (t).Then by Lemma 10.10 we have t = MakePoint (X(t), Y (t)) = MakePoint (t1, t

2).

212 MICHAEL BEESON

But by definition,

t = (v9k, v9k+1)

= MakePoint (v9k, v9k+1)

= MakePoint (t1, t2)

= t as shown just above

Then X(t) = t1 and Y (t) = t2 by Lemma 10.10. That completes the casewhen t is a Point variable.

Now consider the case of a line variable t = ℓk. Then t is a list of length4, giving the coordinates of two specified points on the line; specifically t =(v9k+2, v9k+3, v9k+4, v9k+5). Then according to the definition of x for variablesx, we have

t1 =X(pointOn1 (t))

t1 = Y (pointOn1 (t))

t3 =X(pointOn2 (t))

t4 = Y (pointOn2 (t))

Applying Lemma 10.10, we have

pointOn1 (t) = MakePoint (t1, t1)

pointOn2 (t) = MakePoint (t3, t4)

By Axiom CA1 we have t = Line (pointOn1 (t), pointOn2 (t)), and substitutingthe values we calculated for pointOn1 (t) and pointOn2 (t), we have

t = Line (MakePoint (t1, t1),MakePoint (t3, t

4))

= t

by the definition of u when u is a list of length 4 (see Definition 12.49). Thatcompletes the case of a Line variable.

Now consider the case of a Circle variable t = ck. Then t is a list oflength 3, giving the coordinates of the center and the radius; specifically t =(v9k+6, v9k+7, v9k+8), so t1 is v9k+6, t

2 is v9k+3, and t3 is v9k+4. Then according

to the definition of x for variables x (Definition 12.47), we have

t1 =X(center (t)

t2 = Y (center (t)

t3 = Radius (t)

Applying Lemma 10.10, we have

center (t) = MakePoint (t1, t1).

That completes the case of a Circle variable.Now consider the case when t is constant. Then t is α, β, or γ. Suppose t = α.

Then α = (0, 0), and α1 = 0 = 0, and α

2 = 0 = 0, so MakePoint (α1, α

2) =

MakePoint (0, 0) = 0 = α, proving (i) for the case t = α. Similarly when t = β,with (1, 0) in place of (0,0). Now suppose t = γ. Since γ is not given specificcoordinates the argument is different. Then t = (γ1, γ2), and t1 = X(γ);


similarly t2 = Y (γ). Hence MakePoint (t1, t2) = MakePoint (X(γ), Y (γ)) = γ,

by Lemma 10.10, completing the case t = γ.That completes the basis case of the induction. The induction step has a case

for each function symbol of ECG. Consider the case when t is pointOn1 (L).Then t = (L

1, L2), and

t = (L1, L

2)

= MakePoint (L1, L

2) by the definition of a, b

= MakePoint (L1, L2) by the induction hypothesis

which completes this case. The cases when t is pointOn2 (L), or center (C), orpointOnCircle (C) are treated similarly.

Conside the case when t = Line (a, b). Then t = Append(a, b) = (a1, a2, b

1, b

2)

and

t = (a1, a2, b

1, b

2)

= Line (MakePoint (a1, a2),MakePoint (b1, b

2))

= Line (MakePoint (a1, a2),MakePoint (b1, b2))

by the induction hypothesis

= Line (a, b) by Lemma 10.10

= t

which completes the proof for the case t = Line (a, b).Consider the case when t = Circle (a, b, c). (Remember Circle (a, b) is just an

abbreviation for Circle (a, a, b).) By definition of t we have

t = (a1, a2, |b − c|)

t = Circle (MakePoint (a1, a2),MakePoint (x, y))

where x and y are calculated according to the last clause of Definition 12.49,namely, with r = |b − c|,

x = Add (a1,HilbertMultiply (γ3, r))

y = Add (a2,HilbertMultiply (SquareRoot (Add (1,Minus (γ3))), r))

By the induction hypothesis we have a1 = X(a) and a2 = Y (a). Then

MakePoint (X(a), Y (a)) = a by Lemma 10.10.

It remains to show that Radius (t) = t3. We have

t3 = |b − c|t3 = |b − c|

= SquareRoot (HilbertSquare (Add (b,Minus (c))))

= Distance (b, c)

= Distance (b, c) by the induction hypothesis

= Radius (t) by definition of Radius

214 MICHAEL BEESON

Now we have proved that t and t have the same center and the same radius.Hence by Lemma 12.31, they coincide. Then by Lemma 12.45, they are equal.That completes the case when t is Circle (a, b, c).

Consider the case when t is pointOnCircle (C). Let C = (a, b, c). Then bydefinition, t is (x, y) where

x = a + γ3 · |b − c|

y =√

1 − γ23 · |b − c|

= γ4 · |b − c| where γ4 :=√

1 − γ23

Then t = MakePoint (x, y), and

x = Add (a,HilbertMultiply (γ3,Distance (b, c)))

= Add (a,HilbertMultiply (γ3,Distance (b, c))) by the induction hypothesis

Similarly

y = Add (b,HilbertMultiply (γ4,Distance (b, c)))

γ4 = SquareRoot (Add (1,Minus (HilbertSquare (γ3))))

Let p = pointOnCircle (Circle (0, 1)). Then by definition γ3 = X(p) and wealso have γ4 = Y (p). Putting that in we have

x = Add (a,HilbertMultiply (X(p),Distance (b, c)))

y = Add (b,HilbertMultiply (Y (p),Distance (b, c)))

Using the definition of VectorAdd we then have

t = VectorAdd (center (C),ScalarMultiply (Distance (b, c), p))

= VectorAdd (center (C),ScalarMultiply (Radius (C), pointOnCircle (Circle (0, 1))))

But by Lemma 12.46 the right hand side is equal to pointOnCircle (C). Thatcompletes the case when t is pointOnCircle (C).

Consider the case when t = IntersectLines (L,K). Recall that t is a list oftwo terms (x, y) giving the coordinates of the intersection point in terms ofa = (L

1, L2, b = (L

3, L4), c = (K

1 ,K2 , and d = (K

3 ,K4 ). Then

t = ((x), (y)

= (x, y) by the induction hypothesis

So in order to prove t = t, we have to check in ECG that the algebraic formulafor the solution in terms of a, b, c, and d is correct. That is, first we have used

to write a matrix equation for the coordinates (x, y) of the intersection point,in terms of the coordinates of two points a, b on L and two points c, d on K;that it works is a piece of algebra (linear algebra at that), plugging in x andy and simplifying until the sides are the same. Since we have verified the ringlaws are preserved under φ 7→ φ, the same simplification can be done usingHilbertMultiply and Add in ECG.

Consider the case when

t = IntersectLineCircle1 (L,C) or IntersectLineCircle2 (L,C).


Then L gives the coordinates of two points on the line, namely a = (L1, L

2)

and b = (L3, L

4), and the center of the circle is e = (C

1 , C2 ), and its radius is

r = C3 . By definition, we have

t = a+ λ(b − a)

where λ is given by equation (1), which we repeat here:

λ = − (a− e) · (b− a) ±√

((a− e) · (b − a))2 − (b− a)2((a− e)2 − r2)

(b − a)2

where the plus sign is taken for IntersectLineCircle2 and the minus sign forIntersectLineCircle1 . The term now of interest, t, is given by replacing thearithmetic of EF by geometric arithmetic. Thus

t = Add (a+ HilbertMultiply (λ,Add (b,Minus (a))))

where in turn λ is computed by the formula above, but using geometric arithmeticand with bars over a, b, e, and r. Those inner terms are evaluated like this:

a = (L1, L

2)

= MakePoint (X(pointOn1 (L), Y (pointOn1 (L))))

by the induction hypothesis

= pointOn1 (L) by Lemma 10.10

Similarly b = pointOn2 (L) and e = center (C) and r = Radius (C). Now whathas to be verified is free of any mention of the interpretation: it is just thatthe formulas above, interpreted using geometric arithmetic, do correctly give theintersection points of a line and circle. Looking at the derivation of equation(1), we see that what was used was the formula for a circle, and the fact thatthe points on Line (b, a) have the form a+λ(b− a), plus the laws of field theory,which we have already verified. By Lemma 12.31, the formula for a circle isprovable in ECG. By Lemma 12.36, the points on Line (a, b) are all of the form

x = VectorAdd (a+ ScalarMultiply (λ,Add (b,Minus (a))))

for some λ depending on x, which is exactly what is required. Hence t is apoint lying on C and on L. To complete the argument, we still need to showthat the positive sign for λ corresponds to IntersectCircles2 and the negative signto IntersectCircles1 . That is, for z not on the line, Right (z, p, q)↔Right (z, a, b).Since linear transformations of positive determinant preserve the sign of thecross product, and Right is given by the sign of a cross product, and lineartransformations preserve the equation of a line, we can assume without loss ofgenerality that a = 0 and b lies on Ray (0, 1), i.e. b > 0. The the formula saysx = HilbertMultiply (λ, b). Then let λ+ be the positive value of λ and λ− thenegative value; and let

q = HilbertMultiply (λ+, b)

p = HilbertMultiply (λ−, b)

Then q > 0 and p < 0. Hence by Lemma 9.89, Right (z, a, b) is equivalent toRight (z, p, q). Hence

p = IntersectLineCircle1 (L,C) and q = IntersectLineCircle2 (L,C)

216 MICHAEL BEESON

as claimed. That completes the case when t = IntersectLineCircle1 (L,C) ort = IntersectLineCircle2 (L,C).

The final case to consider is when

t = IntersectCircles1 (C,K) or t = IntersectCircles2 (C,K).

Then C and K are triples of points giving the centers and radii of two circles,and the formulas given in the proof of Theorem 6.1 give the intersection points, ifthey are defined. The definition of t gives an explicit formula for the intersectionpoints. When we compute t we replace the arithmetic operations +, ·, and

√by Add , HilbertMultiply , and SquareRoot , and the induction hypothesis replacese by center (C) for the center e of a circle, and r by Radius (C) for the radius r.Thus what has to be verified is that the formulas given in the proof of Theorem6.1 (and used to define t) are provable in ECG, when geometric arithmeticis used for addition, multiplication, and square root. This is proved by justgoing over the proof of Theorem 6.1, and observing that nothing is used butcalculations involving the laws of field theory. (That is why we gave the proof ofthat theorem in such detail.) That proves that t, when t is defined, lies on bothcircles. It remains to check that t has been correctly defined, in the sense thatthe correct intersection points have been assigned to IntersectCircles1 (C,K) andIntersectCircles2 (C,K). That was done using the cross product in the proof ofTheorem 12.21. But now, we have formalized the connection between Right andCrossProduct , so that argument can now be given in ECG as well. In case thereader has any doubt, we will do so. Namely, by Lemma ??, Right is invariantunder linear transformations of positive determinant. Hence by a translationand rotation we can assume that C has its center at origin and K has its centeron the positive x-axis. In that case IntersectCircles1 (C,K) lies in the (closed)lower half plane (if it is defined) and IntersectCircles2 (C,K) lies in the (closed)upper half plane (by Lemma 9.104 and Axiom H5). So we only need to checkthat in this special case, t2 has a positive sign when t = IntersectCircles2 (C,K)and a negative sign when t = IntersectCircles1 (C,K).


Theorem 12.61. (i) EF+ proves ((φ)↔φ) for every formula φ of EF+.(ii) ECG proves (φ↔φ) for every formula φ of ECGnot containing equality

between terms of type Circle.(iii) The interpretation φ 7→ φ is faithful on φ with no equality between terms

of type Circle.(iv) The interpretation φ 7→ φ is faithful.

Proof. We first show that (i) and (ii) implies (iii). Suppose EF+ proves φ. Thenby the soundness of the interpretation φ 7→ φ, ECGproves φ. By (ii) then ECGproves φ, so the interpretation φ 7→ φ is faithful (on formulas without = betweenterms of type Circle. Similarly, if ECGproves φ, then EF+ proves (φ), so by(i), EF+proves φ. Hence the interpretation φ 7→ φ is faithful.

We turn to the proof of (i). Another fact we need in the proof is that (0 < x)

is equivalent in EF to P (x), where 0 < x is the defined relation in ECG, definedby Left (I, 0, x).


Let φ be a formula of EF. Then φ expresses φ in terms of variables restrictedto Line (0, 1) and using Add and HilbertMultiply for addition and multiplication,and 0 < x for P (x). Then we introduce coordinates X(b) and Y (b) in ECG,and use them to translate back into EF. We must show we get (somethingequivalent to) the original formula. We proceed by induction on the complexityof the formula φ. The base case is when φ is atomic. Then φ is of the form t = sor the form P (t), or of the form t ↓, for terms t and s. Consider the case whenφ is t = s. Then φ is t = s, where t is obtained from t by replacing + and · byAdd and HilbertMultiply . Applying we find that φ is t = s. Thus what hasto be proved is

t = s → t = s.

Suppose t = s. Then t ↓ and s ↓, so by Lemma 12.59 we have t = t and s = s.Hence t = s as desired.

