shah - solovay's arithmetical completeness for provability logic

MA

EGNS

IT A T

MOLEM

UNIVERSITAS WARWICENSIS

Solovay’s Arithmetical Completeness Theorem for

Provability Logic

by

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0906963

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Thesis

Submitted to The University of Warwick

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Supervisors:

Associate Professor A Epstein and Assistant Professor W Dean

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Mathematics Institute

April, 2013

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Contents

1 Introduction 1

1.1 Formal Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 A Word on Modal Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Modal Logic & PA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Incompleteness and Undecidability 4

2.1 Peano Arithmetic, PA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Number Theory in PA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 ⌃1-completeness of PA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Primitive Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Arithmetical Definability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6 Gödel’s Incompleteness Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Modal Logic 29

3.1 The Language of Propositional Modal Logic . . . . . . . . . . . . . . . . . . . . . . . 293.2 Semantics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.3 Axiomatic Proof System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.4 Completeness for K,T,K4,B and S4 . . . . . . . . . . . . . . . . . . . . . . . . . . 363.5 GL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Arithmetical Completeness in Provability Logic 53

4.1 Arithmetical Soundness of GL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Arithmetical Completeness of GL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

SOLOVAY’S ARITHMETICAL COMPLETENESS THEOREM FOR PROVABILITY LOGIC

1 Introduction

How do we create a system in which we could prove (or disprove) some assertion based on the“language” of our system? This is probably best dealt with via an example, so let us pick our targetas basic arithmetic.

We all use arithmetic, whether it be for elementary dealings, say multiplication tables at schoolor monetary exchange, or more intrinsic mathematical proofs. In any case, the axioms we assume aresimple and modest, and the language we use is the addition and multiplication of natural numberswith standard first-order logical apparatus. The structure of our natural numbers is clear: we beginwith zero, then in an infinite sequence of no repetition each number has an unique successor. Howcan we state these axioms? If we leave it at “for each natural number, there is an unique successor”,how can we justify this? As mathematicians, we probably assume this is just how it is. If we begin toquestion this statement, then we question everything built upon it and begin to shake the foundationsof our own beautiful work. So then how is it just true? We can investigate our understanding of thestatement by asking ourselves whether we have some special faculty of intuition1 such that we caninterpret and accept the axioms.

Broadly speaking, formal theories consist of a formal language and some interpretation we haveon that language, and we construct a formal system from this theory by introducing axioms andrules of inference. The formal language is used because, in our everyday speech, we often havevagueness and obscurities alongside sentences that we cannot clearly say are true or false. Theformal language removes these ambiguities of different types and allows us to argue with clarity,coherence and precision. With this system of formal deductions we can lay out arguments in a‘logical’ way and with valid inferences so as to eradicate the potential for misunderstanding andfalse conclusions. By indulging in this formal style, we earn the right to undoubtedly claim that ourconclusions are valid derivations from the declared premises. Thus, if our premises are in fact true,then our conclusions must be true too.

After having chosen our formalised language and deductive system, we then need to consider therigour of our arguments. If we are rigorous enough and leave no space whatsoever for ambiguitythen our propositions, lemmas, theorems and other results will hold on the proviso that our initialfoundational axioms are true. The formalised language provides us with a language in which we cansay explicitly what is in and what is out.

One could well question why we do not lay out proofs in mathematics in a more formalised fashion,but instead settle upon some level of rigour. The answer is quite apparent to a mathematician: inmathematics one would maintain a substantial amount of rigour so that we have confidence inour proofs and that we have not made some fallible deductive/inferential step. However, if ourproofs were written resembling proofs in a formal system then they would be in danger of losingsome explanatory power. Our mathematical proofs are good not just because they make valid stepsthroughout, but because they are also transparent and inform the reader why each deduction followsfrom the last, and hence why the conclusion is true not just that it is.

Mathematical logic has some extensive applications to many areas of mathematics, the most well-known perhaps Kurt Gödel’s Incompleteness Theorems, but others include model-theoretic algebraand modal logic on provability. The purpose of this work is to show how modal logic can be usedto provide very useful information on formal arithmetic. But of all things, why arithmetic?

1Cf. ‘special sort of intuition’ [14, pp. 38-9]

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1.1 Formal Arithmetic

We should have some concern for formal arithmetic because our mathematical world contains a copyof it. This copy is our Natural numbers N2 with the basic operations addition + and multiplication⇥, the less-than relation <, and induction on N as we know it. We will give a definition of PeanoArithmetic (PA) later, but for now it suffices to think of PA as “our” arithmetic in a formalisedsetting (with formal axiomatic proofs). So we know it could be of concern to us but not that it is.Recall Gödel’s First Incompleteness Theorem which tells us that an effectively generated, consistenttheory that can express basic arithmetic cannot prove all arithmetical truths, and the Second whichsays that any such theory cannot prove its own consistency. In particular, Zermelo set theory Z hasenough machinery to interpret3 basic arithmetic, and so Z cannot prove its own consistency. Thistells us that neither Z nor Zermelo-Fraenkel set theory ZF can prove its own consistency.

Now what would it mean if we could derive a contradiction, something like 0 = 1 say, in PA?Well, we should certainly begin to worry for then we could derive a contradiction in Z and ZF.Further, since we can express a lot of mathematics in ZFC, Zermelo-Fraenkel set theory with theAxiom of Choice that is a conservative extension of ZF, this would imply that we could derive acontradiction in mathematics, and hence lead to a collapse of our framework. Luckily, we cannotseem to derive an obvious contradiction in PA and we can prove the consistency of PA, Con

PA

,in stronger theories such as ZF. Thus, to prove ConPA we need to “remove” ourselves from PA

and establish ourselves in ZF, say, but then we cannot prove ConZF

, so then we need to migrateto another and even stronger theory, and so on. Hence, PA is of importance to our other well-known theories, but what about to our informal mathematical world? If there is, as claimed earlier,a copy of PA in mathematics, then proofs of assertions in PA could be translated into informalmathematics in a natural way to give proofs in mathematics. For example, the commutativity ofaddition is provable in PA, and so if we could interpret the symbols of PA to give us a coherentargument for the commutativity of addition in our natural number system then PA would bearat least some importance to us. We may have had an informal argument for the commutativityof addition already, but the interpretation of the proof from PA would be incredibly rigorous andformal because we only allow formal axiomatic proofs using very precise rules within the proof systemof PA. This tells us that there is some reason to study formal arithmetic.

1.2 A Word on Modal Logic

We will ask similar questions of modal logic: firstly, what is (propositional) modal logic? The typeof modal logic that we shall be using is standard propositional logic together with a modal operator⇤. This ⇤ takes a sentence, A say, formulated in the language of modal logic as its argument andqualifies the truth of A. Commonly, ⇤ is interpreted as “it is necessary that...”, or “necessarily...”, analethic modality. Although this is the usual interpretation, ⇤ can be taken to mean various othermodalities. For example, we also have epistemic, deontic, and temporal modalities. For these ⇤would express “it is known that...”, “it is obligatory that...”, and “it has been...”, respectively. Clearly,

2We will adopt the convention that 0 2 N.3PA can be interpreted in Z in the sense that there is some translation t of the language of PA into the language

of Z, such that under t any formula ' has its structure preserved in a way so that if ' is provable in PA, then ' isalso provable in Z. Intuitively, an interpretation t is able to translate formulas from one language to the other butmaintain the meaning of individual formulas, in order to produce an accurate representation of the original theory inthe new one [18, p. 48].

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if we take a sentence, say A = it is raining, then taking different meanings for ⇤ yields differentmeanings for ⇤A, which may not be equivalent. E.g. comparing the alethic it is necessary that itis raining to the temporal it has been raining, we see that it has been raining can only inform us ofwhat has happened before and not what the situation is now.

Upon giving ⇤ a certain interpretation, we can define its dual as well. It takes a little thoughtto notice that ¬⇤A is only the negation of ⇤A and not its dual. We shall deduce the dual to itis necessary that it is raining. If it is not the case that it is necessary that it is raining, then it ispossible that it is not the case that it is raining, i.e. it is possible that it is not raining. Now letB = ¬A =it is not raining, and ⌃ =possibly, then

⌃B , it is possible that it is not raining, it is not necessarily raining, ¬⇤A, ¬⇤¬¬A, ¬⇤¬B

Hence, possibility ⌃ is the dual to necessity ⇤. We will have ⇤ as primitive in our modal logiclanguage and ⌃ as an abbreviation for ¬⇤¬.

Once we have assertions involving modals, we can try to determine when they are true andwhen they are false. Some simple examples include extending the law of the excluded middle witha modal operator at the start. For example, the assertion it is necessary that (it is raining or it isnot raining) (⇤(A _ ¬A)) is true because it is either raining or it is not, and this is always true.However, the statement it is necessarily raining or it is necessarily not raining (⇤A_⇤¬A) is false(at least in the actual world) because if we were in the case of one of the disjuncts, then it wouldalways be raining or it would always be not raining.

Now if we had a statement A which is true in our actual world, then it would be natural tosuppose that A were possible and hence ⌃A would be true. That is, for any sentence A, A ! ⌃Ashould be true. Why is this? Well, if A were true in this - the actual - world then there exists aworld in which A is true, and hence it is possible that A, i.e. ⌃A, is true. In arguing this way, weare assuming that our world actually exists and we know that our world exists. We are assumingsome sort of reflexivity. We can sum up this intuition with the formula A ! ⌃A. Then by settingB = ¬A, we get the following:

B ! ⌃B , ¬A ! ⌃¬A, ¬⌃¬A ! ¬¬A, ⇤A ! A

2A ! A is the T axiom. Now let ⇤ mean “it is provable that...”, then we should think that ⇤A ! Astill holds, for if A is provable (⇤A) is true, then surely there is a proof of A and hence A is true.We are concerned with this interpretation of ⇤ because in mathematics we seek to prove/disprovestatements, a form of provability in an informal system, and since we want to maintain some formalityand rigour provability becomes an important concept.

The second intuitive idea is that of transitivity: ⇤A ! ⇤⇤A. If A is provable is true, then thereshould be a proof of A is provable; that is, it is provable that A is provable should be true, i.e. ⇤⇤A,should be true from knowing ⇤A. We define ⇤A ! ⇤⇤A to be the 4 axiom.

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1.3 Modal Logic & PA

From the discussion above, if we are to formalise our mathematical world into a system of modallogic, L say, then intuitively we should expect all instances of the 4 and T axioms to be true in L.The normal modal logic system S4 satisfies this and so naturally we would want L=S4. However, weshall see later that S4 6=GL, where GL is the modal logic system that will be of greatest importanceto us in this exposition. So why is S4 not quite right? It will turn out that instances of ⇤A ! Aare not theorems of GL (see §3.5), contrary to our intuition.

We call a map ⇤ a realisation if it takes a modal sentence letter to an arithmetical sentence suchthat

p⇤ = ⇤(p) for any sentence letter p?⇤ = ?(A ^B)⇤ = A⇤ ^B⇤

(¬A)⇤ = ¬(A⇤)(⇤A)⇤ = Prov(pA⇤q)

where Prov(x) is a provability predicate in PA asserting that x is provable in PA, and pxq denotesthe Gödel number of x (Prov(x) and the notion of a Gödel number will be treated more preciselylater). So Prov(pA⇤q) means it is provable in PA that A. It happens that ‘A is provable in S4 iff(if and only if) A⇤ is provable in PA’ is not true, but it is true if S4 is replaced by GL. This claimis Solovay’s Arithmetical Completeness for Provability Logic.