Suppose φ has the form t ↓. Then we must prove

t ↓ → t ↓ .By Lemma 12.59 we have t ∼= t. By the definition of ∼=, that means that ifeither side is defined then both are defined and they are equal; in particular ift ↓ then t ↓.

Now we check the case of an atomic formula P (t). Then

(P (t)) ↔ (P (t)

↔ (0 < t)

↔ P (t)

↔ P (t) since t = t

That completes the proof of (i).We turn to the proof of (ii). Let φ be a formula of ECG. Then φ expresses φ

using coordinates and analytic geometry, and φ translates back into ECG usingAdd and HilbertMultiply and 0 < x. We must prove the result is equivalent toφ. We do this by induction on the complexity of φ. Consider the atomic casewhen φ has the form t = s for terms t and s. Then

φ ↔ t = s

↔ t = s

↔ t = s by Lemma 12.60

↔ φ

That completes the case φ is t = s.Now consider the case in which φ is t ↓. Then

φ ↔ t ↓φ ↔ t ↓

↔ t ↓ by definition of t ↓↔ t ↓ by Lemma 12.60

↔ φ

That completes the case in which φ is t ↓.

218 MICHAEL BEESON

Now consider the case when φ is B(a, b, c). Then φ is

a 6= b ∧ b 6= c ∧ |b − a| + |b − c| = |a − b|.and by Lemma 12.58, φ is what we get by dropping the and using HilbertMultiplyto multiply and SquareRoot to take square roots, and Add and Minus to addand subtract. That is, φ is

Add (Distance (a, b),Distance (b, c) = Distance (a, b).

By the induction hypothesis, that is equivalent to

Add (Distance (a, b),Distance (b, c)) = Distance (a, c).

By Lemma 12.44, that is equivalent to B(a, b, c), which is φ. That completes thecase when φ is B(a, b, c).

Now consider the case when φ is E(a, b, c, d), or ab = cd as we usually writeit. Then φ is |a − b|2 = |c − d|2, so

φ ↔ |a − b|2 = |c − d|2

↔ |a − b|2 = |c − d|2↔ Distance (a, b) = Distance (c, d)

↔ Distance (a, b) = Distance (c, d) by the induction hypothesis

↔ E(a, b, c, d)by Lemma ??

That completes all the cases in which φ is atomic. Since both φ and φcommute with the propositional connectives, there is nothing to prove when φis a disjunction, conjunction, implication, or negation.

Consider the case when φ is ∀xψ, where x is a variable of type Point. Thevariables x of ψ are two variables belonging to the same “group” in Defi-nition 12.50, and so they have the same variable x∗ associated to them, andmoreover that variable x∗ is the original variable x of ECG. Hence

φ ↔ ∀x ψ

↔ ∀x∗ψ

↔ ∀x∗ψ by the induction hypothesis

↔ ∀xψ since x∗ is x

↔ φ

That completes the case when φ is ∀xψ. The case when ∀ is changed to ∃ is thesame throughout.

The case when φ is ∀x, ψ, where x is a variable of type Circle, is handled bythe same calculation; the point is that the quantified variables x belong to thesame group and hence have the same x∗, which is the original variable. Also thecase when φ is ∃x, ψ and x of type Circle works the same way.

The case when φ is ∀Lψ, where L is a variable of type Line, offers an additionalwrinkle, because φ restricts the quantifier to the defined predicate L. We have

φ ↔ (∀Lψ)

↔ ∀L(L(L) → ψ)


Here L is a list of four variables belonging to the same “group”, so they havethe same corresponding variable L∗ in the Definition 12.50. Then we have

φ ↔ ∀L∗(L(L) → ψ)

↔ ∀L∗(L(L) → ψ) by the induction hypothesis

By Lemma 12.20, L(L) is provable in ECG, so we can drop it:

φ ↔ ∀L∗ψ)

↔ φsince L∗ is L as before


§13. Classical geometry and constructive geometry compared. Inthis section we will ask and answer the question, what classes of formulas havethe property that if they are classically provable, then they are constructivelyprovable? We will prove that the negative formulas of ECG have this property;that covers formulas of the form, if some points, lines, and circles stand in certainrelations to one another, and you perform a certain construction, then the resultof the construction has certain properties (where the relations and properties inquestion do not involve ∃ or ∨). That covers all of Euclid’s theorems, I believe.

At the end of the section we give an example to show that this result cannotbe extended to formulas containing an existential quantifier. There are relativelysimple cases (the weak Pasch axiom for example) in which a point that classicallyexists does not depend continuously on the parameters, and hence cannot beproved to exist constructively.

13.1. The double negation interpretation. In ??, Godel defined his double-negation interpretation. One can also find it explained in [17]. Here is how itworks: we start with a first-order theory T . It can be a many-sorted theory,and it can use the logic of partial terms. We associate to each formula A ofT , another formula A− (the double-negation interpretation of A). The rules fordefining A− are as follows:

(∃xA)− is ¬∀x¬A−

(A ∨B)− is ¬ (¬A− ∧ ¬B−)

P− is ¬¬P for atomic P

(A → B)− is A− → B−

(¬A)− is ¬A−

(∀xA)− is ∀xA−

Informally, we put a double negation in front of ∃, ∨, and atomic formulas,and leave the rest of the formula unchanged. The actual definition then converts¬¬∃ to ¬∀¬, and similarly for ∨, because it is convenient that A− is always anegative formula (i.e. does not contain ∃ or ∨.

Theorem 13.1 (Godel). If first-order classical logic proves A, then first-orderintuitionistic logic proves A−. If theory T is such that T with intuitionistic logic

220 MICHAEL BEESON

proves A− for every nonlogical axiom A of T , then intuitionistic T proves A−

for every theorem of classical T .

The proof is a routine induction on the length of proofs. (Godel’s genius wasin thinking of the definition and theorem, not in proving it.) Although Godel didnot explicitly mention many-sorted theories, nothing new is required to checkthe validity of the theorem in that setting. We want to apply the theorem totheories in the logic of partial terms; so there is something to be checked. Thatcan also easily be checked directly; but rather than go into the details, we checkit indirectly.

Corollary 13.2. Suppose T is a theory in many-sorted predicate logic withthe logic of partial terms. Supose that T proves φ− for each axiom φ of T , andalso suppose T proves ¬¬ t ↓ → t ↓ for each term t. Then the double-negationinterpretation is also sound for T ; that is, if T classically proves φ then T provesφ−.

Proof. We have already shown in Theorem 8.3 that the logic of partial termsLPT (with one sort or many sorts) can be reduced to the usual first-order logic

(with one sort). The procedure described there assigns a formula φ to eachformula φ. In the case of ECG, we would add three predicate symbols, Point,Line, and Circle. Variables of type Point become variables relativized to theunary predicate Point, and similarly for Line and Circle. For t a term of typePoint, the formula t ↓ becomes Point (t). But there is nothing specific to ECGabout this procedure. It applies to any theory in LPT. The plan of the proof isthis: first make this translation, then apply the double-negation interpretation.Then translate back into the logic of partial terms, replacing ¶(t) by t ↓. Nowwe carry out this plan in detail.

Let T be the first-order theory axiomatized by the φ for φ an axiom of T .Since by hypothesis, T proves ¬¬ t ↓, we will have T proves ¬¬P (t) → P (t).

Now we claim that the φ interpretation and the double-negation interpretation

commute, i.e. if ψ = φ− then ψ = φ−. This is proved by induction on thecomplexity of φ. The only tricky case is when φ is atomic. If t is a term of type

P , and φ is t ↓, then ψ = φ− is ¬¬ t ↓, and ψ is ¬¬P (t). On the other hand φ

is P (t), and φ− is ¬¬P (t), so the conclusion is valid in this case.Then

T classically proves φ

↔ T classically proves φ by Theorem 8.3

↔ T proves (φ)− by Godel

↔ T proves (φ− as proved above

↔ T proves φ− by Theorem 8.3

Note that when the double-negation interpretation is applied in predicate cal-culus, we need ¬¬P (t) → P (t) to be provable in T ; but by hypothesis, we dohave that. That completes the proof of the corollary.


13.2. The double-negation interpretation applied to ECG. We finallyarrive at the double-negation interpretation for ECG. To make it work we neededthese things:

(1) The axioms of ECG are negative, so they imply their double negationinterpretations;

(2) Betweenness and congruence are stable;(3) Definedness is stable, i.e. ¬¬ t ↓ → t ↓. We proved this in Theorem 10.16,

and it required the strong parallel postulate; and in fact it is equivalent to thestrong parallel postulate.

These are the the things that make the following theorem work. Since theseare in hand, the proof is short.

Theorem 13.3 (Double negation interpretation). Suppose ECG with classi-cal logic proves A. Then ECG with intuitionistic logic proves A−.

Proof. First we observe that ¬¬A is equivalent to A for atomic A. This is anaxiom for all atomic formulas not of the form t ↓, and was proved in Theorem10.16 for formulas of the form t ↓. Since the axioms of ECG are quantifier-freeand disjunction-free, it follows that A− is equivalent to A for axioms A of ECG.Now the theorem follows from Corollary 13.2. That completes the proof.

Corollary 13.4. ECG with classical logic is conservative over ECG withintuitionistic logic for negative formulae.

Proof. For negative A, A− is identical to A.A typical theorem of Euclid has the form H → A, where A will be quantifier-

free when formulated in ECG, and H is a collection of hypotheses that certainpoints are distinct, or certain incidence relations hold or do not hold. As Procluspointed out, sometimes this implies a theorem formulated by cases. For example,Euclid I.2 has one proof if A = C and another proof if A 6= C. Using thedouble-negation interpretation, we find a proof that A = C ∨A 6= C implies theconclusion of Euclid 1.2, but without the law of the excluded middle we cannotconclude the “uniform version” of Euclid I.2.

Theorem 13.5. All negative theorems of Tarski’s geometry (formulated withclassical logic and with line-circle continuity) are provable in ECG.

Remark. One may use theory Tarski-B with strict betweenness, or theory Tarski-T with non-strict betweenness T , but then one must replace T with its negativedefinition in terms of B for the conclusion of the theorem to make sense.

Proof. Suppose φ is provable in Tarski-B with classical logic. Then φ is provablein EF+ with classical logic, so φ is provable in ECG with classical logic. Butby Theorem ??, φ is equivalent to φ. Hence φ is provable in ECGwith classicallogic. Hence φ− is provable in ECG. But since φ is negative, φ− is equivalentto φ. Hence φ is provable in ECG. That completes the proof of the theorem.

Recall the example given in Section 2.3 of Euclid’s Prop. I.6, where we showedthat a particular disjunction was eliminable from a proof, because the conclusionwas negative, i.e. disjunction-free and ∃-free. But all the theorems in Euclidare either already negative, or assert the existence of some objects that can be

222 MICHAEL BEESON

constructed using the terms of ECG; when formulated more explicitly, they arenegative in the sense that they say that the result of a certain construction hascertain (quantifier-free) properties. Hence the double-negation interpretationtells us that, once we have formulated the axioms of ECG in a quantifier-free,disjunction-free way, the work of constructivizing Euclid is done: we can thenfreely use classical logic to prove negative theorems, and all of Euclid’s theoremsare negative.

In particular, this gives us considerable freedom in choosing the between-ness and congruence axioms of ECG. As discussed above, considerable effortby Tarski, Gupta, Szmielew, and Schwabhauser went into finding a short andsimple axiom system for geometry; and classical Euclidean geometry is nowwell-founded on this modern axiomatic system, based on equidistance and be-tweenness. Although betweenness does not occur explicitly in Euclid, we havechosen the same formal language, to permit an easy comparison of the classicaland constructive theories. ECG has been designed in such a way that

(i) The axioms are constructively valid.

(ii) The axioms are quantifier-free and disjunction-free.

(iii) Betweenness and congruence are stable.

(iv) Definedness is stable, i.e. ¬¬, t ↓ → t ↓.(v) The axioms are classically equivalent to the system given in [26] (with

line-circle continuity rather than the full first-order continuity schema).

Condition (v) follows immediately from our reduction of ECG to Euclideanfield theory, since it is known that classically the models of Tarski’s theory areplanes over Euclidean fields. (See [26], part II, or [29]).

Our point here is that any axiomatization satisfying these five properties willalso have the property in the theorem, that it proves all negative classical theo-rems of Tarski’s geometry.

§14. The relation of ECG to Tarski’s theories of geometry. In theprevious section, we showed that negative theorems of Tarski’s geometry ( withline-circle continuity) are theorems of ECG. But one can still ask whether theintuitionistic versions of Tarski’s theories, which we have called Tarski-B andTarski-T, are equivalent to ECG. The desired result here is that ECG is con-servative over Tarski-B (for non-experts in logic, that means that if φ is a theo-rem in the common language of the two theories, then if ECG proves φ so doesTarski-B). A relatively trivial result is that Tarski-B and Tarski-T are equiva-lent, since one can define nonstrict betweenness in terms of strict betweennessand vice-versa.