2 Incompleteness and Undecidability

2.1 Peano Arithmetic, PA

As stated earlier, informally Peano Arithmetic (PA) is a formalisation of arithmetic on the standardNatural numbers N, where we have formalised versions of the addition and multiplication operationsand mathematical induction. Firstly, we need to define the language of arithmetic that we will take tobe standard. The language of arithmetic LA = hLA, IAi consists of the syntactical language LA andthe interpretation IA we give the language. LA can be thought of as the union of two alphabets: thelogical alphabet containing the normal Boolean connectives and first-order quantifiers, parentheses,relation symbols R1, R2, . . ., function symbols f1, f2, . . ., a countably infinite number of distinct ofvariables x0, x1, . . ., and the identity symbol =; and the non-logical alphabet {0, 000,+++,⇥⇥⇥} where 0

is a constant and 000,+++,⇥⇥⇥ are functions of PA. We define IA by interpreting the logical symbolsas normal, taking {^,¬,?} to be our primitive propositional truth-functionally complete set ofconnectives with 8 primitive, and 0, 000,+++,⇥⇥⇥ as the constant 0 2 N, the unary successor function, andthe binary operations addition and multiplication on N, respectively.

It is worth spending a moment on understanding the difference between arguing within PA, i.e.using formal language and giving formal axiomatic proofs, and arguing “outside” of the theory but,i.e. in the metatheory. The reason we need to make this difference so apparent is because it isimportant to know what is a theorem of, i.e. provable in, PA and what we are proving about PA.To keep these different possible situations apart we will introduce some notation, and for the sakeof readability we will make some natural abbreviations. We shall let a boldface numeral n denotethe constant 0 followed by n occurrences of 000, in order to avoid deciphering expressions such as

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0

00···00···00···000. One of the most important distinctions to be noted is the difference between a natural numbern 2 N that lives in our mathematical world and the numeral n of PA; n does not make sense inPA because it is not constructed from the syntax of PA, and the numeral n has no meaning in ourmetatheory until we give it an interpretation.

We shall assume that the following object exists:

Definition. The (Kuratowski) ordered pair4 is hx, yi := {{x}, {x, y}}.

Proposition 2.1. hx, yi = ha, bi ) x = a and y = b.

Proof. By assumption, {{x}, {x, y}} = {{a}, {a, b}}, so either {x} = {a} or {x} = {a, b}.If {x} = {a}, then by extensionality5 (the uniqueness of sets) we must have x = a. Now if

{x, y} = {a}, then we obtain {x} = {x, y} = {a}, so {a, b} = {a} and hence x = a = y = b. If{x, y} = {a, b}, then y = b follows from x = a and extensionality.

On the other hand, if {x} = {a, b} then x = a = b. Either {x, y} = {a} or {x, y} = {a, b}, butin both cases we immediately obtain x = a = b = y.

Using the notion of an ordered pair we can construct another object which will be used:

Definition. The ordered triple of x, y and z is hx, y, zi := hx, hy, zii, i.e. the ordered pair of x andhy, zi, respectively.

The uniqueness of the ordered triple follows directly from the uniqueness of the ordered pair.We will also need the notion of a finite sequence which we define below.

Definition. We define a finite sequence to be an object s with a length n such that 8i, 0 6 i < n,there is an object si that is the ith entry of s.

Suppose that s is a sequence of length n such that for each i < n si is the ith entry, then wewill write s = hs0, s1, . . . , sn�1i. We will allow there to be a sequence of length 0, it is unique andwill be known as the empty sequence, denoted hi. We will say that two sequences s = hs0, . . . , sn�1iand t = ht0, . . . , tm�1i of lengths n,m respectively define the same sequence if n = m and si = ti 8i,0 6 i < n. We will also allow the concatenation of sequences; for example, if we want to concatenates and t as defined above, then we extend s to the sequence s ⇤ t = hs0, . . . , sn�1, t0, . . . , tm�1iby recursively defining s0 = s, and 8i, 0 6 i < m, si+1 to be the sequence with si+1

j = sij 8j,0 6 j < n+ i+ 1, and si+1

n+i+1 = ti.We want to be able to string together objects of PA in a coherent way using LA, so now we

shall make a few definitions to make use of the language of arithmetic.

Definition. The set of terms of LA is the smallest set TERMLA such that

(i) 0 2 TERMLA

(ii) xi 2 TERMLA 8i 2 N

(iii) t 2 TERMLA ) t000 2 TERMLA

4One can verify the existence of such objects using basic set theory; more precisely, using an instance of the axiomof separation and the axiom of (unordered) pairing.

5More precisely, by the axiom of extensionality which states that a set is defined completely by its member elements.

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(iv) s, t 2 TERMLA ) (s+++ t), (s⇥⇥⇥ t) 2 TERMLA .

Definition. The set of (well-formed) formulas of LA is the smallest set FORMLA such that

(i) ? 2 FORMLA

(ii) s, t 2 TERMLA ) s = t 2 FORMLA

(iii) ', 2 FORMLA ) (' ^ ) 2 FORMLA

(iv) ' 2 FORMLA ) (¬') 2 FORMLA

(v) '(x) 2 FORMLA ) 8x('(x)) 2 FORMLA where '(x) means x is a variable occurring in '.

If s, t 2 TERMLA , then the term s+++t should be formally written as+++(s, t), similarly with⇥⇥⇥, andwe can view these as ordered triples. So we shall suppose s000, s+++ t, s⇥⇥⇥ t and s = t are h000, si, h+++, s, ti,h⇥⇥⇥, s, ti and h=, s, ti, respectively. For formulas ', we will suppose ¬',' ^ , 8x'(x) are theordered pair h¬,'i and the ordered triples h^,', i, h8, x,'i, respectively. For readability, we shallwrite terms and formulas like s +++ t and 8x'(x), whilst also omitting some pairs of parentheses.However, if the parentheses are needed to verify the scope of a logical symbol then we will becareful to include them. For example, there is a noticeable difference between 8x'(x) ^ (x) and8x['(x)^ (x)]: in the first instance, only occurrences of x in '(x) are bounded by the 8x, but in thesecond all occurrences of x are bounded. Furthermore, in any case where we need many parentheses,we shall use different types of brackets to emphasise the scope of certain connectives or quantifiers.

Definition. We call ' 2 FORMLA an atomic formula if ' ⌘ ? or ' ⌘ s = t for some terms s, t.

Definition. For ', 2 FORMLA and s, t 2 TERMLA , we make the following abbreviations:

(i) ' _ , ¬(¬' ^ ¬ )

(ii) '! , ¬' _

(iii) '$ , ('! ) ^ ( ! ')

(iv) 9x'(x) , ¬[8x(¬'(x))]

(v) s 6= t , ¬(s = t).

Thus whenever we need to prove a result via induction on the form of a formula ', we needonly show the result is true in the following cases: ' is an atomic formula, ' ⌘ ^ �, ' ⌘ ¬ ,or ' ⌘ 8x (x). Further, we shall typically use x, y, z, . . . as variables to help aid clarity in specificcases, but revert back to using x0, x1, . . . if we are in a general setting where the number of variablesis arbitrary.

Now that we have our definitions of term and formula in the language of arithmetic, we canconsider the so-called subterms and subformulas.

Definition. The set of subterms Subterm(t) for a term is defined as follows:

(i) Subterm(0) = {0}

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(ii) Subterm(xi) = {xi} 8i 2 N

(iii) Subterm(t000) = Subterm(t) [ {(t000)} for t 2 TERMLA

(iv) Subterm(s+++ t) = Subterm(s) [ Subterm(t) [ {(s+++ t)} for s, t 2 TERMLA

(v) Subterm(s⇥⇥⇥ t) = Subterm(s) [ Subterm(t) [ {(s⇥⇥⇥ t)} for s, t 2 TERMLA .

Definition. The set of subformulas Sub(') for a formula is defined as follows:

(i) ' 2 FORMLA atomic ) Sub(') = {'}

(ii) Sub(' ^ ) = Sub(') [ Sub( ) [ {(' ^ )} for ', 2 FORMLA

(iii) Sub(¬') = Sub(') [ {(¬')} for ' 2 FORMLA

(iv) Sub(8x('(x))) = Sub(') [ {8x('(x))} for ' 2 FORMLA .

Having defined the subterms of a term, we can view the definition of a term in an equivalentway which will allow us see the uniqueness of a term more easily. An object t of PA is a term ofPA iff there exists a finite sequence of terms of PA such that t is the final entry in the sequence,and each entry of the sequence is either 0, or a variable xj for some j 2 N, or s000 or r+++ s or r⇥⇥⇥ sfor some earlier entries r, s of the sequence. Recalling that we take s000 and r +++ s, r ⇥⇥⇥ s to be anordered pair and ordered triples respectively, it is clear that the term t is uniquely represented. Itmay be that two different terms have the same interpretation, e.g. 0

000000 = h000,0000i = h000, h000,0ii and0

000 +++ 0

000 = h+++,0000,0000i = h+++, h000,0i, h000,0ii are both interpreted as 2 2 N, but as terms 0

000000 is an orderedpair and 0

000 +++ 0

000 is an ordered triple, and hence they are distinct.

Definition. The set of free variables FV (t) for terms is defined as follows:

(i) FV (0) = ;

(ii) FV (xi) = ; 8i 2 N

(iii) FV (t000) = FV (t) for t 2 TERMLA

(iv) FV (s+++ t) = FV (s) [ FV (t) for s, t 2 TERMLA

(v) FV (s⇥⇥⇥ t) = FV (s) [ FV (t) for s, t 2 TERMLA .

Definition. The set of free variables FV (') for formulas is defined as follows:

(i) FV (?) = ;

(ii) FV (s = t) = FV (s) [ FV (t) for s, t 2 TERMLA

(iii) FV (' ^ ) = FV (') [ FV ( ) for ', 2 FORMLA

(iv) FV (¬') = FV (') for ' 2 FORMLA

(v) FV (8x'(x)) = FV (')\{x} for ' 2 FORMLA .

Definition. t 2 TERMLA is called closed if FV (t) = ;. All other terms are called open.

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Definition. ' 2 FORMLA is called a sentence if FV (') = ;, i.e. a closed formula is a sentence.We shall just refer to open formulas as formulas.

The set of sentences SENTLA:= {' 2 FORMLA | FV (') = ;}.

The sentences in the language of arithmetic are particularly important to us because when weinterpret LA in the standard way, we can assign truth values to sentences. Since sentences have nofree variables, whenever we substitute natural numbers in for variables into the interpretation we candetermine whether the arithmetical assertion is true or false. The sentence has a definite truth valuewhen interpreted. However, a formula could have a free variable, so even if we make substitutionsfor all the bound variables we still have at least one variable, and thus the formula does not have acertain truth value. The free variable (possibly) allows us to change the truth value depending onthe substitution we make. This idea is best demonstrated via an example.Example. Let '(x) ⌘ 8x(x+++ 0 = x), and (y) ⌘ y+++ y = y. Note that because we do not have aproof system for PA just yet, we cannot give formal proofs or disproofs of ' or in PA. However,we can still interpret them, giving claims in our natural arithmetic. Since all occurrences of x in 'are bound by 8, ' is a sentence in LA, but has unbound occurrences of y so is just a formula.Using the standard interpretation, ' translates to ‘for all natural numbers x, x + 0 = x’. This isclearly true in natural arithmetic by virtue of what we understand 0 to be. So ' is a true6 sentencein the language of arithmetic. Now if we substitute 2 in for y in , then the left-hand side of yields 2+++2 which translates to 2+2 which is equal to 4. However, the right-hand side yields just 2which is trivially 2. But 2+ 2 6= 2, which suggests could be false. Yet 0+ 0 = 0 which suggests could be true. doesn’t have a constant truth value under the standard interpretation and so wecannot assign it a definite truth value.

In the example above, we used substitution to obtain the concrete assertions 2 +++ 2 = 2 and0+++0 = 0 from and this method gives us an easy way to get from formulas to sentences. The ideaof substitution is common to a mathematician so we just state the notation we will use. Formally,we will say '[t/x] is the result of substituting t 2 TERMLA in for variable x in ' 2 FORMLA .More generally, if '(x0, . . . , xn) 2 FORMLA then the result of substituting in terms t0, . . . , tn in fordistinct variables x0, . . . , xn respectively is '[t0/x0, . . . , tn/xn]. However, we will write '(t0, . . . , tn)instead of '[t0/x0, . . . , tn/xn] when it is clear what is meant. Thus, from the example (2) ⌘ 2+++2 =2 is an atomic formula in PA because 2+++ 2,2 2 TERMLA . Substitution directly gives us a way ofgrouping certain sets of n-tuples: if we are given a formula '(x0, . . . , xn�1) with distinct free variablesx0, . . . , xn�1, then ' defines the n-ary relation containing precisely all n-tuples hm

0

, . . . ,mn�1

i suchthat '(m

0

, . . . ,mn�1

) holds.Definition. (First-order) Peano Arithmetic PA is the first-order theory whose language is thelanguage of arithmetic LA and whose axioms are the following:

(PA1) 8x(0 6= x000)

(PA2) 8x8y(x000 = y000 ! x = y)

(PA3) 8x(x+++ 0 = x)

(PA4) 8x8y(x+++ y000 = (x+++ y)000)

(PA5) 8x(x⇥⇥⇥ 0 = 0)

(PA6) 8x8y(x⇥⇥⇥ y000 = (x⇥⇥⇥ y) +++ x)

6We give a more precise definition of a true sentence of arithmetic once we have introduced a proof system for PA.

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plus any instance of the following form:

InductionSchema : ['(0) ^ 8x('(x) ! '(x000))] ! 8x'(x)

where '(x) is any formula that has x, but possibly more variables, free. When we work throughformal proofs in PA, we shall cite the axioms as they are labeled above when we employ them.

We have now formally introduced the formal theory of Peano Arithmetic so we can talk aboutPA and give results around it, but we cannot derive results in PA without a formal proof system.To this end, we will present and use a Hilbert-type7 axiomatic proof system of classical first-orderlogic with identity for PA. This style of proof system is a synthetic [8, p. 46] one, in that we use themany axioms and few rules we have to construct the formula being proved. In our axiomatic proofsystem, the axioms will be all tautologies, all formulas of the form 8x'(x) ! '(t) for t 2 TERMLA

and all the axioms stated above; and the rules of inference will be modus ponens and universalgeneralisation, which we describe below.

Modus Ponens: From ' and '! we can deduce by modus ponens, i.e. ' '!

, for

', 2 FORMLA .Universal Generalisation: From ' ! (t) we can deduce ' ! 8x (x) by universal generalisa-

tion, i.e.'! (t)

'! 8x (x), for ', 2 FORMLA and parameter t not occurring in ' nor in .

In these cases, we call and ' ! 8x (x) (logical) consequences of the premises ',' ! and'! (t) respectively.

Definition. A proof or derivation D in PA of ' 2 FORMLA is a finite sequence of formulas inLA such that the last entry of the sequence is ' and each entry of D is either a tautology, or anaxiom, or is a logical consequence of a rule of inference. If there exists a proof D of ' 2 FORMLA ,then we say ' is provable or derivable in PA or ' is a theorem of PA, and in this case we writePA ` '.

Definition. We call ' 2 SENTLA true if ' is true under the standard interpretation IA while thevariables of ' range over N.

In order to aid readability in formal derivations, various abbreviations will be made for certainformulas; most abbreviations will be made as they are needed, but it is worth stating some nowwhich are used frequently.

Definition. (i) x < y := 9u(x+++ u000 = y)

(ii) y > x := x < y

(iii) x 6 y := (x < y _ x = y)

(iv) y > x := x 6 y

(v) 8x < y'(x) := 8x(x < y ! '(x))

7Smullyan presents a Hilbert-type system in more generality in his book [18, pp. 80-1].

9


(vi) x will be used to abbreviate x0, . . . , xn�1 for the appropriate n. Similarly, 8x will abbreviate8x08x1 . . . 8xn�1.

(vii) 9!y'(x, y) := 9y['(x, y) ^ 8u('(x, u) ! u = y)]

(viii) 9x 6 y'(x) := 9x(x 6 y ^ '(x)). This is called a bounded occurrence of 9.

(ix) 8x 6 y'(x) := 8x(x 6 y ! '(x)). This is bounded occurrence of 8.

2.2 Number Theory in PA

We have seen above that PA has a set of tools which may allow us to prove some assertions wecan make in the language of arithmetic, but we have not yet seen any concrete examples of whatPA can actually prove formally. In this section we will see in fact that within PA we can do muchnumber theory. Our aim in this section will to be to show that we can prove the commutativity ofaddition formally. Moreover, the infinitude of primes is also formally provable - i.e. provable in PA

- but the proof of this is particularly involved and not very enlightening.

Lemma 2.2. PA ` 8x(x+++ 0

000 = x000)

Proof. By PA4 and PA3, respectively, PA ` 8x(x+++ 0

000 = (x+++ 0)000) and PA ` 8x(x+++ 0

000 = x000).

Lemma 2.3. [2, p. 21]

(i) For n, l,m 2 N, if l +m = n then PA ` l+++m = n.

(ii) For n, l,m 2 N, if l ⇥m = n then PA ` l⇥⇥⇥m = n.

Proof. (i) We do this by induction on m. Suppose l+m = n and m = 0 2 N, then l = n and m = 0.So we have PA ` l+++m = l+++ 0, but by PA3 PA ` l+++ 0 = l and hence PA ` l+++m = l. But sincel = n, PA ` l = n and so PA ` l+++m = n. Now suppose if l + k = n then PA ` l+++ k = n. Wewant to show the result is true if l+ (m+1) = n. Assume l+ (m+1) = n, then l+m = k for somek 2 N. Thus, by the induction hypothesis PA ` l+++m = k. Since l +m = k and l + (m+ 1) = n,we have that k + 1 = n and so k

000 = n because PA ` k+++ 1 = (k+++ 0)000 = k

000 by Lemma 2.2. SoPA ` l+++m

000 = (l+++m)000 by PA4, and since PA ` l+++m = k we also have PA ` l+++m

000 = k

000. Hence,PA ` l+++m

000 = n.(ii) Again we proceed by induction on m. Firstly, suppose l ⇥ m = n and m = 0 2 N. Then

m = 0, and n = l ⇥ m = l ⇥ 0 = 0 so n = 0. PA ` l⇥⇥⇥ m = l⇥⇥⇥ 0 = 0 where the last equalityfollows from PA5. Thus, PA ` l⇥⇥⇥m = n as n = 0. Now suppose l⇥ k = n ) PA ` l⇥⇥⇥k = n andl⇥ (m+1) = n. Note n = l⇥ (m+1) = (l⇥m) + l = k+ l for some k 2 N. Then, by the inductionhypothesis PA ` l⇥⇥⇥m = k and from (i) PA ` k+++ l = n. So PA ` l⇥⇥⇥ (m+++1) = l⇥⇥⇥m

000 by Lemma2.2; then PA ` l⇥⇥⇥ (m+++ 1) = (l⇥⇥⇥m)+++ l by PA6, which is just PA ` l⇥⇥⇥ (m+++ 1) = k+++ l and thusPA ` l⇥⇥⇥ (m+++ 1) = n and we are done.

Proposition 2.4. [2, p. 21] Suppose t 2 TERMLA is closed and denotes n 2 N, then PA ` t = n.

Proof. We prove this by induction on the form of t. If t ⌘ 0, then trivially 0 = 0 is a logical truthand so PA ` t = 0.

10


Suppose t ⌘ r+++s and r and s denote l 2 N and m 2 N, respectively. Note that l+m = k for somek 2 N. Then by the induction hypothesis and by Lemma 2.3, we have PA ` r = l, PA ` s = m

and PA ` r+++ s = l+++m, so PA ` t = l+++m.Similarly, if t ⌘ r⇥⇥⇥ s then the result holds by the induction hypothesis and Lemma 2.3.Finally, suppose that t ⌘ r000, r000 denotes l + 1 2 N and PA ` r = l. Since PA ` r +++ 1 = r000 by

Lemma 2.2 and PA ` r+++ 1 = l+++ 1, we have that PA ` r000 = l+++ 1. Thus, PA ` t = l+++ 1.

Corollary 2.5. [2, p. 22] Suppose s, t 2 TERMLA are closed, and denote m,n 2 N, respectively.If m = n is true, then PA ` s = t.

Proof. We know by Proposition 2.4 that PA ` s = m and PA ` t = n. Since m = n is true, wemust have that m,n denote the same numeral, i.e. PA ` m = n, so PA ` s = t.

Lemma 2.6. [2, p. 20] PA ` 8x[x = 0 _ 9y(x = y000)]

Proof. Let '(x) ⌘ x = 0 _ 9y(x = y000). PA ` 0 = 0 and PA ` (0 = 0) ! '(0) because 0 = 0and 0 = 0 ! (0 = 0 _ 9y(0 = y + 1) are logical truths. So we have PA ` '(0), the base case.Now suppose PA ` '(x). We need to show PA ` '(x000). Note that trivially PA ` x000 = x000 andPA ` (x000 = x000) ! 9y(x000 = y000), so PA ` 9y(x000 = y000). Hence, PA ` x000 = 0 _ 9y(x000 = y000)because PA ` 9y(x000 = y000) ! (x000 = 0 _ 9y(x000 = y000)). Thus, PA ` '(x) ! '(x000) and by universalgeneralisation we obtain PA ` 8x('(x) ! '(x000)). PA ` '(0) and PA ` 8x('(x) ! '(x000)) )PA ` '(0) ^ 8x('(x) ! '(x000)) ) PA ` 8x[x = 0 _ 9y(x = y000)] by the induction schema of PA.

Lemma 2.7. PA ` 8x(¬(x < 0))

Proof. Recall that x < 0 ⌘ 9u(x+++ u000 = 0). By PA4, PA ` 8x8u(x+++ u000 = (x+++ u)000) and, by PA1,PA ` 8x8u((x+++ u)000 6= 0). Thus, PA ` 8x8u(x+++ u000 6= 0), i.e. PA ` 8x(¬x < 0).

Lemma 2.8. [2, p. 22] PA ` 8x8y[x < y000 $ (x = y _ x < y)]

Proof. PA ` x < y000 , PA ` 9u(x+++ u000 = y000) , PA ` 9u((x+++ u)000 = y000) , PA ` 9u(x+++ u = y) ,PA ` x+++ 0 = y _ 9w(x+ w000 = y) by using Lemma 2.6 on u , PA ` x = y _ x < y.

Definition.

Wi<n

x = i is ? if n = 0 and otherwise is the disjunction of all formulas x = i for i < n.

Proposition 2.9. [2, p. 22] PA ` 8x(x < n $Wi<n

x = i)

Proof. We prove this by induction on n. Firstly, assume n = 0, then (Wi<0

x = i) ⌘ ?. But

PA ` ¬x < 0, so PA ` x < 0 $Wi<0

x = i. Now suppose PA ` x < n $Wi<n

x = i for some n 2 N.