We have already interpreted ECG into EF+ and faithfully, so it only remainsto interpret EF+ faithfully into (intuitionistic versions of) Tarski’s theories. Theideas used to interpret ECG into EF+ can also be used to interpret ECGdirectly into Tarski-B, i.e. circles can be represented as triples of points, and linesas pairs of points; but there seems to be no extra increment of understandingto be gained by writing out the details. Instead, we just interpret EF+ intoTarski-B. Of course, the geometric definitions of multiplication and addition and


square root are essentially the same as for ECG, but Tarski-B has no functionsymbols, and does not have the logic of partial terms, so there are some detailsto attend to.

Theorem 14.1. ECG can be faithfully interpreted into intuitionistic Tarski-B.

Proof sketch. Since we have already interpreted ECG faithfully in EF+, itsuffices to interpret EF+ faithfully in Tarski-B. This is done as follows. Weassign to each term t of EF+, a formula At of Tarski-B, with one more freevariable than t. Let x = x1, . . . , xn and let t have free variables x. ThenAt(y, x) is intended to express y = t, i.e. t is defined and has the value y.

If t = x·y, then At(z, x, y) should say that z is the value of HilbertMultiply (x, y).That is, there is a circle passing through z, I, x, and y. That is,

Strong Parallel Axiom2

We set aside three variable α, β, γ to serve as the interpretations of theconstants α, β, and γ. These variables will not be used otherwise. Lines areinterpreted as pairs of points; specifically Line (a, b) is interpreted as (a, b). Avariable L of type Line is interpreted as a pair of point variables L1 and L2.The terms pointOn1 (L) and pointOn2 (L) are interpreted as L1 and L2. Theterm pointOn1 (Line (t, s)) is interpreted as (t, s), where t is the interpretationof t. Circles are also interpreted as pairs of points, and center(Circle 3(a, b, c)) isinterpreted as a, while pointOnCircle (Circle (a, b, c)) is interpreted as a point bcalong the canonical line parallel to Line (α, β) that passes through a, and in thesame direction from a as β is from α. The construction of the canonical paralleland the explanation of “same direction” will come later on; this is only a “proofsketch.”

Incidence of points on lines is interpreted this way: on(p,Line (a, b) is inter-preted as

¬¬(B(p, a, b) ∨ p = a ∨ B(a, p, b) ∨ p = b ∨ B(a, b, p))

This is, of course, equivalent to a negative formula obtained by pushing thedouble negation in.

For each term t with free variables x = x1, . . . , xn, there is a formula At

defining the interpretation of t, At(x, y), expressing that t is defined and has thevalue y.

The interpretation of the formula t ↓ is ∃y At(x). To interpret a formula notof the form t ↓, one first writes the formula in the form φ(y := t), where yis a list of variables and t a corresponding list of terms, and y := t indicatessimultaneous substitution, and φ contains no function symbols or constants.Then the interpretation of φ(y := t) is given by

∀y(At(x, y) → φ)

On formulas not containing function symbols and constants, the interpretation

φ 7→ φ preserves logical connectives. Quantifiers are handled as follows. Theinterpretation of ∀xφ(y := t) is

∀x∃y (At(x, z, y) ∧ φ)

224 MICHAEL BEESON

Note that x can occur free in At since some of the terms t may contain x; z is alist of other free variables of φ and t. The interpretation of ∃zφ(y := t) is

∃z, y (At(x, z, y) ∧ φ.That brings us to the interpretations of atomic formulas. The interpretation

of t = s is

∃y, z(At(x, y) ∧As(x, z) ∧ y = z)

where x on the left is the list of free variables of t and x on the right is the list offree variables of s, which may of course not be the same list. The interpretationof B(t, s, r) is

∃y, z, w(At(x, y) ∧As(x, z) ∧Ar(x,w) ∧ B(y, z, w)

where as before, the list x of free variables may be different at each occurrence.Similarly for the interpretation of E(t, s, p, q).

We have to exhibit the formulas At and Bt corresponding to terms with onefunction symbol and variables for arguments. Without yet having done this, wecan already prove a basic lemma:

Lemma 14.2. If a target theory T in ordinary first-order logic proves the in-terpretations of the axioms of a source theory S in the logic of partial terms,then T proves the interpretation of every theorem of S.

Proof. We check the axioms and rules of inference of the logic of partial terms.Consider the axiom

∀w φ ∧ t ↓ → φ[w := t].

Here t and φ may have other variables x not explicitly shown. The interpretationof this formula, in the special case when φ contains no compound terms, is

∀wφ ∧ ∃yAt(x, y) → ∃y(At(x, y) ∧ φ ∧ w = y)

which is immediate. In the general case where the formula contains other terms,the axiom takes the form

∀y φ[z := s] ∧ t ↓ → φ[z := s, y := t]

where φ has no compound terms. The interpretation of this formula isConsider the axiom

φ[y := t] ∧ t ↓ → ∃yφ)

The interpretation of this, in the special case when φ contains no compoundterms, is

(∃y At(x, y) ∧ φ ∧ ∃y At(x, y) → ∃y φwhich is again immediate.

Consider the rule of inference, from B → A infer B → ∀xA, with x not freein B. We write B = ψ(y := t) and A = φ(z := s), where ψ and φ do not containfunction symbols or constants, and t and s are (lists of) terms. x does not occurfree in ψ and is not among the variables z, but it could occur in As or Bs. Tocarry out induction on the lengths of proofs, we assume that the interpretationof B → A is provable in T . That is

(∃y At(w, y) ∧ ψ) → ∀z(As(q, z) → φ)


where q are other variables besides x that may occur in A. From this we mayinfer

∀y(At(w, y) → ψ) → ∀x∀z(As(q, z) → φ)

since x does not occur free in the antecedent. But this is the interpretation ofB → ∀xA, the conclusion of the rule.

The rule of inference, from A → B infer ∃xA → B is treated similarly.That completes the proof of the lemma.

FINISH THIS

14.1. Interpretation of Tarksi’s theory in ECG. We will show in thissection that the axioms of ECG suffice (at least with classical logic) to interpretTarski’s axioms, as listed in [1], p. 745, where they are referenced to [30].30

Tarski’s B cannot be interpreted as the strict betweenness of ECG, because ofhis segment-extension axiom that says, given a segment qa and another segmentbc (either of which might be a degenerate, one-point segment), there is an x suchthat B(q, a, x) ∧ ax = bc. Indeed, if both are degenerate segments, then we willneed B(x, x, x) to be true, so Tarski’s B cannot even be interpreted as non-strictbetweenness on nondegenerate intervals. We must allow not only B(a, a, b) andB(a, b, b) but also B(a, a, a).

Therefore we take as the interpretation of Tarski’s betweenness

T(a, b, c) := ¬¬(a = b ∨ b = c ∨B(a, b, c))

Note that if T(a, b, a) then ¬¬(a = b ∨B(a, b, a)

Theorem 14.3. If B(a, b, c) is replaced by T(a, b, c), then every axiom ofTarski’s geometry is provable in ECG plus classical logic.

Remark. Since Tarski’s theory uses classical logic, this theorem implies, but isstronger than, the claim that ECG plus classical logic proves all the theoremsof Tarski’s theory.

Proof. We take Tarski’s axioms one at a time.

Equidistance axiom E1. ab = ba. This is our Axiom 42.

Equidistance axiom E2. ab = pa ∧ ab = rs → pq = rs. This is our Axiom 43.

Equidistance axiom E3. ab = cc → a = b. This is our Axiom 44.

Segment construction axiom (SC): ∃x(T(qax) ∧ (ax = bc)).

If we knew that q and a were distinct, we could take

x = IntersectLineCircle2 (Line (q, a),Circle (a, b, c)).

And of course if q = a then x can be any point on Circle (a, b, c). So we canverify the double negation of SC, but not SC itself.

Five-segment axiom 5S.

a 6= b ∧ T(a, b, c) ∧ T(p, q, r) ∧ ab = pq ∧ bc = qr

∧ ad = ps ∧ bd = qs → cd = rs

30See page 190 of [30]; the system in question is EG2 with the continuity axiom 11 replacedby line-circle continuity.

226 MICHAEL BEESON

Pasch axiom. T(a, p, c) ∧ T(q, c, b) → ∃x(T(a, x, q) ∧ T(b, p, x).Let L = Line (a, q). If q = c then L has the same points as Line (a, c), and sowe can take x = p. If q = b then we can take x = b. Hence we may assumeB(b, q, c). We apply the version of the Pasch axiom in ECG to the three pointsb, c, and p, and the line L. Then L meets bp in some point x, as desired.

Lower two-dimension axiom 2L. ∃a, b, c(¬T(a, b, c) ∧ ¬T(b, c, a) ∧ ¬T(c, a, b)).Take a, b, c to be α, β, γ. Suppose T(α, β, γ); we must derive a contradiction.By definition of B, we have (classically) α = β ∨ β = γ ∨ B(α, β, γ). By AxiomD4 we have α 6= β and β 6= γ. Hence B(α, β, γ). Then by Axiom B1-i, we haveon(β,Line (α, γ)), contradicting axiom D1. That proves ¬T(α, β, γ). Similarlyone proves ¬T(β, γ, α) and ¬T(γ, α, β), using Axioms D2 and D3, respectively.

Upper two-dimension axiom 2U.

¬(a = b) ∧ ∧3i=1xia = xib → T(x1, x2, x3) ∨ T(x2, x3, x1) ∨ T(x3, x1, x2)

This says (using Axiom C5) that if three circles meet in two distinct points,then their centers are collinear. That is exactly the upper dimension axiom D5of ECG.

Parallel postulate PP.

T(a, d, t) ∧ T(b, d, c) ∧ a 6= d → ∃x, y(T(a, b, x) ∧ T(a, c, y) ∧ T(y, t, x)

First we deal with the degenerate cases: if b = a or c = a then Line (a, b) andLine (c, d) have the same points, so since B(b, d, c), the point d is also on thatline, by Axiom B1-i. Then we can take x = y = t. If d = b then we take x = t,and y can be taken to be any point on Line (a, c) with T(a, c, y), for example,y = c will do. Similarly if d = c. If d = t then we take x = b and y = c. That isall pairs of possibly equal points, so we may assume the points mentioned in thehypothesis are distinct and the betweenness relations given are strict. Moreover,we may assume that a, b, and c are not collinear, since if they are, then Line (a, x)and Line (a, y) coincide, and since d is between b and c, then by Axiom B1-i, dis also on that line, so Line , a, d) coincides with the other two, and we can takex = y = t.

We claim point t is not on Line (b, c). To prove this, suppose t is on Line (b, c).Since t 6= d, then Line (b, c) and Line (a, t) have the same points. Then a, b, andc are collinear, contradiction. Hence t is not on Line (b, c).

Now (without using a parallel axiom) we can construct line L through t parallelto Line (b, c). We use the parallel axiom this way: from the parallel axiom wecan prove the transitivity of the relation of being parallel. Now Line (b, c) is notparallel to Line (a, b), since they meet at b. Then L is not parallel to Line (a, b).Hence L meets Line (a, b) in some point x. Consider triangle atx; note thatLine (b, c) meets side at at d, but does not meet side xt, and is not equal tox, since Line (b, c) is parallel to Line (x, t). Then by Pasch, either x = b orLine (b, c) meets ax. But the meeting point is b; hence B(a, b, x). Similarly, Lmeets Line (a, c) in some point y, and B(a, c, y). That completes the verificationof Tarski’s PP in ECG.

Line-circle continuity LC

ax = ax′ ∧ az = az′ ∧ T(a, x, z) ∧ T(x, y, z) → ∃y′(ay = ay′ ∧ T(x′, y′, z′)


Let L = Line (x′, y′) and C = Circle (a, y). Take y′ = IntersectLineCircle2 (L,C).Then y′ is defined, by Axiom Cont 1, and by Axiom I7, we have on(y′, L) andOn(y, C). It remains to show T(x′, y′, z′). We have ax = ax′ and ay = ay′

and az = az′. Therefore by Lemma ??, we have x′y′ = xy and y′z′ = yz andx′z′ = xz. It follows that T(x′, y′, z′).

FINISH THIS

§15. Interpretation of ECG in Hilbert’s and Tarski’s theories.

15.1. Right and left handedness. Since ECG, following Euclid, has nodisjunctive axioms (and hence no disjunctive theorems), we cannot prove thattwo points p not on L are either on the same side or the opposite side of L; thatis constructively all right since to be given a point p “on a given side of L(a, b)”means that we are told either that abp is a right turn, or that it is a left turn.Nevertheless we can, in constructive geometry, reduce the question of whetherp and q are on the same sides of L to the case when L = Line (α, β) and q isthe point γ, where α, β, and γ are the three fixed constants mentioned in thehandedness axioms H1 and H2. The idea is that handedness is invariant undertranslations, rotations, and dilations, and using these operations we can bring Lto coincide with Line (α, β).

Lemma 15.1. (In ECG)Let ABC be any triangle. Then we can determinethe handedness of the turn ABC, in the following sense. Let α, β, and γ be thethree fixed non-collinear points mentioned in the axioms of ECG, so that αβγis a left turn by definition. Let L = Line (α, β). Then we can construct a pointR such that ABC is a left turn if and only if R is on the same side of L as γ,and a right turn if and only if R is on the opposite side of L from γ.

Proof. The idea of the proof is this: by a series of “moves” (applications ofthe axioms for Left and Right , which correspond to translations, dilations, androtations), the triangle ABC is “reduced” to either triangle αβγ or triangle αγβ,preserving the handedness of the turns; but Axioms 61 and 62 directly specifythe handedness of the turns αβγ and αγβ. We now give the details.

By a “move”, applied to a triple of non-collinear points PQR, we mean aconstruction of a new triple UVW such that, according to the axioms for Leftand Right , if PQR is a left turn then so is UVW , and if PQR is a right turnthen so is UVW . This latter condition we describe for short by saying thatthe move “preserves handedness”. The axioms describe several types of movesthat preserve handedness, specifically, moving P along the ray QP , moving Ralong the ray QR, rotating PR or QR in such a way that the points P ,Q, andR never become collinear, and translating the whole triple. We first show thatthe number of moves (applications of these axioms) required to perform a givenrotation is bounded by a fixed constant. Consider the following procedure: Firstmove P to decrease the angle to less than a right angle. Then we can rotatePQR by any angle up to a right angle, using two moves (one moves PQ and onemoves QR). An arbitrary rotation can be performed by performing at most fourrotations of less than ninety degrees. Thus in four or fewer rotations (requiringeight or fewer moves) we can bring one of its sides onto the desired (“target”)ray. Then we can move P by the same amount as in the first step resulting in

228 MICHAEL BEESON

the desired rotation. Hence ten or fewer applications of the above axioms sufficeto perform any rotation.

Now consider three non-collinear points P , Q, and R. We give a procedure fordetermining whether PQR is a left turn or a right turn. First translate PQR sothat Q coincides with the point β (given by a constant of ECG). Then rotateit so that P lies on Ray (β, α). Then move P to α. By the axioms above, allthese steps preserve the handedness of PQR. Now, if R is on the same side ofLine (α, β) as γ, then by Axioms 65 and 69, PQR is a left turn, since αβγ is a leftturn by definition. And if R (after the moves described) is on the opposite sideof Line (α, β), then we claim that PQR is a right turn. To see this, let γ′ be apoint on the same side of Line (α, β) as γ, and on the same side of Line (Q,R) asα. (Such a point can be constructed by bisecting the angle formed by Ray (β, α)and the opposite ray to Ray (Q,R).) Then we can move P to γ′ without changingthe handedness of PQR, and then we can move R to α without changing thehandedness of PQR. But now PQR coincides with γβα, which by definition isa right turn. That completes the proof of the lemma.

Remark. With classical logic, we could prove that ABC is either a right turn ora left turn. To reach that conclusion, we would need to know that if point R isnot on L = Line (α, β), then either R is on the same side of L as γ or on theopposite side. The constructive status of this statement is discussed below.

Lemma 15.2. The predicates Right (A,B,C) and Left (A,B,C) are definablein Greenberg’s theory G, relative to an arbitrary choice of Left (α, β, γ) andRight (α, γ, β) for some triple of non-collinear points α, β, and γ. This caneven be done with intuitionistic logic.

Proof. It will suffice to define the relation T (A,B,C, P,Q,R) with the meaning“ABC and PQR have the same handedness.” First we note that it is possibleto define the notion of one triangle being a translation of another; namely, ABCis a translation of PQR if the two triangles are congruent and AP = BQ = CR.It is also possible to define the notion of ABC being a rotation of PBQ (whenthe two angles share vertex B). This requires twenty variables to express, so it istoo complex to write down intelligibly, but the definition in question says thereexist twenty points representing ten “moves” according to the ECG axioms forrotations given above. Then ABC and PQR have the same handedness if thereexist P ′ and Q′ such that P ′BR′ is a translation of PQR and there is a rotationP ′′BR′′ of PQR with R′′ on Ray (B,C) and P ′′ is on the same side of Line (B,C)as A.

15.2. Apartness. The absence of disjunction in Euclid is crucial to our ax-iomatization. In order to apply the methods of proof theory, it will be veryhelpful that ECG is axiomatized in a quantifier-free, disjunction-free way. Thatis possible, while still providing the ability to prove all Euclid’s theorems, becausenone of Euclid’s theorems involves disjunction in an essential way.

Here is an example, due to Mandelkern [19], of a theorem that requires dis-junction to state and to prove, which some might consider constructively valid.Given two lines L and K that are not parallel, and a third line M , then eitherM meets L or M meets K. This theorem cannot be proved in ECG, sincethe intersection point of M with one of the two lines K and L will not depend


continuously on line M . For the same reason, the intersection point cannot beconstructed by ruler and compass. It is not hard to see that Mandelkern’s theo-rem (which he takes as an axiom) is essentially a version of “apartness”. Statedarithmetically, the principle of apartness is that for all numbers x we have x < 1or x > 0. That principle is constructive, if we think that we can compute anynumber to any accuracy; but that “computation” cannot be done with ruler andcompass.

Unlike previous intuitionistic geometries, ECGdoes not have apartness. Un-like previous algebraic theories of geometric constructions, it does not have atest-for-equality construction. In both these respects, ECG is like Euclid.

In order to arrive at such a metatheorem, we first formulate the principle inquestion in the language of ECG, which does not contain < as a primitive. Ourformulation is as follows: If two unequal points B and C are both between Aand D, then either B is between A and C or C is between A and B. Formallythat is

B 6= C ∧ B(A,B,D) ∧ B(A,C,D) → B(A,B,C) ∨ B(A,C,B) (Axiom 75)

The point D is a matter of convenience; the axiom is really about A, B, andC and their positions on a ray emanating from A, but ECG does not have raysas primitive, so we need point D to express this in ECG.

We defined AB < CD in Definition 48. We write AB > CD to mean CD <AB.

Lemma 15.3. (in ECG) Axiom 75 implies that if AB 6= CD, then eitherAB > CD or AB < CD.

Proof. Given AB and CD, let Q be the intersection point mentioned in thedefinition of AB > CD. Then according to Axiom 75, if Q 6= B then either Q isbetween A and B or B is between Q and A. In the first case we have AB > CD.In the second case, B is inside the circle of radius CD about A. It follows thatD is outside the circle of radius AB and center C. That completes the proof ofthe lemma.

The theory ECG has quantifier-free, disjunction-free axioms. It follows (aswe will prove in Theorem 16.3) that no non-trivial disjunction can be provedin ECG. That is, if P is negative and ECG proves ∀xP (x) → A(x) ∨ B(x),then ECG proves ∀xP (x) → A(x) or ECG proves ∀xP (x) → B(x). Hence,Axiom 75 is not provable in ECG.

When we add Axiom 75, we will also introduce a new construction term,which we write if (AB > CD,P,Q). This abbreviates the official version, whichis if (A,B,C,D, P,Q). Provided AB 6= CD, this term constructs a point whichis equal either to P or to Q, depending on whether AB > CD or AB < CD .The axiom expressing this is

(AB > CD → if (AB > CD,P,Q) = P ) (Axiom 76)∧(AB < CD → if (AB > CD,P,Q) = Q)

Note that Axiom 76 does not contain disjunction, but that by Axiom 75, wehave

AB 6= CD → if(AB > CD,P,Q) ↓ .

230 MICHAEL BEESON

Definition 15.4. The theory ECGD is ECG plus the new function symbol“if ”, with Axioms 75 and 76.

Remark. The “D” in ECGD is for “disjunction”.

The following lemmas give two appealing theorems of ECGD that cannot beproved in ECG (because they are non-trivial disjunctions).

Lemma 15.5. (in ECGD) Let P be point not on line L. Then any point Qis either on the same side of L as P , or on the opposite side.

Proof. Drop a perpendicular K from P to L, meeting L at point R. ProjectingQ to point Q′ on K. Extend segment PR past R by the amount RQ′ twice,arriving at point D on K. Then both R and Q′ are between P and D, so byAxiom 75, either R is between P and Q′ or Q′ is between P and R. In the firstcase, Q′ and P are on the same side of L, and in the second case, they are onopposite sides. But Q and Q′ are on the same side of L. Hence Q and P areon the same side, or on opposite sides, of L. That completes the proof of thelemma.

Lemma 15.6. (in ECGD) In any triangle ABC, either ABC is a left turnor ABC is a right turn.

Proof. In Lemma 15.1, we have already shown how to construct a point Q noton line L = Line (α, β), such that ABC is a left turn or a right turn accordingas Q lies on the same side of L as γ, or on the opposite side. In ECGD byLemma 15.5, Q must lie on one side or the other of L. Hence ABC is either aleft turn or a right turn. That completes the proof.

Terms of ECGD that involve the new symbol if represent geometrical con-structions that can proceed by cases, with comparisons between constructed(unequal) lengths determining the next construction steps. Given Euclid’s cav-alier approach to case splits, the fact that such constructions are not explicitlymentioned in Euclid does not necessarily mean that they are not required togive a correct and complete version of Euclid. The question thus arises, whetherAxiom 75 (or more generally, disjunctive axioms of any kind) are required toformalize Euclid. But because the theorems of Euclid do not mention disjunc-tion in any essential way, we can simply take the double-negation interpretation,and eliminate Axioms 75 and 76, as will be shown below. Thus what happenedin the example of Proposition I.6 happens necessarily in all examples of similarlogical form

§16. Metatheorems.

16.1. Things proved to exist in ECG can be constructed. In thissection we take up our plan of doing for ECG what cut-elimination and recur-sive realizability did for intuitionistic arithmetic and analysis, namely, to showthat existence proofs lead to programs (or terms) producing the object whoseexistence is proved. In the case of ECG we want to produce geometrical con-structions, not just recursive constructions (which could already be produced byknown techniques, since ECG is interpretable in Heyting’s arithmetic of finitetypes). Terms of ECG correspond in a natural way to straightedge and compassconstructions.


Theorem 16.1 (Geometric constructions extracted from intuitionistic proofs).(i)Suppose ECG proves P (x) → ∃y φ(x, y) where P is negative (does not con-tain ∃ or ∨). Then there is a term t(x) of ECG (representing a geometricconstruction) such that P (x) → (φ(x, t(x)) is also provable in ECG.

(ii) Same as (i) but with ECGD in place of ECG.

(iii) Let ECG + DE be ECG, augmented with a constant D and the axiomsaying D is a test-for-equality function. Then the analogue of (i) holds for ECG+ DE.

Proof. We use cut-elimination.31 Since our axiomatization is quantifier-free, ifψ → ∃y φ is provable in constructive ECG, then there is a list Γ of quantifier-freeaxioms such that Γ, ψ ⇒ ∃y φ is provable by a cut-free (hence quantifier-free)proof. Since our axiomatization is disjunction-free, by [18] (or rather, by itsadaptation to multi-sorted logic with the logic of partial terms) we can permutethe inferences so that the existential quantifier is introduced at the last step.Then we obtain the desired proof just by omitting the last step of the proof.That completes the proof of part (i). All the work was in arranging the axiomsystem to be quantifier-free and disjunction-free. Part (iii) is proved in the sameway, noting that the axioms for D are also disjunction-free.

Part (ii) requires a bit more work. Since Axiom 75 contains a disjunction, thereis an issue about permuting the inferences in a cut-free proof. Suppose that wehave a cut-proof of Γ ⇒ ψ → ∃y φ, where Γ is a list of axioms (or subformulasof axioms) of ECGD. Among Γ there may be occurrences of Axiom 75, whichcontains a disjunction. We prove by induction on the number of disjunctions inΓ that there exists a term t of ECGD and a list ∆ of axioms of ECGD suchthat Γ,∆ ⇒ φ(t) is provable. The basis case, when there are no disjunctions,is part (i) of the theorem. Now for the induction step. If the ∃ on the right isintroduced at a lower level (nearer the end-sequent) than the lowest introductionof disjunction on the left, then we can complete the proof as above, since theline just before the ∃ is introduced will contain the desired term, and the ∃-introduction can just be postponed until the end. Otherwise there is a part ofthe proof that looks like this:

AB > AC,Γ1 ⇒ ∃y φ AB < AC,Γ2 ⇒ ∃y φAB > AC ∨AB < AC,Γ1,Γ2 ⇒ ∃y φ

By induction hypothesis, there are terms t1 and t2 such that AB > AC,Γ1 ⇒φ(t1) is provable and AB < AC,Γ2 ⇒ φ(t2) is provable. Let t be the termif (AB > AC, t1, t2). Then ECGD proves AB > AC → t = t1 (by Axiom 76),and ECGD proves AB < AC → t = t2. Hence, for some list ∆ of axiomsof ECGD, there is a cut-free proof of AB > AC,∆,Γ1 ⇒ φ(t), and a cut-freeproof of AB < AC,∆,Γ2 ⇒ φ(t). These two proofs can then be combined asfollows:

AB > AC,∆,Γ1 ⇒ φ(t) AB < AC,∆,Γ2 ⇒ φ(t)

AB > AC ∨AB < AC,∆,Γ1,Γ2 ⇒ φ(t)

31In fact, we use cut-elimination for many-sorted logic with the logic of partial terms. Thedetails of the cut-elimination theorem for such logics have not been published, but they arenot significantly different from Gentzen’s formulation for first-order logic.