Note PA ` x < n

000 $ (x = n _ x < n) by Lemma 2.8 , PA ` x < n

000 $ (x = n _Wi<n

x = i) by the

induction hypothesis , PA ` x < n

000 $W

i<n+1x = i.

Now we will show that PA can prove that addition commutes, and we will show this in thefollowing way:

11


Theorem 2.10.

(i) PA ` 8x[0+++ x = x]

(ii) PA ` 8x[0000 +++ x = (0+++ x)000]

(iii) PA ` 8x8y8z[x+++ (y+++ z) = (x+++ y) +++ z]

(iv) PA ` 8x8y[x000 +++ y = (x+++ y)000]

(v) PA ` 8x8y[x+++ y = y+++ x]

Proof. In each of the parts (i)-(v), we will make use of the induction schema of PA. Moreover, sothat the proofs do not look like meaningless statements, we will not argue completely within PA

but we shall maintain the rigour by justifying each step with either a use of an axiom, previouspart or the induction hypothesis. Therefore, statements like PA ` 0

000 +++ x000 =PA3

0

000 +++ (x+++ 0)000 =IH

· · ·will appear where ‘ =

PA3

’ means that PA3 justifies the equality and ‘=IH

’ means that the inductionhypothesis justifies the equality.

(i) Base: PA ` 0+++ 0 = 0 by PA3. Suppose IH : PA ` 0+++ x = x as the induction hypothesis.Then PA ` 0+++ x000 =

PA4

(0+++ x)000 =IH

x000. So by the induction schema, we have PA ` 8x[0+++ x = x].

(ii) Base: PA ` 0

000 +++ 0 =PA3

0

000 =(i)

(0+++ 0)000. Suppose IH : PA ` 0

000 +++ x = (0+++ x)000. Then

PA ` 0

000 +++ x000

=PA4

(0+++ x000)000

=PA4

((0+++ x)000)000

=(i)

(x000)000

=(i)

(0+++ x000)000.

Thus, by the induction schema PA ` 8x[0000 +++ x = (0+++ x)000].(iii) For this part we shall show PA ` 8z[x+++(y+++z) = (x+++y)+++z], then by universal generalisation

on both x and y we will have our result. Base: PA ` x+++(y+++0) =PA3

x+++y =PA3

(x+++y)+++0. SupposeIH : PA ` x+++ (y+++ z) = (x+++ y) +++ z. Then

PA ` x+++ (y+++ z000)=

PA4

x+++ (y+++ z)000

=PA4

(x+++ (y+++ z))000

=IH

((x+++ y) +++ z)000

=PA4

(x+++ y) +++ z000.

Thus, by the induction schema we have PA ` 8z[x+++ (y +++ z) = (x+++ y) +++ z], and using universalgeneralisation on x, y we get PA ` 8x8y8z[x+++ (y+++ z) = (x+++ y) +++ z].

12


(iv) Base: PA ` 0

000 +++ y =(ii)

(0+++ y)000. Now suppose IH : PA ` x000 +++ y = (x+++ y)000. Then

PA ` x000000 +++ y=

PA3

(x+++ 0)000000 +++ y

=PA4

(x+++ 0

000)000 +++ y

=PA4

(x+++ 0

000000) +++ y

=(iii)

x+++ (0000000 +++ y)

=(i)

x+++ ((0+++ 0

000)000 +++ y)

=(ii)

x+++ ((0000 +++ 0

000) +++ y)

=(iii)

x+++ (0000 +++ (0000 +++ y))

=(ii)

x+++ (0000 +++ (0+++ y)000)

=(iii)

(x+++ 0

000) +++ (0+++ y)000

=(i)

(x+++ 0

000) +++ y000

=PA4

(x+++ 0)000 +++ y000

=PA3

x000 +++ y000

=PA4

(x000 +++ y)000.

Therefore, by the induction schema we get PA ` 8x[x000 +++ y = (x +++ y)000], and finally by universalgeneralisation on y we are done.

(v) Base: PA ` 0+++ y =(i)

y =PA3

y+++ 0. Suppose PA ` x+++ y = y+++ x. Then

PA ` x000 +++ y =(iv)

(x+++ y)000 =IH

(y+++ x)000 =PA4

y+++ x000. Thus, by the induction schema we have

PA ` 8x[x+++y = y+++x], and by universal generalisation on y we have that PA ` 8x8y[x+++y = y+++x].

So PA can prove formally many elementary facts about arithmetic. Notice that we relied onthe induction schema of PA in parts of the proof of Theorem 2.10, and so if we went to a weakertheory without the induction schema, such as Robinson Arithmetic8

Q, then it would be hard if notimpossible to prove the commutativity of addition. Q is just like PA with the induction schemaremoved but with an additional axiom 8x(x 6= 0 ! 9y(x = y000)), and it is known that Q cannotprove 8x8y[x+++ y = y+++ x]. Hence, why we have chosen PA as our candidate for formal arithmetic.

Definition. Prime(x) ⌘ (0000 < x) ^ 8y < x8z < x(y⇥⇥⇥ z 6= x)

Theorem 2.11. PA ` 8x9y[x < y ^ Prime(y)]9

8See [16, §8.3, pp. 55-6]9For a proof of this, see [10, §5.3, pp. 65-7]

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2.3 ⌃1-completeness of PA

Definition. [2, p24] Let '(x, y) 2 FORMLA , then we say '(x, y) is a pterm (with respect to y) ifPA ` 9!y'(x, y).

Notice that a pterm F (x, y) 2 FORMLA defines a well-defined n-ary function, f say, becausePA ` 9!yF (x, y). The uniqueness part in the definition of a pterm gives us that 8x9!y(f(x) = y),so it is not ambiguous to write just f(x) for the value of f at x instead of y where f(x) = y. As aresult of this we shall sometimes simply refer to f(x) instead of F (x, y).

Definition. We call ' 2 FORMLA bounded if only bounded occurrences of quantifiers appear in'.

The reason we care about such bounded formulas is because we will introduce the arithmeticalhierarchy of formulas and a class of functions that can be defined by a certain type of formulaappearing in the hierarchy.

Definition. The arithmetical hierarchy [9, p. 13] is defined for ' 2 FORMLA as follows:

(i) ' is ⌃0 and ⇧0 if ' is bounded

(ii) ' is strict ⌃n+1 if ' ⌘ 9x where is ⇧n

(iii) ' is strict ⇧n+1 if ' ⌘ 8x where is ⌃n.

In the above definition we allow that the first appearing quantifier to be unbounded, but allother quantifiers in the formula must be bounded.

Definition. A formula ' is said to be (generalised) ⌃1 if ' is built up from �0 formulas by con-junction, negation, bounded universal quantification and unbounded existential quantification. IfFV (') = ; and ' is ⌃1 then we call ' a ⌃1 sentence.

Definition. ' is �n just in case ' is both ⌃n and ⇧n, and a formula of the form s = t or s < t,where s, t 2 TERMLA , is said to be atomic �0.

Recall that an atomic formula ' is either identically ? or s = t for some terms s, t. Thus, wecan construct another formula that is logically equivalent to ' using conjunction and existentialquantification from the following types of formula: u = v, 0 = v, u000 = v, u+++ v = w, and u⇥⇥⇥ v = w[2, p. 25]. Hence, any atomic formula is also a ⌃1 formula. Furthermore, it can be shown that¬u = v is logically equivalent to u < v _ v < u, so ¬u = v is also ⌃1. That is, the negation of anyatomic formula is also ⌃1, and therefore all atomic formulas are �1 as well as �0. We omit theproof to the next result as it is routine.

Lemma 2.12. [2, p. 25] The class of ⌃1 formulas is closed under conjunction, disjunction, boundeduniversal quantification, and bounded and unbounded existential quantification. Note it is not closedunder negation.

Proposition 2.13. [2, p. 27] If '(x, y) 2 FORMLA is both ⌃1 and a pterm, then it is �1.

Proof. '(x, y) is a pterm, which means PA ` 9!y'(x, y), so ¬'(x, y) ⌘ 9u[¬u = y ^ '(x, y)]. Note¬'(x, y) is ⌃1 because '(x, y),¬u = y are ⌃1 and the conjunct is thus ⌃1 by Lemma 2.12.

14


Proposition 2.14. [2, p. 27] If F (x, y) is a ⌃1 pterm and A(y) is �1, then 9y(F (x, y) ^ A(y)) is�1.

Proof. 9y(F (x, y) ^ A(y)) and is ⌃1 by Lemma 2.12 since F (x, y) and A(y) are both ⌃1. SinceF (x, y) is a pterm, 9y(F (x, y) Â(y)) is logically equivalent to 8y(F (x, y) ! A(y)). Therefore,¬9y(F (x, y) Â(y)) $ ¬8y(F (x, y) ! A(y)) $ 9y(F (x, y) ^ ¬A(y)) which is ⌃1. Hence,9y(F (x, y) Â(y)) is �1.

Theorem 2.15. (⌃1-completeness of PA) ' 2 SENTLA is a true ⌃1 sentence of arithmetic ,PA ` '.

Proof. (() All the axioms of PA are true, so any theorem of PA proved from these axioms will betrue as well. Thus, PA ` ') ' is true.

()) By induction on the form of '. If ' ⌘ ? then PA 0 ? and ' is not true. If ' ⌘ s = t forsome terms s, t, and ' is true, then by Corollary 2.5 PA ` s = t, i.e. PA ` '.

Suppose ' ⌘ ^ � for some ,� 2 SENTLA and that ,� is true, respectively, ) PA ` ,PA ` �, respectively. If ' is true, then ^ � is true so must be true and � must be true. Thus,PA ` and PA ` � so PA ` ^ �, i.e. PA ` '.

Now suppose ' ⌘ ¬ is true; that is, ¬ is true. Assume for contradiction that PA 0 ¬ , thenPA [ { } is consistent. Since ¬ is true, is not true. If PA ` then we would have that istrue, so PA 0 . Note that PA [ { } ` and PA [ { } ◆ PA, i.e. PA [ { } is a conservativeextension of PA, so we must have PA ` because is expressed in the language of PA. Thus,PA 0 and PA ` . Hence, PA ` ¬ .

If ' ⌘ 8x < n (x) is true, then PA ` (i) for each i < n by the induction hypothesis. Thus, foreach i < n we have PA ` x = i ! (x). Note that from Proposition 2.9, PA ` x < n $

Wi<n

x = i,

and so PA ` x < n ! (x), i.e. PA ` 8x < n (x).Finally, suppose ' ⌘ 9x (x) is true. Therefore, for some k 2 N we must have (k) is true. So

PA ` (k) for some k 2 N, and hence PA ` 9x (x).

Corollary 2.16. For any ' 2 SENTLA which is �1, either PA ` ' or PA ` ¬'.

Proof. ',¬' are both ⌃1 and at most one can be true. Thus, PA proves the true one.

Thus, whenever we know that a ⌃1 sentence in the language of arithmetic is true under thestandard interpretation, we know that it is formally provable in PA. However, it is important torealise that we do not have that PA can disprove false ⌃1 sentences of arithmetic; that is, if weknew a sentence ' were false, then there is no way of knowing if ¬' were provable in PA.

2.4 Primitive Recursive Functions

This section is dedicated to a particular set of important functions: the primitive recursive functions.These functions turn out to be an even nicer subset of the already quite amiable set of recursivefunctions.

Definition. The primitive recursive basis functions are the following:

(i) the constant zero function z : N ! N, given by z(x) = 0 8x 2 N

15


(ii) the successor function s : N ! N, given by s(x) = x+ 1 8x 2 N

(iii) the projection functions, for n 2 N and 8i 6 n, idni : Nn ! N given by idni (x1, . . . , xn) = xi.