232 MICHAEL BEESON

That completes the induction step, and with it, the proof of the lemma.

Theorem 16.2 (Geometric constructions extracted from classical proofs).Suppose ECG with classical logic proves

P (x) → ∃y φ(x, y) where P is quantifier-free and disjunction-free.

Then there are terms t1(x), . . . , tn(x) of ECG such that

P (x) → φ(x, t1(x)) ∨ . . . ∨ φ(x, tn(x))

is also provable in ECG with classical logic.

Proof. This is a special case of Herbrand’s theorem.

Example 2. Euclid’s proof of Book I, Proposition 2 provides us with two suchconstructions, t1(A,B,C) = C and t2(A,B,C) the result of Euclid’s construc-tion of a point D with AD = BC, valid if A 6= B. Classically we have∀A,B,C ∃D(AD = BC), but we need two terms t1 and t2 to cover all cases.

Example 3. Let P and Q be distinct points and L a given line, and A, B, andC points on L, with A and B on the same side of C. Then there exists a pointD which is equal to P if B is between A and C and equal to Q if A is between Band C. The two terms t1 and t2 for this example can be taken to be the variablesP and Q. One term will not suffice, since D cannot depend continuously on Aand B, but all constructed points do depend continuously on their parameters.This classical theorem is therefore not constructively provable.

16.2. Disjunction properties. We mentioned above that ECG cannot proveany non-trivial disjunctive theorem. That is a simple consequence of the factthat its axioms contain no disjunction. We now spell this out:

Theorem 16.3 ( ECG cannot prove a nontrivial disjunctive theorem). Sup-pose ECG proves H(x) → P (x)∨Q(x), where H is negative. Then either ECGproves H(x) → P (x) or ECG proves H(x) → Q(x). (This result dependsonly on the lack of disjunction in the axioms of ECG.)

Proof. Consider a cut-free proof of Γ, H(x) → P (x)∨Q(x), where Γ is a list ofsome axioms of ECG. Tracing the disjunction upwards in the proof, if we reacha place where the disjunction was introduced on the right before reaching a leafof the proof tree, then we can erase the other disjunct below that introduction,obtaining a proof of one disjunct as required. If we reach a leaf of the proof treewith P (x) ∨ Q(x) still present on the right, then it occurs on the left, where itappears positively. Its descendants will also be positive, so it cannot participatein in application of the rule for proof by cases (which introduces ∨ in the leftside of a sequent); and it cannot reach left side of the bottom sequent, namelyΓ, H(x), as these formulas contain no disjunction. But a glance at the rulesof cut-free proof, e.g. on p. 442 of [18], will show that these are the onlypossibilities. That completes the proof.

Theorem 16.4 (Disjunction Properties for ECG and ECGD). Suppose ECGDproves

H(x) → P (x) ∨Q(x)


where H is negative. Then there is a term t(x) of ECGD such that ECG proves

H(x) → t(x) = α ∨ t(x) = β

and ECGD proves

t(x) = α → P (x) ∧ t(x) = β → Q(x).

(Here α and β are two constants of ECGD, with α 6= β an axiom.)

Proof. Suppose ECGD proves H(x) → P (x)∨Q(x). Then also ECGD proves

H(x) → ∃y ((y = α → P (x)) ∧ (y = β → Q(x))).

The formula on the right is disjunction-free, so by Theorem 16.1, there is a termt as required.

§17. Independence results for the parallel axioms. By the interpreta-tions given in the previous section, questions about the logical relations betweendifferent parallel axioms are reduced to the corresponding questions of logicalrelations between different axiom systems for Euclidean fields. For example,those results explain why Axiom 58 implies Euclid’s Postulate 5: it is just thatif reciprocals of non-zero elements exist, then of course reciprocals of positiveelements exist. In fact, technically the interpretations given above already provethat Axiom 58 implies Euclid’s Postulate 5, but we still felt it worthwhile toinclude the direct proof in the previous section.

One might also ask what the field-theoretic version of Playfair’s axiom is.Recall that Playfair says, if P is not on L and K is parallel to L through P , thatif lineM through P does not meet L thenM = K. Since ¬¬M = K → M = K,Playfair is just the contrapositive of Axiom 58, which says that if M 6= K thenM meets L. Hence it corresponds to the contrapositive of x 6= 0 → 1/x ↓;that contrapositive says that if x has no multiplicative inverse, then x = 0.Thus Playfair geometries correspond to ordered fields in which elements withoutmultiplicative inverses are zero.

Thus, constructively, Axiom 58 implies Euclid 5 implies Playfair; we wish toshow that neither implication is reversible. Since classically, the implications arereversible, we cannot hope to give counterexamples. In terms of field theory,we won’t be able to construct a Euclidean ring in which positive elements havereciprocals but nonzero elements do not. We must use some tools of logic.

Among the possible techniques for proving that a constructive implication isnot reversible is the method of Kripke models. We shall not try to explain thistechnique in its full generality, but only in the case of theories based on (ordered)ring theory. A full explanation can be found in [32], but our presentation hereis self-contained.

17.1. Kripke models of ring theory. It will be essential to understand thenotions “term” and “formula” (of ordered ring theory) as logicians use them. Aterm is, intuitively, an expression meant to denote an element of a ring. Avariable (such as x, y, etc.) is a term, and so are the constants 0 and 1. If t ands are terms then so are (t · s) and (t+ s), as well as (−t) and (1/t). In informalusage, many parentheses are left unwritten, according to the usual conventions.

234 MICHAEL BEESON

For example, (1/((x+1) · (y+1))) is a term. Note that 1/0 is technically a term;not every term necessarily denotes something (“is defined”).

Next we explain the notion “formula”. This notion is defined recursively:formulas are built up by combining smaller formulas using the logical connectives∧ (and), ∨ (or), ¬ (not), and → (implies), as well as the quantifiers ∀x and∃x, according to some rules which we will not spell out in detail. The base caseof this definition is the “atomic formula”, which is either of the form P (t) forsome term t, or of the form t = s for some terms t and s, or of the form t ↓ forsome term t. The formula t ↓ is read “t is defined.”

An (ordered) ring is a model of (ordered) ring theory; a Kripke model of(ordered) ring theory is a slightly more complicated thing. It is a collection ofrings, Rα, where the subscripts α come from some partial ordering (K,<). EachRα must have a notion of “positive”, i.e. a subset of so-called positive elements tointerpret the predicate symbol P (x), but it is not required that Rα be an orderedring. The ordering on the index set K has nothing to do with the ordering onthe ring, which is given by a predicate P (x) defining the positive elements. It isusually required that if α ≤ β, then Rα is a subring of Rβ . It will be convenientto generalize this requirement by allowing Rα to be embedded in Rβ by meansof a one-to-one function jαβ (which would be the identity if Rα ⊆ Rβ). Thesefunctions must compose according to the law jαβjβγ = jαγ . Using some abstractnonsense, we could replace each Rα by a suitable copy to ensure that the jαβ areall the identity and Rα ⊆ Rβ , but it will convenient not to require that. Thereis also a requirement of “persistence”: if x is positive in Rα, then jαβx has tobe positive in Rβ. But note, there can also be positive elements in Rβ that donot arise in that way.

The Kripke model R is technically the function α 7→ Rα with domain K,though one often thinks of it as the collection of the Rα. Usually the indexset K has a least element, the “root”. The elements of K are called “nodes”.Usually (and in all the models we will use) the set K is a tree, i.e., the set ofnodes less than any given α is linearly ordered. What we have defined so faris a “Kripke structure”; to be a Kripke model of (ordered) ring theory, or ofordered field theory, the structure must “satisfy the axioms” of the theory. Wenext define that concept.

We consider valuations σ; these are functions that assign an element xσ ofRα to each variable x. (Logicians usually write valuations on the right as xσ,rather than σ(x).) If σ is a valuation, it starts out as a function defined onvariables, but is easily extended to a function defined on terms. For example,(t+ s)σ = tσ+ sσ, where the + on the right is addition in Rα, and the + on theleft is just a symbol. This extended function is a partial function, because (1/t)σis undefined if tσ is 0. If t belongs to the domain of the extended valuation σthen we say t is defined in Rα. If α < β and σ is a valuation into Rα and τ is avaluation into Rβ then we say τ agrees with σ at x if xσ = (jαβx)τ .


We next define the notion “Rα satisfies formula A under valuation σ”, whichis written Rα |=σ A. The rules for this definition are as follows:

Rα |=σ P (x) if and only if xσ is positive in the ordered ring Rα

Rα |=σ t = s if and only if terms tσ and sσ are equal (equivalent) elements of Rα

Rα |=σ t ↓ if and only if tσ is defined in Rα

Rα |=σ (A ∧B) if and only if Rα |=σ A and Rα |=σ B

Rα |=σ (A ∨B) if and only if Rα |=σ A or Rα |=σ B

Rα |=σ (¬A) if and only if for all β ≥ α and valuations τ on Rβ ,

if τ agrees with σ then not Rβ |=σ A

Rα |=σ (A → B) if and only if for all β ≥ α, Rβ |=σ A implies ∃γ ≥ βRγ |=σ B

Rα |=σ (∃xA) if and only if for some τ that agrees with σ except perhaps on x, Rα |=τ A

Rα |=σ (∀xA) if and only if for all β > α, if x ∈ Rβ and

τ is a valuation on Rβ that agrees with σ except perhaps on x, then Rβ |=τ A

A Kripke model of a theory T is one such that all the axioms of T are satisfiedat every node of the model, i.e. Rα |=σ A for every axiom A and every valuation.The Kripke completeness theorem says that the formulas A that are provablefrom T with intuitionistic logic are exactly the formulas satisfied in all Kripkemodels of T . In particular, if one can construct a Kripke model of T in whichsome formula B is not satisfied, then B is not a consequence of T with intu-itionistic logic.32 Such a model is called a Kripke countermodel to B. We willapply this technique to settle the question of the reversibility of the implicationsbetween the three parallel postulates.

17.2. Euclid 5 does not imply Axiom 58. The following concepts willbe used in the next proof, because we will make Kripke models in which the“points” are functions.

Definition 17.1. A function f from R to R is called positive definite iff(x) > 0 for every x.f is called positive semidefinite if f(x) ≥ 0 for all x.f is called strongly positive semidefinite if it is positive semidefinite and

is not zero on any open interval.

Definition 17.2. A Pusieux series in t is a power series in a rational powerof t, convergent in a neighborhood of 0.

A Pusieux series at a in t is a power series in a rational power of (t− a),convergent in a neighborhood of a.

A generalized Pusieux series in t is a Pusieux series in t or a power seriesin a rational power of t, times |t|, convergent in a neighborhood of 0.

A generalized Pusieux series in t at a is a Pusieux series in (t − a) ora power series in a rational power of (t − a), times |t − a|, convergent in aneighborhood of a.

32See [32] for a proof of the Kripke completeness theorem, and Exercise 4, p. 99 of [2] forthe extension to the logic of partial terms.

236 MICHAEL BEESON

A Pusieux series at ∞ is a Pusieux series convergent in a neighborhood of∞; similarly for a generalized Pusieux series at ∞.

For example,√t3 = |t|t has a generalized Pusieux series at 0, but not a Pusieux

series.

Theorem 17.3. (i) If the strong parallel axiom SPP in ECG is replaced byEuclid’s Postulate 5, then SPP is not provable.

(ii) In Euclidean field theory, the existence of reciprocals of positive elementsdoes not imply the existence of reciprocals of nonzero elements, i.e. Axiom EF7does not imply Axiom EF1 (using axioms EF0 and EF2-EF6).

Proof. We first show that (ii) implies (i). Suppose Euclid 5 plus neutral ECGdoes imply SPP. Then, by Theorem 12.21, the theory of weakly Euclidean fields(i.e., EF with Axiom EF1 replaced by EF7, that positive elements have recipro-cals) proves the interpretation SPP of SPP. Now we argue that nonzero elementshave reciprocals. Assume m 6= 0 and consider the line through (0, 1) with slopem. Then by SPP, the line meets the x-axis. The meeting point (a, 0) lies onthe line, so 1 = am. But then m has multiplicative inverse a. Hence nonzeroelements have reciprocals. But that contradicts (ii). Hence (ii) implies (i) asclaimed.

We now proceed to part (ii). Let K be the field of “constructible numbers”,which is the least subfield of R closed under taking square roots. Let A0 be thering of polynomial functions from R to R with coefficients in K. A0 is not a field,since for example 1/t does not belong to A0. For each nonnegative integer n, wedefine the ring An+1 to be the least set of real-valued functions containing An

together with all sums and differences of members of An, together with all squareroots of positive semidefinite members of An, and reciprocals of all stronglypositive semidefinite members of An. These square roots and reciprocals aredefined on dense subsets of R, as we will show below. For example, the functions√

1 + t2 and 1/(1 + t2) are in A1, and√

√

1 + t2 +√

1 + t4 +1

1 + t2

is in A2. Also√x2 = |x| is in A2, and |x| − x is in A3; that function is zero on

the positive real axis. Now define A to be the union of the An. Then A is a ringof functions from R to R. Another way of describing A is to say that it is theleast ring of functions containing K[x] and closed under taking square roots andreciprocals of positive semidefinite functions.