From these we will be able to build the whole bank of primitive recursive functions using themethods we introduce next.

Definition. Let f be an m-ary function, and suppose g1, . . . , gm are n-ary. Then the compositionof f, g1, . . . , gm is the n-ary function Cn[f, g1, . . . , gm] that is defined by

Cn[f, g1, . . . , gm](x1, . . . , xn) = f(g1(x1, . . . , xn), . . . , gm(x1, . . . , xn)).

Definition. Let f be an n-ary function and g an (n+ 2)-ary function. Then by Pr[f, g] we denotethe function h defined by primitive recursive from f and g, where h is the (n+1)-ary function givenby

h(x1, . . . , xn, 0) = f(x1, . . . , xn)h(x1, . . . , xn, s(y)) = g(x1, . . . , xn, y, h(x1, . . . , xn, y)).

Definition. The set of primitive recursive functions PRIMREC is the smallest class of functions, whichis a subset of

Sn2N

{f | f : Nn ! N}, such that

(i) z, s 2 PRIMREC

(ii) 8n 2 N and 8i 6 n, idni 2 PRIMREC

(iii) if m-ary f and n-ary g1, . . . , gm are in PRIMREC, then Cn[f, g1, . . . , gm] 2 PRIMREC

(iv) if f is n-ary and g is (n+ 2)-ary, then Pr[f, g] 2 PRIMREC.

This definition for PRIMREC is quite abstract and although we have the zero, successor andprojection functions as concrete examples of primitive recursive (p.r.) functions, these are notparticular deserving of our attention.

Example. sum(x, y) = x+ y is p.r.: it is defined bysum(x, 0) = id11(x) = xsum(x, s(y)) = Cn[s, id33](x, y, sum(x, y)) = s(sum(x, y)).Therefore, sum := Pr[id11, Cn[s, id33]] is p.r..

Example. prod(x, y) is p.r.:prod(x, 0) = z(x) = 0prod(x, s(y)) = Cn[sum, id31, id

31](x, y, prod(x, y)) = sum(x, prod(x, y))

Thus, prod := Pr[z, Cn[sum, id31, id31]].

Example. fact(x) = x! is p.r.:fact(0) = Cn[s, z](x) = s(z(x)) = s(0) = 1fact(s(x)) = Cn[prod, Cn[s, id21], id

22](x, fact(x)) = prod(s(x), fact(x))

Hence, fact = Pr[Cn[s, z], Cn[prod, s, fact]].

Example. pred (modified predecessor) [3, pp. 69-70] is defined by pred(0) = 0 and pred(s(x)) = x,and is hence p.r..

16


Example. monus (modified subtraction) [3, pp. 69-70] is defined by monus(x, 0) = x andmonus(x, s(y)) = pred(monus(x, y)), and is again p.r..

Example. Define sg [3, pp. 69-70] by sg(x) = monus(1, (monus(1, x)), which is p.r. as it is thecomposition of p.r. functions. That is, sg(0) = 0 and sg(x) = 1 for any x > 0.

As displayed by these examples, there are some very important functions which are p.r.. Fromnow we shall be slightly less formal when showing functions are p.r.; for example, we could write

fact(x) :=

(1 x = 0

x⇥ fact(x� 1) x > 0is p.r. as it defined in a primitive recursive way.

Definition. Let f be an (n+ 1)-ary function, then we define Mn[f ] by

Mn[f ](x1, . . . , xn) =

(y if f(x, y) = 0 and 8z < y[f(x, z) is defined and f(x, z) 6= 0]

undefined if there is no such y.

Example. Mn[prod](x) = 08x 2 N and Mn[sum](x) =

(0 if x = 0

undefined else.

Definition. The set of recursive functions REC is the smallest class of functions such that

(i) PRIMREC ✓ REC, and

(ii) if f 2 REC then Mn[f ] 2 REC.

We can now extend these properties to sets and relations of tuples of natural numbers. In thefollowing definition, observe that if k = 1 then we will obtain a set of natural numbers as opposedto a set of k-tuples.

Definition. Let R ✓ Nk be a relation for some k 2 N. Then the characteristic function�R : Nk ! {0, 1} ⇢ N is the function given by

�R(x) =

(1 if x 2 R

0 if x /2 R.

We call the relation R (primitive) recursive if �R is (primitive) recursive.

Lemma 2.17. The relation < is p.r..

Proof. Note that for any x, y 2 N: x < y ) monus(y, x) > 0 ) sg(monus(y, x)) = 1 andx > y ) monus(y, x) = 0 ) sg(monus(y, x)) = 0. Thus, the relation < is characterised byx < y , sg(monus(y, x)) = 1, i.e. it has a characteristic function that is the composition of p.r.functions.

Corollary 2.18. = is p.r..

17


Proof. Since x < y ) ¬(x > y), then only one these will ever hold so the functionsum(sg(monus(y, x)), sg(monus(x, y)))

has maximum value 1. Ifsg(monus(y, x)) = sg(monus(x, y)) = 0

then ¬x < y and ¬y < x, so we must have x = y as x < y _ y < x _ x = y is true in arithmetic.And clearly

x = y ) sg(monus(y, x)) = sg(monus(x, y)) = 0

because monus(y, x) = monus(x, y) = 0, so = is characterised by

x = y , 1� sum(sg(monus(y, x)), sg(monus(x, y))) = 1.

Proposition 2.19. [3, p. 74] If C1, . . . , Cm are (primitive) recursive (n + 1)-ary relations thatare mutually exclusive and g1, . . . , gm are (primitive) recursive (n+ 1)-ary total functions, then the(n+ 1)-ary function

f(x1, . . . , xn, y) =

8>><

>>:

g1(x, y) if C1(x, y)...gm(x, y) if Cm(x, y)

is (primitive) recursive.

Proof. For each i 6 m, let ci be the characteristic function of Ci and define recursively hi(x, y, 0) = 0,hi(x, y, s(n)) = gi(x, y). Then each hi is (primitive) recursive. Now set Fi(x, y) = hi(x, y, ci(x, y))for each i 6 m, so Fi(x, y) equals 0 if ci(x, y) = 0 and gi(x, y) is ci(x, y). Since the Ci are mutuallyexclusive, only one holds for any one n-tuple, so we can see that f(x, y) = F1(x, y) + · · ·+Fn(x, y).Hence, f is (primitive) recursive as it is obtained from (primitive) recursive functions by composition,primitive recursion and the sum function which we have seen is primitive recursive.

Corollary 2.20. The minimum function

min(x, y) =

8><

>:

x if x < y

x if x = y

y if y < x

and the maximum function

max(x, y) =

8><

>:

y if x < y

x if x = y

x if y < x

are both p.r. functions.

Lemma 2.21. Let f be an n-ary p.r. function then the functionsyP

i=0f(x, i) and

yQi=0

f(x, i) are p.r.

as well.

18


Proof.0P

i=0f(x, i) = f(x, 0) and

s(y)Pi=0

f(x, i) = sum

✓yP

i=0f(x, i), f(x, s(y))

◆.

0Qi=0

f(x, i) = f(x, 0) ands(y)Qi=0

f(x, i) = prod

✓yQ

i=0f(x, i), f(x, s(y))

◆.

Theorem 2.22. [3, pp. 76-7] Let R,S be (primitive) recursive relations. Then

(i) the complement ¬R = {x 2 Nk | ¬R(x)} is (primitive) recursive

(ii) if R and S are of the same arity, then R\S = {x | R(x)^S(x)} and R[S = {x | R(x)_S(x)}are both (primitive) recursive

(iii) if R is (k+1)-ary, then 8z 6 yR(x, z) and 9z 6 yR(x, z) are both (primitive) recursive

(iv) if R is k-ary, and f1, . . . , fk are n-ary (primitive) recursive total functions, then the relationdefined by substituting f1, . . . , fn into R is also (primitive) recursive

(v) the graph of a (primitive) recursive total function is again (primitive) recursive.

Proof. Suppose r is the characteristic function of R and s of S.(i) The characteristic function of ¬R is just monus(1, r).(ii) R \ S and R [ S are characterised by x 2 R \ S , min(r(x), s(x)) = 1 and x 2 R [ S ,

max(r(x), s(x)) = 1.(iii) Note that for 8z 6 yR(x, z) to fail, it takes only the existence of some z 6 y for which

R(x, z) is not true, i.e. for which r(x, z) = 0. Thus, the characteristic function for the relation

8z 6 yR(x, z) isyQ

i=0r(x, i). If for each z 6 y R(x, z) is true then for each z 6 y r(x, z) = 1, so

yQi=0

r(x, i) = 1.

For 9z 6 yR(x, z) it takes only the existence of some z 6 y such that R(x, z) is true, i.e. forwhich r(x, z) = 1. However, we cannot just say the the characteristics function for 9z 6 yR(x, z) isyP

i=0r(x, i), because if there is more than just one z 6 y for which R(x, z) is true then we will have

yPi=0

r(x, i) > 1 andyP

i=0r(x, i) would not be a characteristic function . Recall that sg(x) = 0 if x = 0

and sg(x) = 1 if x > 0. Therefore, we can set the characteristic function of 9z 6 yR(x, z) to be

sg

✓yP

i=0r(x, i)

◆.

(iv) Let T be the relation obtained from substituting f1, . . . , fn into R. Thus, for anyx1, . . . , xn 2 Nn we have

hx1, . . . , xni 2 T , hf1(x1), . . . , fn(xn)i 2 R

and the characteristic function t of T is given by

t(x1, . . . , xn) = r(f1(x1), . . . , fn(xn)).

Since t is obtained from composition of (primitive) recursive functions, we have that t is also (prim-itive) recursive, and hence T is (primitive) recursive.

19


(v) Let f be an n-ary (primitive) recursive function. The graph relation of a f is the relation Gdefined by:

G(x1, . . . xn, y) , f(x1, . . . , xn) = y.

Therefore, the graph relation can be obtained from the = relation by substituting in f , and henceby (iv) is (primitive) recursive.

Corollary 2.23. [3, pp. 77-8] Let R be an n-ary (primitive) recursive relation. Then

Min[R](x1, . . . , xn, z) =

8><

>:

the smallest y 6 z for which if such y existsR(x1, . . . , xn, y) holds

z + 1 else

and

Max[R](x1, . . . , xn, z) =

8><

>:

the largest y 6 z for which if such y existsR(x1, . . . , xn, y) holds

0 else

are (primitive) recursive total functions.

Proof. Min[R]: Let c(x, i) = �8t6i¬R(x,t) be the characteristic function of the (primitive) recursiverelation 8t 6 y ¬R(x, t). Suppose that there exists minimal y 6 z such that R(x, y) holds. Thennote c(x, 0) = c(x, 1) = · · · = c(x, y � 1) = 1, but c(x, y) = c(x, y + 1) = · · · = c(x, z) = 0 by theminimality of y and the fact that R(x, y) is true. Thus,

zX

i=0

c(x, i) =y�1X

i=0

c(x, i) +zX

i=y

c(x, i) = y + 0 = y.

If no such y exists then c(x, i) = 1 for all i 6 z and sozP

i=0c(x, i) = z + 1. Hence,

Min[R](x, z) =zP

i=0c(x, i) and is (primitive) recursive because R(x) (primitive) recursive ) ¬R(x)

(primitive) recursive ) 8i 6 z¬R(x, i) (primitive) recursive.Max[R]: Let d(x, i) = �9t6z[i<t^R(x,t)] be the characteristic function of the (primitive) recursive

relation9t 6 z[i < t^R(x, t)]. Suppose that there exists a maximal y 6 z such that R(x, y) is true.Then d(x, 0) = · · · = d(x, y� 1) = 1 and d(x, y) = · · · = d(x, z) = 0, by the maximality of y, and so

zX

i=0

d(x, i) =y�1X

i=0

cd(x, i) +zX

i=y

d(x, i) = y + 0 = y.

If no such y exists then d(x, i) = 0 for al i 6 z sozP

i=0d(x, i) = 0. Hence, Max[R](x, z) =

zPi=0

d(x, i).

Example. In this example, we shall show that the function which finds the nth prime is in factprimitive recursive. Recall Prime(x) ⌘ (0000 < x) ^ 8y < x8z < x(y ⇥⇥⇥ z 6= x). If we can showPrime(x) is p.r. then we can use this in our definition for np(x), the next prime function.

20


We have seen above that = is a p.r. relation and thus by Theorem 2.22 so is 6=, the complement of=. The relation y⇥⇥⇥z = x is the graph of p.r. function prod and so y⇥⇥⇥z 6= x is p.r. as the complementof a p.r. function. Then, by two applications of (iii) from the Theorem, 8y < x8z < x(y⇥⇥⇥ z 6= x) isp.r.. But note 0

000 < x is obtained from the relation y < x by substituting in s(z(y)) and id11(x), twop.r. functions, so 0

000 < x is p.r. and by (ii) of the TheoremPrime(x) ⌘ (0000 < x) ^ 8y < x8z < x(y⇥⇥⇥ z 6= x)

is p.r..We want to define the next prime function by np(x) = the least y such that x < y and Prime(y).

Above we found that the minimisation maintains the primitive recursiveness provided we can boundthe search, but we know from Euclid’s proof of the infinitude of primes that the bound is x! + 1.Thus, np(x) = the least y 6 x! + 1 such that R(y) holds = Min[R](x, x! + 1) where y 2 R ,x < y ^ Prime(y). Hence, np is primitive recursive.

Now define the prime finder function ⇡ by ⇡(0) = 2 and ⇡(s(x)) = np(⇡(x)). Thus, the function⇡ is primitive recursive.

2.5 Arithmetical Definability

Arithmetical definability is a key concept when discussing relations and functions on the naturalnumbers that can be “described” by a formula in the language of arithmetic. We will see by theend of this section that any recursive function can be “described” by a ⌃1 formula. What is allthis leading us to? We will use recursive functions to essentially code our language of arithmetic,thereby allowing us to treat formulas and finite sequences of formulas completely arithmetically andformally in PA.

Definition. A k-ary relation R is arithmetically definable just in case there exists a formula'R(x1, . . . , xk) 2 FORMLA such that

hn1, . . . , nki 2 R , 'R(n1

, . . . ,nk

) holds.

A k-ary function f is arithmetically definable iff there exists a formula 'f (x1, . . . , xk, y) 2 FORMLA

such that 8hn1, . . . , nki 2 Nk

f(n1, . . . , nk) = m , 'f (n1

, . . . ,nk

,m) holds.

Definition. [2, p. 28] Rem(x, d, r) ⌘ (d = 0 ^ r = x) _ (r < d ^ 9q 6 x[x = (q⇥⇥⇥ d) +++ r])

Definition. [3, p. 200] Quo(x, d, q) ⌘ (d = 0 ^ q = 0) _ (9r < d[x = (q⇥⇥⇥ d) +++ r])

Rem(x, d, r) is ⌃1 and arithmetically defines the modified remainder function rem(x, d), whichif d = 0 returns r = x and if d 6= 0 returns the remainder r when x is divided by d. Quo(x, d, q) is⌃1 and defines the quotient function, which if d = 0 returns q = 0 and if d 6= 0 returns the unique qsuch that q · d divides x but (q + 1) · d does not. These are both in fact ⌃1 pterms.

Definition. The p.r. exponential function exp is defined as follows

exp(x, 0) = 1, exp(x, s(y)) = prod(x, exp(x, y)).

We will abbreviate by writing xy to mean exp(x, y).

21


Lemma 2.24. (Gödel’s �-function Lemma) [3, p. 203] 8k 2 N and 8a0, . . . , ak 2 N, there exists, t 2 N such that 80 6 i 6 k we have ai = �(s, t, i) := rem(s, t(i+ 1) + 1).

Proof. We omit the proof for this as it is a number-theoretical proof and follows from the ChineseRemainder Theorem and the fact that there are infinitely many primes.

Definition. [3, p. 12] The ordered pair hs, ti codes the sequence ha0, . . . , aki iff for every 0 6 i 6 kai = ent(s, t, i) = �(s, t, i) = rem(s, t(i+ 1) + 1).

Since we have Gödel’s �-function Lemma, we know that for any sequence we can find a pair whichcodes it. Furthermore, observe that ent(s, t, i) is defined by 'ent(s, t, i, y) ⌘ Rem(s, t(i+++ 1)+++ 1, y).

Theorem 2.25. [3, p. 201] If f(x1, . . . , xk) is a k-ary recursive function, then there exists a formula'f (x1, . . . , xk, y) 2 FORMLA which arithmetically defines f .

Proof. We prove this claim by induction on the form of f . First, we handle the basis functions.The zero function z is defined by 'z(x, y) ⌘ y = 0.The successor function s is defined by 's(x, y) ⌘ y = x+++ 1.The projection function idni is defined by 'idni

(x1, . . . , xn, y) ⌘ y = xi.For composition, we prove the case for when f is the composition of two arithmetically defined

recursive unary functions g, h. Let g, h be defined by 'g,'h, respectively and suppose f = gh. Letl 2 N be arbitrary, m = h(l) and n = g(h(l)) = f(l). Since 'h defines h and 'g defines g, we know'h(l,m) holds and 'g(m,n) holds. Therefore, 'h(l,m) ^ 'g(m,n) holds ) 9y['h(l, y) ^ 'g(y,n)]holds, so take 'f (x, z) ⌘ 9y['h(x, y) ^ 'g(y, z)]. Now suppose 'f (l,n) ⌘ 9y['h(l, y) ^ 'g(y,n)]holds, so 'h(l,m)^'g(m,n) holds for some m 2 N ) 'h(l,m),'g(m,n) hold for some m 2 N. 'h

defines h implies that h(l) = m and 'g defines g implies g(m) = n, and thus g(m) = (g(h(l)) = f(l).Primitive recursion and minimisation are similar but use 'ent, so a little more needs to be done

and we omit those proofs.

Corollary 2.26. [3, p. 204] Every recursive relation is arithmetical.

Proof. Let R be a recursive relation, and r its characteristic function. Then r is recursive, so isarithmetically definable by 'R(x, y) say. Then R is definable by 'R(x,1).

Theorem 2.27. [3] Every recursive relation and function is arithmetically definable by a ⌃1 formula.

Proof. The basis functions are all definable by a ⌃1 formula. In the full proof of Theorem 2.25, weonly do operations which are ⌃1-closed, so all formulas are in fact ⌃1.

2.6 Gödel’s Incompleteness Theorems

The purpose of this section is to introduce some key incompleteness results that can be formalised inPA. That is, not only do these metatheoretical results apply in our metatheory, but because we willshow that we can formalise our metalanguage we will have formalised versions of our metatheoreticalresults. We will employ a technique, called Gödel numbering, which Gödel introduced in his 1931paper.

We begin by assigning the natural numbers, called Gödel numbers, 1, 3, 5, 7, 9, 11, 13, 15 and 17to the symbols ?,^,¬, 8,=,0, 000,+++, and ⇥⇥⇥ of LA, respectively. For a variable xi we associate the

22


number 2i + 19. Each primitive symbol of PA now has a Gödel number assigned to it and thatnumber is odd.

Now let x, y be two objects which could be symbols, ordered pairs, or ordered triples, then weshall define the Gödel number of the ordered pair hx, yi to be 2(i + j)(i + j) + i + 1, where x hasGödel number i and y has Gödel number j. Recall from §2.1 that any term or formula is eitherprimitive or is an ordered pair, so now all terms and all formulas have Gödel numbers. Moreover,by our assignment of Gödel numbers, all ordered pairs have even Gödel numbers as opposed to ourodd primitive symbols. Our next task is formalise our treatment of ordered pairs, ordered triplesand finite sequences so that we can utilise these objects in the formal setting of PA

10.

Definition. Pair(x, y, z) ⌘ z = 2⇥⇥⇥ [(x+++ y)⇥⇥⇥ (x+++ y) +++ x+++ 1]

We will abbreviate x⇥⇥⇥y to xy, so we have pair(x, y) = 2[(x+++y)(x+++y)+++x+++1]. And rather thanwrite pair(x, y) each time for the pair hx, yi, we shall just write (x, y) = 2[(x+++ y)(x+++ y) +++ x+++ 1].From this coding of an ordered pair, it is clear that the uniqueness of an ordered pair is provablein PA and therefore this is a suitable coding, which works, of pairs of natural numbers as singlenumber.

Definition. Ft(z, x) ⌘ 9y < [z(x, y) = z] _ (¬9w, y < z[(w, y) = z ^ x = 0]), thusPA ` ft((x, y)) = x.

Definition. Sd(z, y) ⌘ 9x < z[(x, y) = z] _ (¬9x,w < z[(x,w) = z ^ y = 0]), thusPA ` sd((x, y)) = y.

These two functions give us a way of pulling out elements of an ordered pair. But as we hadbefore, we now need the notion of an ordered triple within PA:

Definition. (x, y, z) = (x, (y, z))

We use the following functions to extract elements of an ordered triple:

Definition. (i) fst(w) = ft(w), thus PA ` fst((x, (y, z))) = x

(ii) snd(w) = sd(ft(w)), thus PA ` snd((x, (y, z))) = y

(iii) trd(w) = sd(sd(w)), thus PA ` trd((x, (y, z))) = z.

Accordingly, we use two letters for the functions that act on ordered pairs and three letters forfunctions acting on ordered triples. We now turn our attention to finite sequences of length k, say.

Definition. FinSeq(s) ⌘

9a < s9b < s9k < s[s = ((a, b), k) ^ 8c < s8d < s[(c, d) < (a, b) ! 9i < k�(c, d, i) 6= �(a, b, i)]]

Definition. lh(s) = sd(s)

Definition. val(s, i) = �(ft(ft(s)), sd(ft(s)), i)

10We follow [2, pp. 39-44] in our treatment of Gödel numbering, which is quite concise.

23


Thus, lh(s) gives us the length of the sequence coded by s and val(s, i) finds the ith entry of thesequence. Instead of val(s, i), we shall write si where hs0, . . . , slh(s)�1i is the sequence coded by s.Notice also that FinSeq(s) is �1, and lh and val can be arithmetically defined by ⌃1 formulas. Sincea sequence is defined entirely by its entries, the uniqueness of sequences can be proved formally inPA. We also have the unique empty sequence hi := ((0,0),0). We saw in §2.1 that we can truncateand concatenate sequences, and we can formalise these operations in PA. Here we give a slightlymore generalised version of truncation, in which we can cut-off sequence entries from the beginningas well as the end of the sequence. However, concatenation works just as it did previously.

Definition. Trunc(s, e, j, t) ⌘

(¬e 6 j < lh(s) ^ t = hi) _ (e 6 j < lh(s) ^ FinSeq(t) ^ lh(t) = j � e^8i < j � e[val(t, i) = val(s, i+ e)])

trunc(s, e, j) returns the code of the finite sequence hse, se+1, . . . , sji of length j � e, which hasbeen found from the sequence

hs0, . . . , se, . . . , sj , . . . , slh(s)i.