We claim that each f in An (i) is defined at all but a finite number of pointsin R, and has a Pusieux series at all but a finite number of points a, and also hasa Pusieux series at ∞, and a generalized Pusieux series at the remaining pointsof its domain, and (ii) f is zero on a finite number of closed intervals and a finitenumber of isolated points.

For example |x| =√x2 does not have a Pusieux series at 0; but such points

occur only at the zeroes of functions belong to An−1, as we will show in detailbelow.

We prove (i) and (ii) simultaneously by induction on n. First we show that(i) at n follows from (ii) at n − 1. Case 1, f is a square root, f =

√g and


g(x) ≥ 0 for all x. Then at points where g(x) > 0, f has a Pusieux series, and atthe finitely many points (by (ii) for n− 1) where g(x) = 0, f has a generalizedPusieux series; and f is everywhere defined and has a Pusieux series at ∞. Case2, f is a reciprocal of a strongly positive semidefinite function g in An−1. Thenby (ii) for n − 1, g has only finitely many zeroes (since there are no intervalsof zeroes because g is strongly positive definite), so the domain of f omits onlyfinitely many points, and at each of those points where g is positive, f is definedand has a Pusieux series, and at the finitely many points where g is zero, it isundefined; so (i) for n follows from (ii) for n − 1 also when f is a reciprocal.When f is a sum or difference, i for n follows from (i) for n− 1. Now we showthat (ii) at n follows from (i) at n. By (i) at n, f in An has a Pusieux seriesat a for all but finitely many points a; and at the points where f has a Pusieuxseries, if f(a) = 0 then the zero is isolated, or the function is identically zero inan interval about a. The endpoints of these intervals of zeroes will be among thefinitely many points where f does not have a Pusieux series, so there are finitelymany of these intervals. Hence the zeroes of f consist of (at most) finitely manypoints where f does not have a Pusieux series, plus finitely many intervals, plusfinitely many points where f does have a Pusieux series–altogether finitely many.So (ii) for n does follows from (i) for n, as claimed. That completes the inductiveproof of (i) and (ii).

We call a zero of f “half-isolated” if it is the endpoint of one of the intervalson which f is zero. By (i) and (ii), there is a countable set of reals that includesall the isolated or half-isolated zeroes and all the singularities of all the functionsin A. Define Ω to be the complement of that set; thus for each f in A, if f(x)is zero for any x in Ω, then f is identically zero on some interval about x. Notethat, since the complement of Ω is countable, Ω is dense in R.

We will exhibit a Kripke model in which positive elements have reciprocals,but the model does not satisfy that all nonzero elements have reciprocals. Thepartially ordered set K is 0∪Ω, ordered so that 0 is the root node, and 0 < αfor each α ∈ Ω, and the different elements of Ω are incomparable. The root nodeof this model is the ring A, with P (x) interpreted to mean “x is strongly positivesemidefinite.” Like A0, A is not a field because 1/t is not in A; but also A isnot even an ordered ring, because there are polynomials x such that neither xnor −x is strongly positive semidefinite. But we can still use A as the root ofour Kripke model.

For α ∈ Ω, the structure Aα at the node α is the quotient field of A, with P (x)interpreted to mean that x(α) > 0, and x = y interpreted to mean that x andy are equal on some neighborhood of a. In other words, the elements of A(α)are equivalence classes of quotients of members of A, where f/g is equivalent tou/v if vf and ug agree in some neighborhood of α. Since Ω does not include anysingularities or zeros of members of A, x(α) is defined for each x in Aα.

We note that Aα is isomorphic to the least Euclidean subfield of R containingα. The isomorphism is given by x 7→ x(α). It is an isomorphism because itskernel is trivial, since x(α) = 0 only if x is identically zero on some neighborhoodof α. Moreover, if P (x) holds in A, then x is strongly positive semidefinite andnot identically zero; hence for α in Ω, we have x(α) > 0, since x(α) 6= 0 for α inΩ. Hence P (x) holds in Aα.

238 MICHAEL BEESON

The ring axioms are satisfied in this Kripke structure, since all these structures(A and the Cσ) are rings. The ordered field axioms are satisfied at the leaf nodesAα, since these structures are classical ordered fields. We therefore only need toverify the reciprocal and order axioms at the root node A.

Consider Axiom EF2, which says that sums and products of positive elementsare positive. This holds at A since the sum and product of strongly positivesemidefinite functions are also strongly positive semidefinite.

Consider Axiom EF3,which says that not both x and −x are positive. Supposeboth x and −x are strongly positive semidefinite members of A. Then for eachα ∈ Ω we have x(α) > 0 and −x(α) > 0, since x(α) is not zero for α in Ω. Butthat is a contradiction; hence Axiom EF3 holds at A.

Consider Axiom EF4, which says that if both x and −x are not positive, thenx is zero. Suppose both x and −x are satisfied at A to be not positive. Thatmeans that for every node Aα, x and −x are not positive at Aα. That meansthat for every α ∈ Ω x(α) ≤ 0 and −x(α) ≤ 0. Hence, x(α) = 0 for every αin Ω. But if x is not identically zero, then x(t) 6= 0 for some t, and since allthe functions in A are continuous on their domains, xα) 6= 0 for some α in Ω,contradiction. Hence A satisfies Axiom EF4.

Consider Axiom EF5, which says that if −x is not positive, then x has asquare root. If x is identically zero there is nothing to prove, so we may assumethat x is not identically zero. If A satisfies that −x is not positive, that meansthat −x is not positive in any Aα; that is, −x(α) ≤ 0 for all α ∈ Ω. Thenx(α) ≥ 0. Since this is true for every α ∈ Ω, and since Ω is dense in R, and x

is continuous, it follows that x is positive semidefinite. Hence√

x(σ) belongs toA, by construction of A. Hence A satisfies Axiom EF5.

We now consider Markov’s principle EF6. Suppose that ¬¬P (x) is satisfiedat the root node A. Then for every α in Ω, P (x) is satisfied at the leaf nodeAα; that means that x(α) > 0 for each α in Ω. As shown in the verification ofE5, this implies that x is positive semidefinite; and it is not identically zero on aneighborhood of α, since x(α) > 0. Hence P (x) is satisfied at the root node A.

Now we consider Axiom EF7, which says positive elements have reciprocals.Suppose P (x) is satisfied at the root node; then x is strongly positive semidefiniteand not identically zero, and x belongs to some An; so 1/x belongs to An+1.

Therefore the Kripke structure K is a model of all the axioms EF2-EF7. Wewill now show that it does not satisfy EF1, which says that all nonzero elementshave reciprocals. Consider the element of A given by the identity function,i(t) = t. Suppose A satisfies i ·y = 1, where 1 is the constant function with value1. Then for each real number t we have ty(t) = 1. But this is a contradictionwhen t = 0. Hence A does not satisfy EF1.

That completes the proof.

Remark. In the Kripke model K, the line parallel to the x-axis, passing throughthe point (0, 1) with slope t, where t is the element of A representing the identityfunction, will intersect the x-axis if t is positive or negative, but if we only assumethat the slope t is nonzero, we cannot construct the intersection point. Indeed,in different extensions Cσ of A, t might be positive, or it might be negative, sothe intersection point (for small t) will be a point on the x-axis, far from the


origin, but it will be a large positive value in some of the Cσ and a large negativevalue in others.

17.3. Playfair does not imply Euclid 5. Recall that “Playfair field the-ory” is the theory of ordered fields satisfying Axioms EF0, EF2-EF6, and EF9-EF10. That is, without the axiom of reciprocals (EF7), but with the axiom(EF9) that elements without reciprocals are zero, and the axiom (EF10) thatsays if x is greater than a positive invertible element, then x is invertible.

Theorem 17.4. (i) If the parallel axiom (Axiom 58) in ECG is replaced byPlayfair’s Axiom, then Euclid’s Proposition 5 is not provable, and (hence) Axiom58 is not provable.

(ii) Playfair field theory does not imply that reciprocals of positive elementsexist.

Proof. By the results of the previous section, part (ii) implies part (i). Wetherefore proceed to part (ii). To say that ¬∃y (y · x = 1) holds at a node Aof a Kripke model is to say that no node above A contains an inverse of x. Ifone of the leaf nodes above A is a (classical) field, then x must be zero in thatfield, and hence in A also. Hence the axiom that elements without reciprocalsare zero will hold in any Kripke model, all of whose leaf nodes are fields. Whatwe need, then, is a Kripke model in which all the leaf nodes are ordered fields,and the root node has a positive element without a reciprocal.

We say that x is “almost positive definite” if it is positive semidefinite, and itis not zero on any open interval. We construct a model similar to the one in thepreceding proof, except that when constructing A, we throw in square roots ofpositive semidefinite functions, but reciprocals only of functions bounded belowby functions that already have reciprocals, instead of reciprocals of all positivesemidefinite functions. More precisely, Let A0 be the ring of polynomial functionsfrom R to R with coefficients in K. For each nonnegative integer n, we definethe ring An+1 to be the least set of real-valued functions containing An togetherwith all sums and differences of members of An, together with all square rootsof positive semidefinite members of An, together with 1/f for each functionf in An such that for some function g with both g and 1/g in An, we havef(x) ≥ g(x) ≥ 0 for all x in the domain of both f and g. These square rootsand reciprocals are defined on dense subsets of R, as we will show below. Nowdefine A to be the union of the An. Then A is a ring of functions from R toR. Another way of describing A is to say that it is the least ring of functionscontaining K[x] and closed under taking square roots of positive semidefinitefunctions, and closed under the rule that if g ≤ f and 1/g is in A then 1/f is inA.

The only invertible elements in A0 are the nonzero constants, soA1 will containthe reciprocals of functions that are bounded below by some positive constant,and A2 will contain the reciprocals of functions bounded away from zero by somepositive constant.

As in the previous proof, all the members of A have Pusieux series exceptat finitely many points, and their zero sets have the same structure as before,so we can again find a countable set containing all the singularities and all theisolated and half-isolated zeroes. let Ω be the complement of this countable set,

240 MICHAEL BEESON

and define a Kripke model as before, with index set 0,Ω and R0 = A and forσ ∈ Ω, Rσ = Cσ. In this model, we claim that all members of A are continuouson the real line. To prove this, we proceed by induction on n to prove that themembers of An are continous on the whole real line. That is true for n = 0 sincethe elements of A0 are polynomials. The square root of a positive semidefinitefunction continuous on R is also continuous on R, as are sums and differencesof such functions. Suppose f(x) ≥ g(x) ≥ 0 and f and g and 1/g are in An.Then 1/f will be in An+1. By induction hypothesis, both g and 1/g are definedeverywhere; hence g(x) is everywhere positive. Hence f(x) is also everywherepositive; hence 1/f is everywhere defined and continuous. That completes theproof that all members of A are continuous on R.

We define P (x) to hold at the root node if and only if x is strongly positivedefinite. The verification of axioms E2 through E6 is the same as in the previousproof.

Now we verify Axiom EF10. It is automatic at the leaf nodes since there thestructures are strong Euclidean fields. We check the root node. Suppose P (g) issatisfied at the root; then g is strongly positive definite. Suppose also ¬P (g− f)is satisfied; then f(x) ≥ g(x) for all x. Suppose g has a reciprocal in the model.Then for some n, f , g, and 1/g belong to An. Then by construction of themodel, 1/f belongs to An+1. Hence Axiom EF10 holds.

It only remains to show that EF7 is not satisfied at the root node. If EF7were satisfied at A, then every strongly positive definite element of A would haveits reciprocal in A. To refute this, we must exhibit a strongly positive definite

element of A whose reciprocal is not in A. Take f(x) = |x|. Since |x| =√x2,

this function is in A1. But 1/f is not defined at x = 0, and we have provedabove that all members of A are defined and continuous on the whole real line.Hence 1/|x| is not in A. That completes the proof.

17.4. Independence of Markov’s principle.

Theorem 17.5. Markov’s principle is independent of the other axioms of ECG.

Proof. We construct a model similar to the ones used above. This time, letA be the least class of functions containing the polynomials, and closed undersquare roots of positive semidefinite functions and reciprocals of not-identically-zero functions. Interpret P (x) at the root A as “x is positive definite”, insteadof as “x is positive semidefinite and not identically zero.” Then the functionx(t) = t2 is not satisfied to be positive at the root node, but since 0 is not in Ω,we have x(α) > 0 for all α in Ω, so P (x) is satisfied at every leaf node. Hence¬¬P (x) is satisfied at the root node. Hence ¬¬P (x) → P (x) is not satisfied atthe root. That completes the proof.

§18. Apartness. What happens if we add apartness?

18.1. Constructions and Apartness. Mandelkern’s line chooser; otherapartness constructors;

18.2. Euclidean fields and apartness.

18.3. Apartness and the parallel axioms. In particular are the parallelaxioms still independent?


Lemma 18.1. ECGD proves x 6= 0 → x > 0 ∨ x < 0.