Definition. Concat(s, t, s ⇤ t) ⌘

FinSeq(s ⇤ t) ^ lh(s ⇤ t) = lh(s) + lh(t) ^ 8i < lh(s)[(s ⇤ t)i = si] ^ 8i < lh(t)[(s ⇤ t)lh(s)+i = ti]

Our next definition is the tool that formalises Gödel numbering in PA.

Definition. Let p·q : {?,^,¬, 8,=,0, 000,+++,⇥⇥⇥} ! {1,3,5,7,9,11,13,15,17} be the function givenby p?q, p^q, p¬q, p8q, p=q, p0q, p000q, p+++q, p⇥⇥⇥q are 1,3,5,7,9,11,13,15,17, respectively.

This is an important place to make the distinction between our metatheory and our formaltheory. The Gödel number of p?q is 1 but the Gödel numeral is the formal object 1 ⌘ 0

000 whichlives in PA and not in our metamathematical world. We now have a way of “talking” about theprimitive symbols in PA but what of terms, formulas, or even proofs? For this we need to be ableto determine when a bold-faced numeral defines a variable, term, atomic formula, formula, and evena rule of inference.

Definition. V ariable(v) ⌘ 9i < v[v = (2⇥⇥⇥ i) +++ 19]

Definition. Term(t) ⌘

9s[FinSeq(s) ^ lh(s) > 0 ^ slh(s)�1 = t ^ 8i < lh(s)[si = p0q _ V ariable(si)_9j, k < i[si = (p000q, sj) _ si = (p+++q, sj , sk) _ si = (p⇥⇥⇥q, sj , sk)]]]

Term(t) is visibly ⌃1 but it can be proved11 that Term(t) is actually �1. However, we shall notshow that here but take that fact for granted. Now we give a predicate to characterise ? and theidentities, i.e. the atomic formulas.

Definition. AtForm(x) ⌘ x = p?q _ 9s < x9t < x[Term(s) ^ Term(t) ^ x = (p=q, s, t)]

Recall that we took {?,^,¬, 8} as our primitive connectives and quantifiers, so we obtain thefollowing formula to characterise the formulas of LA.

11For a proof see [2, p. 42]

24


Definition. Formula(x) ⌘

9s[FinSeq(s) ^ lh(s) > 0 ^ slh(s)�1 = x ^ 8i < lh(s)[AtForm(si) _ 9j < i[si = (p¬q, sj)]_9j, k < i[si = (p^q, sj , sk)] _ 9j < i9v[V ariable(v) ^ si = (p8q, v, sj)]]]

Term(t) being �1 gives that AtForm(x) is �1, and in Formula(x) it can be shown that s canbe bounded by 6 x, and thus Formula(x) is also �1. From this definition of Formula(x) we canalso prove formally sentences such as 8x8y[Formula(x)^Formula(y) ! Formula((p^q, x, y))], sowe have formally all possible well-formed formulas. As in [2, p. 43], we suppose the existence of apredicate Axiom(x) asserting that x is the Gödel number of an axiom of PA, and the existence of a⌃1 pterm which correctly defines sub(t, i, x) the operation of substituting in the term that has Gödelnumber t in for the ith variable in the formula with Gödel number x. We now give two formulasto hold whenever the formula with Gödel number x is the conclusion of an application of modusponens and universal generalisation, respectively.

Definition. ConseqMP (x, y, z) ⌘

Formula(x) ^ Formula(z) ^ y = pz ! xq

Definition. ConseqUG(x, y) ⌘

9v < x[Formula(y) ^ V ariable(v) ^ x = (p8q, v, y)]

Now we have all the necessary predicates to establish when y codes a proof of the formula 'x

with Gödel number x, i.e. a finite sequence of formulas which has final entry 'x. ConseqMP andConseqUG are both �1 so the next formula we give is also �1.

Definition. Pf(y, x) ⌘

FinSeq(y) ^ ylh(y)�1 = x ^ 8i < lh(y)� 1[Ax(yi) _ 9j, k < i[ConseqMP (yi, yj , yk)]_9j < iConseqUG(yi, yj)]

Then the ⌃1 formula, which is not �1, which asserts that theres exists a proof (in PA) withGödel number y for a formula with Gödel number x is:

Definition. Prov(x) ⌘ 9yPf(y, x)

We shall show that this very important formula Prov(x) will satisfy some derivability conditions,and hence can be used in Löb’s Theorem for PA which is yet to come. The next few steps to finishthis section will be to prove the Generalised Diagonal Lemma, which is used to prove Löb’s Theoremwhich in turn is used to prove Gödel’s Second Incompleteness Theorem.

Theorem 2.28. [2, p. 53] (Generalised Diagonal Lemma) Suppose that x0, . . . , xn, y1, . . . , ymare distinct variables in LA and that we have '0(x,y),. . . , 'n(x,y) 2 FORMLA such thatFV ('i) ✓ {x0, . . . , xn, y1, . . . , ym} 8i 6 n. Then there exist 0(y), . . . , n(y) 2 FORMLA withFV ( i) ✓ {y1, . . . , ym}, and such that

PA ` 0(y) $ '0(p 0(y)q, . . . , p n(y)q,y),...

PA ` n(y) $ 'n(p 0(y)q, . . . , p n(y)q,y).

25


Proof. Let Subst(w, z0, . . . , zn, y) be a ⌃1 pterm for the function subst(w, z0, . . . , zn) which is definedby: subst(a, b0, . . . , bn) = the Gödel number of the formula which is obtained by substituting thenumerals b0, . . . ,bn in for the variables z0, . . . , zn, respectively, in the formula with Gödel numbera. The fact that there is a ⌃1 pterm for this function follows from the fact that there is a ⌃1 ptermfor the function sub(t, i, x) defined above.

Now, 80 6 i 6 n define

ki := p'i(subst(z0, z0, . . . , zn), . . . , subst(zn, z0, . . . , zn),y)q

and i(y) := 'i(subst(k0,k0, . . . ,kn), . . . , subst(kn,k0, . . . ,kn),y).

It remains to show that PA ` i(y) $ 'i(p 0(y)q, . . . , p n(y)q,y) for each i 6 n, so it suf-fices to show PA ` subst(ki,k0, . . . ,kn) = p i(y)q for each i. Consider subst(ki,k0, . . . ,kn); bydefinition of subst, subst(ki,k0, . . . ,kn) is the Gödel number of the resulting formula when wesubstitute numerals k0, . . . ,kn in for the variables z0, . . . , zn in the formula with Gödel numberki = p'i(subst(z0, z0, . . . , zn), . . . , subst(zn, z0, . . . , zn),y)q. So

subst(ki,k0, . . . ,kn) = p'i(subst(k0,k0, . . . ,kn), . . . , subst(kn,k0, . . . ,kn),y)q

and that is justsubst(ki,k0, . . . ,kn) = p i(y)q.

subst(ki,k0, . . . ,kn) = p i(y)q is ⌃1 and is true, so we know that for each i 6 n

PA ` subst(ki,k0, . . . ,kn) = p i(y)q.

Therefore, PA ` i(y) $ 'i(p 0(y)q, . . . , p n(y)q,y) for each i 6 n.

Corollary 2.29. [2, p. 54] (Diagonal Lemma) Let '(x0) 2 FORMLA with at most one freevariable x0. Then there exists a sentence , i.e. with no free variables, such that PA ` $ '(p q).

Proof. This is an immediate consequence from the Generalised Diagonal Lemma (Theorem 2.28)where we set n = m = 0.

Now consider the formula with one free variable ¬Prov(x) 2 FORMLA . By the DiagonalLemma, there exists sentence G := G

PA

such that PA ` G $ ¬Prov(pGq). This G is our so-calledGödel sentence of PA.

Definition. We call PA !-inconsistent if there exists a formula '(x) such that PA ` 9x'(x) butPA ` ¬'(0),¬'(1), . . .. PA is !-consistent otherwise.

Lemma 2.30. For PA, !-consistency ) consistency.

Proof. Suppose PA is !-consistent and inconsistent. Since PA is inconsistent, anything of ourchoosing can be derived in PA. Thus, PA ` 9x'(x) and PA ` ¬'(0),¬'(1), . . ., i.e. PA is!-inconsistent. This is a contradiction. Hence, PA must have been consistent.

We now have our first limitative theorem for PA.

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Theorem 2.31. (Gödel’s First Incompleteness Theorem for PA)

(i) If PA is consistent, then PA 0 G.

(ii) Further, if PA is !-consistent, then PA 0 ¬G.

Proof. (i) Suppose PA is consistent and that PA ` G, then since PA ` G $ ¬Prov(pGq) we wouldhave PA ` ¬Prov(pGq). But since PA ` G, we know that Prov(pGq) = 9yPf(y, pGq) is a true ⌃1

formula of arithmetic, so PA ` Prov(pGq). Therefore, PA ` Prov(pGq) ^ ¬Prov(pGq). Hence,PA 0 G.

(ii) Suppose PA is !-consistent and PA ` ¬G. PA is !-consistent implies PA is consistent,so from (i) we know PA 0 G. That is, for each natural number n 2 N, never is n the Gödelnumber of a proof in PA of G; that is, all of ¬Pf(0, pGq),¬Pf(1, pGq), . . . are true. Thus, bythe ⌃1-completeness of PA, we know that PA ` ¬Pf(0, pGq),¬Pf(1, pGq), . . .. Since we assumedPA ` ¬G, we also have PA ` Prov(pGq), i.e. PA ` 9yPf(y, pGq), and hence PA is !-inconsistent.

We now present the (Hilbert-Bernays-Löb) derivability conditions for a formal theory extendingPA. They are named the derivability conditions, because provided they hold for an arbitrary con-sistent, axiomatizable theory T ◆ PA and arbitrary formula B(x) in the language of the theory,then one can derive Gödel’s Second Incompleteness Theorem in T .

Definition. Let T,B(x) be as above and suppose ', are arbitrary formulas in the language of T ,then the derivability conditions for T are as follows:

(P1) if T ` ', then T ` B(p'q)

(P2) T ` B(p'! q) ! (B(p'q) ! B(p q))

(P3) T ` B(p'q) ! B(pB(p'q)q).

If B(x) satisfies (P1)-(P3) as above, then we call B(x) a provability predicate for T.

Lemma 2.32. [2, p. 46] For any ⌃1 formula ', PA ` '! Prov(p'q).

Proof. We omit the proof of this as it is long and detailed. We do not lose anything by omitting it.

Proposition 2.33. Prov(x) is a provability predicate for PA.

Proof. Let ', 2 FORMLA be arbitrary formulas.(P1) Suppose that PA ` ', i.e. ' is provable in PA, i.e. there exists a proof in PA of '.

Thus, 9yPf(y, p'q) is true, i.e. Prov(p'q) is true. Since Prov(x) is a ⌃1 formula, Prov(p'q)is a ⌃1 sentence which is true. Therefore, by Theorem 2.15 (⌃1-completeness of PA), we knowPA ` Prov(p'q) as required.

(P2) Suppose PA ` Prov(p' ! q), i.e. PA ` 9xPf(x, p' ! q), and PA ` Prov(p'q), i.e.PA ` 9yPf(y, p'q). If x, y are the numbers which are the Gödel numbers of proofs of '! and ,respectively, then we can concatenate these to get x ⇤ y. Then adding the sequence < > of length1 to the end of x ⇤ y, we obtain a proof for where the final entry is justified by an application of

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modus ponens. Thus, we have PA ` 9zPf(z, p q), which is just PA ` Prov(p q) and the claim istrue.