Proof. Since x, 0, and I are distinct and not collinear, xyI is either a left turnor a right turn, as shown in [??], Lemma 9.3. Note that Axiom 75 is required forthat lemma, so we need ECGD, not just ECG, in the theorem. That completesthe proof.

242 MICHAEL BEESON

§19. Appendix A: Roads not taken. In the course of developing ECG, wemade several decisions, in some cases exploring alternative formulations to somedepth. We chose a rigid compass (i.e. construction of circles from center andradius); we chose to allow degenerate circles; we rejected a primitive operation ofprojection on a line; we chose to have a term center(C) for the center of a circle.In this appendix we justify, or at least explain, those choices, and mention a fewrelated problems, some but not all of which we solve.

19.1. Rigid compass undefinable from the collapsible compass. BookI, Proposition 2 has been discussed in the introduction. The question it addressesconcerns the constructor Circle3 (A,B,C), which constructs a circle with centerA and radius BC. As discussed in the introduction, Euclid gives a term thataccomplishes this aim under the assumptions, not only that B 6= C, but also thatA 6= C ∨B 6= C. We consider the stronger theorem “uniform Euclid I.2”, whichasserts that for every A and BC with B 6= C, there exists a D with AD = BC.The “uniformity” refers to the missing assumption A 6= B. To “realize” uniformEuclid I.2, we would need a term e(A,B,C) that produces D uniformly, whetherA = B or not. If we had such an e, then of course we could define

Circle3 (A,B,C) = Circle (A, e(A,B,C)).

Conversely, such an e can be defined from Circle3 like this:

e(A,B,C) = pointOnCircle (Circle3 (A,B,C))

But the mere fact that Euclid’s own construction does not suffice to defineCircle3 does not show that some other construction won’t do the job. In thissection we will prove that it is really true that no other construction can do thejob:

Theorem 19.1. Circle3 is not definable in the standard plane R2 from the(other) elementary constructions.

We will work up to this theorem gradually, beginning by showing that thecollapsible compass cannot simulate the rigid compass if degenerate circles arenot allowed, and then elaborating the proof to allow degenerate circles.

For that we consider the model R−2, which is the same as R2 except thatCircle (A,A) is undefined.

Theorem 19.2 (with Freek Wiedijk). No total unary point-valued function(other than the identity) is definable in the standard model R2− (with degen-erate circles undefined) from the elementary geometrical constructions excludingCircle3 . More precisely, let t be a compound term containing exactly one variableA of type Point, and no other variables, but possibly containing some constantsα, β, . . . of type Point. Suppose that t does not contain Circle3. Let α be aconstant contained in t, or if t has no constants, let α be any constant. Let theconstants be interpreted as certain (fixed) distinct points α, β, . . . in the standardplane, and let t be the interpretation of t when the variable A is interpreted asα. Then t is undefined.

Proof. We start by eliminating the “overloaded” versions of the elementary con-structions from t. For instance, if t contains a subterm IntersectLines (a, b, c, d),


we replace it by a term using the “fundamental” form of the construction,IntersectLines (Line (a, b),Line (c, d)). The result of such replacements is a termwith the same value as t under any assignment of a value to the variable A, andcontaining no variables of types other than Point .

We proceed by induction on the complexity of terms t as in the theorem, butalso containing only the fundamental versions of the elementary constructions(no overloaded versions).

Since the theorem only applies to compound terms, the basis case occurs whent has only variables or constants for arguments. We note that Circle (A,A),Line (A,A), Ray (A,A), Segment (A,A), Arc (A,A,A) are undefined. But wealso have to consider the possibility that the constants β or γ occupy one of theargument places. For example, Circle (A,α) is undefined when A takes the valueα. In the rest of the proof Similarly, when A takes the value α, Circle (α,A),Line (A,α) and Line (α,A) are all undefined. Note that this argument does notwork for Circle3 , since Circle3 (A, β, γ) is always defined, but by hypothesis, tdoes not contain Circle3 .

Since t does not contain any overloaded constructions, the basis case is fin-ished, as there is no fundamental construction that takes only arguments of typePoint. Specifically, there are just five fundamental constructions for producingthe intersections of lines and lines, lines and circles, or circles and circles, andthey each need arguments of type Line or Circle.

Now consider the induction step. If the main symbol of t is a constructor, suchas Line, then t has the form Line (a, b). One of a or b must contain a variable,and hence be somewhere undefined. Hence t is also somewhere undefined, andindeed the same assignment of a value to A that makes a or b undefined will work.Similarly for the other constructors (since Circle3 , to which this argument doesnot apply, is not allowed).

Next consider the case when t is IntersectLines (p, q). Then p and q have typeLine. The only terms of type Line are those of the form Line (a, b), or variablesof type Line. But t is not allowed to contain variables of type Line. Therefore tmust have the form,

IntersectLines (Line (a, b),Line (c, d))

Since t contains a variable, one of s = Line (a, b) or s = Line (c, d) must contain avariable, and by induction hypothesis the term sσ is undefined when substitutionσ assigns the variable of s to one of the constants in s (or any constant if s hasno constants). Hence tσ is also undefined.

Next consider the case when t is IntersectLineCircle1 (p, q). Since t containsthe variable A, then one of the two arguments of t contains A. By induc-tion hypothesis, whichever it is, its interpretation (be it p or q) is undefined,and that makes t undefined as well. Similarly when t is IntersectLineCircle2 ,IntersectCircles1 , or IntersectCircles2 . There is also the possibility that t hasthe form f(p), where f is one of the access functions pointOnCircle , pointOn1 ,or pointOn2 ). Then p must contain the variable A, and by induction hypothesis,p is undefined. That completes the proof.

The following corollaries show that the rigid compass is not definable fromthe collapsible compass, at least if the collapsible compass cannot be used to

244 MICHAEL BEESON

draw circles of zero radius. The first corollary shows the serious limitations ofgeometry with only a collapsible compass and no equality-test constructor.

Corollary 19.3. Let e(A) be any term built up from the elementary con-structions, not containing Circle3, having type Point, and containing exactlyone variables A of type Point. Then it is not the case that in R2−, we alwayshave e(A) 6= A.

Proof. According to the theorem, e(A) has to be undefined for some value of thevariable A.

Corollary 19.4. Let e(A,B,C) be any term built up from the elementaryconstructions, not containing Circle3, having type Point, and containing exactlythree variables A, B, and C of type Point. Then it is not the case that in R2−,whenever B 6= C then e(A,B,C) is defined and is a point D such that AD = BC.

Proof. Let us invent two constants β and γ, and interpret them as two distinctpoints β and γ (fixed for the rest of the proof). Then let f(A) = e(A, β, γ).Suppose, for proof by contradiction, that e(A,B,C) is defined whenever B 6= Cand is a point D such that AD = BC. Then f(A) is defined for all A and isalways different from A, contradicting the previous lemma.

Corollary 19.5. Circle3 is not definable in the model R2− from the (other)elementary constructions.

Proof. Circle3 is a term e fulfilling the hypotheses of the previous corollary.

19.2. Rigid compass not definable, even with degenerate circles.Next we will extend these results to the standard model R2, that is, we willremove the restriction that circles must have positive radius.

Theorem 19.6. let t be a compound term containing exactly one variable Aof type Point, and no other variables, but possibly containing some constantsα, β, . . . of type Point. Let α, β, . . . be the interpretstions of these constants.Suppose that t does not contain Circle3. Let t be the interpretation of t in R2.Let α be any constant contained in t, or if t does not contain any constant, letα be any constant. Let A be interpreted as α. Then either

(i) t is undefined, or

(ii) t is of type Point, and t is equal to α; or

(iii) t is of type Circle, and t is a degenerate circle with center equal to α.

Remark. The restriction that the term not contain Circle3 is certainly necessary;not only does the proof depend on it, but we will show in the next section thatcircles of zero radius are actually useful in connection with Circle3 , while thistheorem shows that they are essentially useless if we only have the collapsiblecompass constructor Circle .

Proof. As in the proof of the previous theorem, we proceed by induction onthe complexity of terms t with variables only of type Point, and built up usingpoints, lines, and circles only (not rays and arcs). Since the theorem only appliesto compound terms, the basis case occurs when t has only variables or constantsfor arguments. Now Circle (A,A) is no longer undefined; but it is a degenerate


circle, and when A is interpreted as α, then the conclusion of the theorem is true,that Circle (A,A) is interpreted as a degenerate circle with center α. We alsohave to consider Circle (α,A) and Circle (A,α), but both of these degenerate asrequired when A = α. We do not have to consider Circle (α, β) since t is requiredto contain at least one variable. Line (A,A) is undefined, as before, no matterhow A is interpreted. Line (α,A) and Line (A,α) become undefined when A isinterpreted as α. Line (α, β) does not contain a variable and so does not need tobe considered. That completes the basis case.

Now consider the induction step. If the main symbol of t is Line, then t hasthe form Line (a, b). One of a or b must contain the variable A of t, say a. Sincea has type Point, the interpretation a of a is either undefined, or equal to α. Ifit is undefined then so is Line (a, b) and we are done. So we may suppose a = α.Similarly, we may suppose b = α. Then Line (a, b) is interpreted as Line (α, α,which is undefined.

Next consider the case when t is IntersectLines (p, q). Then t must have theform,

IntersectLines (Line (a, b),Line (c, d))

Since t contains a variable, one of s = Line (a, b) or s = Line (c, d) must contain avariable, and by induction hypothesis the term s is undefined, since alternatives(ii) and (iii) cannot apply to a term of type Line. But then t is also undefined,since it has an undefined argument.

Next consider the case when t is IntersectLineCircle1 (Line (a, b),Circle (c, d)).If Line (a, b) contains the variable A, then its interpretation is undefined, andhence t is undefined. So we may assume that Line (a, b) does not contain A; thenCircle (c, d) contains A. By induction hypothesis its interpretation is undefined(in which case we are done with this case) or it is a degenerate circle withcenter α. In that case, the intersection points of Line (a, b) are equal and bothof them are equal to α, so alternative (ii) of the theorem holds. Similarly forIntersectLineCircle2 .

Next consider the case when t is IntersectCircles1 (Circle (a, b),Circle (c, d)).If the interpretation of either argument is undefined, then t is undefined tooand we are done with this case. So we may assume that the arguments aretwo degenerate circles with the same center α. But the intersection points oftwo circles with the same center are undefined (whether or not the circles aredegenerate), so t is undefined. Similarly for IntersectCircles2 .

That takes care of the five constructors; but there are also the access functionsto consider. Consider the case when t is pointOnCircle (Circle (a, b)). By induc-tion hypothesis, the interpretation of Circle (a, b) is either undefined, in whichcase we are done with this case, or a degenerate circle with center α. In thatcase the interpretation of pointOnCircle (Circle (a, b)) is also α, as that is theonly point on the degenerate circle. Hence alternative (ii) of the theorem holds.

Consider the case when t is pointOn1 (Line (a, b). The interpretation of thisterm is a, which by induction hypothesis is either undefined or α, so we are donewith this case. Similarly for pointOn2 (Line (a, b)). That completes the proof.

246 MICHAEL BEESON

Corollary 19.7. Without Circle3 , we cannot even define a construction fsuch that for each point A, f(A) is a point different from A, even if we allowdegenerate circles.

Proof Any definable function f will either be undefined somewhere, or will fix aconstant α that is used in the definition.

Finally we reach the theorem announced at the beginning of this section:

Theorem 19.8. Circle3 is not definable in the standard plane R2 from the(other) elementary constructions.

Proof. If it were definable, then let α and β be distinct constants, and definef(A) = Circle3 (A,α, β). But according to the above theorem, f(A) is undefinedor a degenerate circle when A is interpreted as α. That is a contradiction, sinceCircle 3(A,α, β) is defined and not degenerate, since α 6= β. That completes theproof.

The fact that Circle3 is not definable in R−2 or in R2 means that, if we donot include Circle3 as a primitive construction, we shall not be able to defineit in any axiomatic theory that has R2− or R2 for a model. It seems clearthat Euclid’s book does have R−2 as a model. Although Euclid does not usedegenerate circles, we have now proved that they cannot help; so the uniformversion of Book I, Proposition 2 is essentially non-constructive, using Euclid’snon-rigid compass. We therefore add Circle3 as a fundamental construction(rendering I.2 a triviality) and give a constructive theory that works for the restof Euclid. In the next section we give further justifications for adding the rigidcompass Circle3 .

19.3. Circles of zero radius, the rigid compass, and projection. Inthis section, we address the two issues of whether our basic theory ought toinclude the rigid compass, represented by Circle3 , and whether we ought toallow circles of zero radius. We conclude that both should be allowed. Ourapproach is to consider what is required to achieve a formalization of Euclid usingintuitionistic logic, with no test-for-equality construction needed. We exhibitthree important constructions that are possible with Circle3 , and at least oneof them (projection on a line) is definitely not possible without Circle3 .

Projection is absolutely necessary in order to reduce geometry to algebra. Wewant to pick a line, call it X , and erect a perpendicular Y to X , and projecteach point P onto its coordinates x = Project(P,X) and y = Project(P, Y ).We show below that we can define projection using Circle3 , if circles of zeroradius are allowed. This is the main reason why we want to allow circles of zeroradius. We considered replacing Circle3 by projection, but Circle3 seems morefundamental.

Here are some possible constructions we wish to consider.Euclidean Extension. One of the Euclidean axioms says that we can extend a

given segment AB by a segment CD. More precisely, we can construct a pointP = Extend (A,B,C,D) such that BP = CD and B is between A and P . Theassumptions here are that A 6= B and C 6= D, but it is not assumed that B 6= Cor B 6= D.

Strong Extension. Extend (A,B,C,D) is the unique continuous extension ofExtend (A,B,C,D) that is defined when A 6= B, i.e. without assuming C 6= D.


When C = D, we have

Extend (A,B,C,D) = B.

We will show below how to define Extend in terms of Circle3 .The following lemma helps make the case for allowing Circle3 , since we need

projection. The lemma shows that we can’t define projection without Circle3 ,so the alternative to Circle3 would be to take projection itself as primitive.

Lemma 19.9. In R2− (where circles of zero radius do not exist), as well as inR2 (where they do exist) projection is not definable in terms of the elementaryconstructions without Circle3.

Proof. Let α and β be constants, whose values will be two distinct (fixed)points. Let t(A) be the term Project(A,Line (α, β)). Then t(A) is defined for allvalues of A. By Theorem 19.2, t(A) is not definable in terms of the elementaryconstructions without Circle3 , in the model R2−. That completes the proof.

19.4. Constructing the center of a circle. We chose to include an acces-sor function center producing the center of a circle. There is a question whetherincluding center is faithful to Euclid, since Euclid III.1 provides a way to con-struct center(C) from two points on circle C. Since circle C could only havebeen drawn using its center, in III.1 Euclid appears to tell us, if somehow thecenter of C got accidentally rubbed out of the diagram, here is how you canrecover it. In any event, the existence of Prop. III.1 may be taken to indicatethat we could omit the function symbol center(C) and replace it with the termfor the point constructed in III.1? There are two answers to this suggestion. Thefirst one is that the proof of III.1 is erroneous: it implicitly requires the use ofIII.2, that a chord of a circle lies inside the circle, in order to know that point Ein III.1 exists. But the proof of III.2 requires III.1, that the center of the circlebe known. However, as Heath’s commentary on this proposition [8] shows, thereis another way to construct the center, due to de Morgan, which is correct. Thesecond answer is that Euclid’s construction in III.1 requires two points A and Bon circle C. One point, say A, can be obtained as pointOnCircle (C). But thereis no good (uniform) way to obtain a second point, without knowing the centerof C. The second point can’t be obtained as the intersection of some line throughA with C, because we have no good candidate for a second point on that line.Could we obtain it as the intersection of a circle with center A with C? To knowthat circle K with center A meets C we would have to know that K contains apoint inside C. That is hard to do without having the center of C. Hence, evenif Euclid III.1 were correctly proved, which it is not, we still would need the termcenter(C), or a way to get a second point. Finally, Heath’s commentary showsthat de Morgan’s method requires three distinct points on C. The conclusionis this: we could eliminate the function symbol center, but we would then needthree point-access functions guaranteed to produce three distinct points on a cir-cle. The question would naturally arise, how do we construct these three pointsin a uniform way? We feel that it is more natural to have the symbol center,and regard (de Morgan’s repaired) version of Euclid III.1 as a reconstruction ofthe center from three points, rather than a construction from a circle given deusex machina.

248 MICHAEL BEESON

We can pose technical questions in this connection: Is center definable interms of the rest of the elementary constructions? In other words, if we deletethe function center, is there a term t(C) with C of type Circle such that t(C) isalways the center of circle C? We conjecture there is no such term. We know byde Morgan that there is a term t(p, q, r, C) such that if p, q, and r are distinctpoints on C then t(p, q, r, C) is the center of C. What about with two pointsinstead of three? We conjecture it is not possible.

§20. Appendix B: Possible reductions in the primitive constructionmechanisms. While it is natural to take as primitives the operations of inter-secting a line and a circles, and intersecting two circles, it is known that this isnot a primitive set of constructions. In this section we summarize the knownresults, and show that in one case, the known result cannot be improved.

20.1. The Mohr-Mascheroni theorem. In the “algebraic framework”, whereinstead of logical theories one just considers “algebras” with operations for thegeometric constructions, one can ask for a minimal set of primitives. The Mohr-Mascheroni theorem shows that, in classical geometry, “every geometric con-struction carried out by compasses and ruler can be done without ruler.” Moreprecisely,

Theorem 20.1 (Mohr-Mascheroni). All points constructible using the five el-ementary constructions can be constructed using Circle3 , IntersectCircles1 andIntersectCircles2 only.

Proof. See [16], which also explains the history of this theorem, which goes backto the time of Galileo.

But the proof of the Mohr-Mascheroni theorem uses a case distinction as towhether a line passes through the center of a circle or not. Hence, it will notproduce a single term in the algebra of constructions that will defineIntersectLineCircle1 , but rather, two terms, one of which does the job when theline passes through the center of the circle, and the other when it does not. Thatimplies that the proof is non-constructive.

Is the theorem non-constructive, or only the known proof? One approach tomaking this question precise is to ask whether there is a single term of ECGthat produces the same result as IntersectLineCircle1 ; or a single term i(L,C)that always produces an intersection point of a line L and circle C if they haveone. Another approach is to remove the axioms for IntersectLineCircle1 andIntersectLineCircle2 from ECG and ask if the resulting theory can prove theline-circle continuity axiom of Tarski’s theory. The answers to these questionsare presently unknown.

20.2. Strommer’s theorem. The Mohr-Mascheroni theorem shows us howto eliminate IntersectLineCircle1 and IntersectLineCircle2 in favor ofIntersectCircles1 and IntersectCircles2 , at least classically. On the other hand,it is also possible to eliminate IntersectCircles1 and IntersectCircles2 in favorof IntersectLineCircle1 and IntersectLineCircle2 , as Strommer showed in [28].Apparently this fundamental result was not known until this 1973 publication!while the Mohr-Mascheroni theorem has 450 years of history. Strommer’s con-struction is uniform, i.e., the same construction term applies in all cases, and


works even when the two intersection points of the line and circle coincide, en-abling us to construct the common tangent of the circles.

REFERENCES

[1] Jeremy Avigad, Edward Dean, and John Mumma, A formal system for euclid’s ele-ments, Review of Symbolic Logic, vol. 2 (2009), pp. 700–768.

[2] Michael Beeson, Foundations of constructive mathematics, Springer-Verlag, BerlinHeidelberg New York, 1985.

[3] , Constructive geometry, Proceedings of the tenth asian logic colloquium,

kobe, japan, 2008 (Singapore) (Tosiyasu Arai, editor), World Scientific, 2009, pp. 19–84.[4] Errett Bishop, Foundations of constructive analysis, McGraw-Hill, New York,

1967.[5] Karol Borsuk and Wanda Szmielew, Foundations of geometry: Euclidean and

bolyai-lobachevskian geometry, projective geometry, North-Holland, Amsterdam, 1960,translated from Polish by Erwin Marquit.

[6] Max Dehn, Die grundlegung der geometrie in historischer entwicklung, B. G. Teub-ner, 1926, Appendix in [24].

[7] Rene Descartes, The geometry of rene descartes, with a facsimile of the first

edition, published in 1637 as an appendix to discours de la methode., Dover, New

York, 1952, Original title, La Geometrie. Facsimile with English translation by David EugeneSmith and Marcia L. Latham.

[8] Euclid, The thirteen books of the elements, Dover, New York, 1956, Three volumes.Includes commentary by the translator.

[9] , The bones: A handy, where-to-find-it pocket reference companion to eu-

clid’s elements, Green Lion Press, Santa Fe, New Mexico, 2007, Statements and diagramsonly, no proofs.

[10] , Euclid’s elements, Green Lion Press, Santa Fe, New Mexico, 2007, All thir-teen books in one volume, without commentary.

[11] Kurt Godel, Zur intuitionistischen arithmetik und zahlentheorie, Kurt godel, col-

lected works volume i, Oxford University Press, 1933, with English translation, pp. 286–295.[12] Marvin Jay Greenberg, Euclidean and non-euclidean geometries: Development

and history, fourth ed., W. H. Freeman, New York, 2008.[13] Haragauri Narayan Gupta, Contributions to the axiomatic foundations of geometry,

Ph.D. thesis, University of California, Berkeley, 1965.[14] Robin Hartshorne, Geometry: Euclid and beyond, Springer, 2000.[15] David Hilbert, Foundations of geometry (grundlagen der geometrie), Open

Court, La Salle, Illinois, 1960, Second English edition, translated from the tenth Germanedition by Leo Unger. Original publication date, 1899.

[16] Norbert Hunberbuhler, A short elementary proof of the mohr-mascheroni theorem,The American Mathematical Monthly, vol. 101 (1994), no. 8, pp. 784–787.

[17] Stephen C. Kleene, Introduction to metamathematics, van Nostrand, Princeton,1952.

[18] , Permutability of inferences in Gentzen’s calculi LK and LJ, A.M.S. Memoirs,vol. 10, pp. 1–26, A.M.S. Memoirs, American Mathematical Society, Providence, R. I., 1952,pp. 1–26.

[19] Mark Mandelkern, Constructive coordinatization of desarguesian planes, Beitrage

zur Algebra und Geometrie ( Contributions to Algebra and Geometry), vol. 48 (2007),no. 2, pp. 547–589.

[20] Elena Anne Marchisotto and James T. Smith, The legacy of mario pieri in

geometry and arithmetic, Birkhauser, Boston, Basel, Berlin, 2007.[21] Julien Narboux, Automated deduction in geometry: 6th international workshop, adg

2006, pontevedra, spain, august 31-september 2, 2006, revised papers, (Francisco Botana andTomas Recio, editors), Lecture Notes in Artificial Intelligence, Springer, 2008, pp. 239–156.

250 MICHAEL BEESON

[22] Victor Pambuccian, Axiomatizations of hyperbolic and absolute geometries, Non-

euclidean geometries, janos bolyai memorial volume. papers from the international

conference on hyperbolic geometry held in budapest, july 6–12, 2002 (New York) (AndrasPrekopa and Emil Molnar, editors), Mathematics and Its Applications, vol. 581, 2006, pp. 119–

153.[23] Moritz Pasch, Vorlesung uber neuere geometrie, Teubner, Leipzig, 1882.[24] Moritz Pasch and Max Dehn, Vorlesung uber neuere geometrie, B. G. Teubner,

Leipzig, 1926, The first edition (1882), which is the one digitized by Google Scholar, does notcontain the appendix by Dehn.

[25] Mario Pieri, Della geometria elemetare come sistema ipotetico deduttivo: Monografia

del punto e del moto, Memorie della Reale Accademia delle Scienze di Torino (series 2),vol. 49 (1900), pp. 173–222, Reprinted in [??], pp. 183–234.

[26] W. Schwabhauser, Wanda Szmielew, and Alfred Tarski, Metamathematische

methoden in der geometrie: Teil i: Ein axiomatischer aufbau der euklidischen geome-

trie. teil ii: Metamathematische betrachtungen (hochschultext), Springer, 1983.[27] Dana Scott, Dimension in elementary euclidean geometry, The axiomatic method

with special reference to geometry and physics, proceedings of an internation sympo-

sium, held at the university of california, berkeley, december 26, 1957-january 4, 1958.

(Amsterdam) (Leon Henkin, Patrick Suppes, and Alfred Tarski, editors), Studies in logic andthe foundations of mathematics, North-Holland, 1959, pp. 53–67.

[28] J. Strommer, Konstruktionen mit dem rechten und schiefen zeichenwinkel in der

absoluten geometrie, Periodica Mathematica Hungarica, vol. 6 (1975), no. 1, pp. 87–95.[29] Alfred Tarski, What is elementary geometry?, The axiomatic method, with special

reference to geometry and physics. proceedings of an international symposium held at

the univ. of calif., berkeley, dec. 26, 1957–jan. 4, 1958 (Amsterdam) (Leon Henkin, PatrickSuppes, and Alfred Tarksi, editors), Studies in Logic and the Foundations of Mathematics,North-Holland, 1959, Available as a 2007 reprint, Brouwer Press, ISBN 1-443-72812-8, pp. 16–29.

[30] Alfred Tarski and Steven Givant, Tarski’s system of geometry, The Bulletin of

Symbolic Logic, vol. 5 (1999), no. 2, pp. 175–214.[31] Anne Troelstra, Metamathematical investigation of intuitionsitic arithmetic

and analysis, Lecture Notes in Mathematics, no. 344, Springer, Berlin, Heidelberg, New York,1973.

[32] Dirk van Dalen, Intuitionistic logic, The handbook of philosophical logic, vol. 3,Reidel, 1986, pp. 225–339.

foundations of euclidean constructive geometry - citeseerx

Documents