(P3) Suppose PA ` Prov(p'q). Then from Lemma 2.32 above,PA ` Prov(p'q) ! Prov(pProv(p'q)q) and therefore PA ` Prov(pProv(p'q)q) and we are done.

Corollary 2.34. PA ` '! ) PA ` Prov(p'q) ! Prov(p q).

Proof. Suppose PA ` ' ! , then by (P3) we know PA ` Prov(p' ! q) and by (P2) we getPA ` Prov(p'q) ! Prov(p q).

Theorem 2.35. [2, pp. 56-8](Löb’s Theorem) Suppose ' is a sentence of LA. If PA `Prov(p'q) ! ', then PA ` '.

Proof. Suppose PA ` Prov(p'q) ! '. Set (x) ⌘ Prov(x) ! ', then we can apply the DiagonalLemma to this, because it is just a formula of LA, and we obtain a sentence A such thatPA ` A $ (pAq), i.e. PA ` A $ (Prov(pAq) ! '). In particular, PA ` A ! (Prov(pAq) ! ').Then by Corollary 2.34, PA ` Prov(pAq) ! Prov(p(Prov(pAq) ! ')q). We know by (P2) that

PA ` Prov(p(Prov(pAq) ! ')q) ! [Prov(pProv(pAq)q) ! Prov(p'q)].

Thus, by propositional logic

PA ` Prov(pAq) ! [Prov(pProv(pAq)q) ! Prov(p'q)].

But by (P3),PA ` Prov(pAq) ! Prov(pProv(pAq)q)

and therefore PA ` Prov(pAq) ! Prov(p'q). Since we are assuming PA ` Prov(p'q) ! ', wethen have PA ` Prov(pAq) ! ' (⇤). From PA ` A $ (Prov(pAq) ! ') we get

PA ` (Prov(pAq) ! ') ! A,

then by modus ponens we also know PA ` A. So by (P1) we have PA ` Prov(pAq), and hencefrom (⇤) we have PA ` ' as desired.

Remark. We note here that Prov(p?q) asserts that there is a proof (in PA) of a contradiction inPA, so ¬Prov(p?q) asserts that there is no such proof and hence that PA is consistent.

Theorem 2.36. (Gödel’s Second Incompleteness Theorem for PA) If PA consistent, thenPA 0 ¬Prov(p?q), i.e. PA cannot prove its own consistency.

Proof. Suppose PA is consistent, i.e. PA 0 ?, and PA ` ¬Prov(p?q). That is,PA ` Prov(p?q) ! ?. Then immediately from Löb’s Theorem (Theorem 2.35), we know thatPA ` ?. Hence, PA 0 ¬Prov(p?q).

It is worth observing that Gödel’s theorems apply for any recursively axiomatizable theoryextending Q, Robinson Arithmetic. However, as we stated earlier, we cannot prove (in Q) somenumber theory which is true in our metamathematical world.

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3 Modal Logic

12

In this section we shall introduce and explore propositional modal logic, providing the necessaryinformation we need to start matching up Peano Arithmetic and the modal logic system GL. Wewill start by considering some standard modal logics and conclude the section by focusing on GL,the modal logic system of greatest importance to us in this essay.

3.1 The Language of Propositional Modal Logic

As mentioned in §1.2, propositional modal logic is simply classical propositional logic with theaddition of a modal operator. More formally, we shall set the language of propositional modallogic L⇤ to consist of the countably infinite set of propositional letters p0, p1, . . ., the Booleanconnectives, parentheses and the modal operators ⇤,⌃. As before, we will take {^,¬,?} to be ourtruth-functionally complete set of Boolean connectives and ⇤ as our primitive modal operator. Fornow we shall refer to ⇤ as necessarily, but recall from §1.2 that changing our meaning of ⇤ canaffect the truth of a sentence where ⇤ has been used. However, in this section we shall define how⇤ should be used within our setting and our results will be applicable to whether we interpret ⇤ tomean necessarily or provably.

Definition. The set of atomic propositions At is the smallest set containing ? and all propositionalletters p0, p1, . . ..

Definition. The set of (well-formed) formulas of L⇤ is the smallest set FORML⇤ such that

(i) ? 2 FORML⇤

(ii) p0, p1, . . . 2 FORML⇤

(iii) A,B 2 FORML⇤ ) (A ^B) 2 FORML⇤

(iv) A 2 FORML⇤ ) (¬A) 2 FORML⇤

(v) A 2 FORML⇤ ) (⇤A) 2 FORML⇤ .

We allow all the previous abbreviations concerning Boolean connectives stated in §2.2 as well asthe following:

Definition. ⌃ := ¬⇤¬ and > := ¬?

Based on these definitions, we will refer to ⌃ as possibly and > as tautology ; just as ? is meanta constant falsehood, > will be meant as a constant tautology. We call ⌃ the dual to ⇤ using thedefinition of ⌃ above. This was first argued by Aristotle in De Interpretatione and gives rise to theModal Square of Opposition:

12§§3.1-3.4 follow [6] and [8] in presentation and terminology, but with input from the author.

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It is necessary that A It is necessary that not A⇤A ⇤¬A

- %. &

It is not necessary that not A It is not necessary that A¬⇤¬A ¬⇤A

Figure 1: The Modal Square of Opposition [8, p. 7]

3.2 Semantics

Now that we have a notion of necessity, one can ask when a proposition is necessary. This leads usto a discussion of possible world semantics. Informally, we can say that a statement is necessarilytrue if it is true under any interpretation we take. Moreover, a statement is true if it is true in everypossible world, where a possible world is one of many worlds or situations that could occur. So eventhough we are in the actual world as described in §1.2, other possible worlds could happen and inthose worlds things may be very different to what we experience or think conceivable in our world.However, we need to be careful: not every world may be accessible from every other world. Thecrude definition of a necessary statement just given is the one argued by Leibniz, but it does nottake into consideration that a world may not be accessible from another world. For example, forpossible worlds w, x define an accessibility relation B by wBx iff w occurs no later than x wherewe order the possible worlds temporally [8, p. 9]. That is, a world w can get to world x iff x isbrought into existence after (time-wise) w. Now suppose we have worlds w0 and w1 with w0Bw1, soif statement ⇤A were true at w0 then statement A would be true at w1. However, if ⇤A were trueat w1 then we cannot deduce that A would be true at w0, because w0Bw1 but not w1Bw0. Notethat this accessibility relation B is reflexive, i.e. if ⇤A is true at world x then A is true at x andthat is because trivially x is in existence as soon as x is in existence. However, not all accessibilityrelations will be reflexive; in fact, the relation that we will be interested in will be irreflexive. Thereare various properties a relation can have, and for accessibility relations different properties implydifferent modal formulas hold. When we say ‘hold’ we really mean true at all worlds in our set ofpossible worlds, and so now we need to develop a theory about these possible worlds.

Definition. We call the ordered pair F := hW,Ri a frame where W is a non-empty set andR ✓ W ⇥ W is a relation of W . We shall refer to the elements of W as possible worlds and R asthe accessibility relation. For w, x 2 W , we write wRx if hw, xi 2 R and read this as x is accessiblefrom w, or w has access to x.

Definition. A (Kripke propositional modal logic) model is a structure M := hW,R,�i such thatF := hW,Ri is a frame and �✓ W ⇥ At is a relation between the possible worlds in W and thepropositional letters. We say that M is based on F .

A model is thus a kind of universe where we have various possible worlds and (possibly) differentpropositions true at different worlds, as well as some accessibility relation between the worlds. Itmay be that from one world we can get to all the other possible worlds, or it may be that each worldcan only get to itself or even to no other world, but these potential accessibility relations depend onthe model we find ourselves working with. Therefore, it is necessary that we make a fundamentaldefinition: the definition of truth in a model.

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Definition. (Truth in a Model) Let M := hW,R,�i be a model and w 2 W . ForA 2 FORML⇤ , we define M, w � A inductively as follows:

(i) Not M, w � ?; we will abbreviate to M, w 1 ?

(ii) M, w � pi , hw, pii 2�

(iii) M, w � (B ^ C) , M, w � B and M, w � C

(iv) M, w � (¬B) , M, w 1 B; read B is false at w

(v) M, w � (⇤B) , 8x 2 W such that wRx we have M, x � B

(vi) M, w � (B _ C) , M, w � B or M, w � C

(vii) M, w � (B ! C) , M, w 1 B or M, w � C

(viii) M, w � (⌃B) , 9x 2 W such that wRx and M, x � B.

Typically, � is read as “forces” and we will write ‘A is true at w’ or ‘w forces A’ to informallymean ‘M, w � A’ where M is appropriately understood. For the sake of completeness it was worthmentioning above this once the definition for truth in model for the other connectives. However, wecould have stopped at the end of (v), which would have given us a truth-functionally complete set ofdefinitions along with the definition for ⇤. The definition just given describes truth at a particularworld in the model, but we will encounter sentences that will true at all possible worlds in a model,or even true in all models where the accessibility relation has a certain property. We will call a setof frames sharing property P, the class of P frames; e.g. the class of reflexive frames.

Definition. We say that a formula A 2 FORML⇤ is

(i) valid in (model) M := hW,R,�i iff 8w 2 W we have M, w � A

(ii) valid in (frame) F := hW,Ri iff for all models M based on F , A is valid in M

(iii) valid in (the class of frames) C iff for all frames F 2 C , A is valid in F

(iv) valid iff A is valid in all frames.

There are some characteristics of relations that will come up frequently, and we can extend theseproperties to frames themselves. The next set of definitions will allow us to prove some results aboutclasses of frames and make solid the foundation we need to begin handling the modal logic systemGL in the next section.

Definition. We say that the relation R in the frame F := hW,Ri is:

(i) reflexive iff 8w 2 W we have wRw

(ii) irreflexive iff for no w 2 W do we have wRw

(iii) transitive iff 8w, x, y 2 W we have wRx, xRy ) wRy

(iv) symmetric iff 8w, x 2 W we have wRx ) xRw

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(v) finite iff W is finite.

We will often refer to the frame F = hW,Ri being P whenever R is P. We will show in whichframes the following modal logic axioms are valid.

Definition. Modal Axioms:

(i) K : ⇤(A ! B) ! (⇤A ! ⇤B)

(ii) T : ⇤A ! A

(iii) 4 : ⇤A ! ⇤⇤A

(iv) B : A ! ⇤⌃A

Theorem 3.1. The K axiom is valid in the class of all frames.

Proof. Let M := hW,R,�i be an arbitrary model based on an arbitrary frame F and w 2 W anarbitrary possible world. We want to show that M, w � ⇤(A ! B) ! (⇤A ! ⇤B). Suppose thatM, w � ⇤(A ! B) and M, w � ⇤A. Now it suffices to show that for every x 2 W with wRx, wehave M, x � B because we will then have M, w � ⇤B. Let x 2 W be arbitrary such that wRx.Since M, w � ⇤(A ! B), M, w � ⇤A and wRx, we have that M, x � A ! B and M, x � A.Thus, by modus ponens M, x � B and so M, w � ⇤A ! ⇤B because x was arbitrary, which inturn implies M, w � ⇤(A ! B) ! (⇤A ! ⇤B). Since M and w were arbitrary, we have that⇤(A ! B) ! (⇤A ! ⇤B) is valid in F .

Theorem 3.2. The T axiom is valid in the class of reflexive frames.

Proof. Let M := hW,R,�i be an arbitrary model based on a reflexive frame F and w 2 W anarbitrary world. Suppose M, w � ⇤A. R is reflexive, since M is based on a reflexive frame, so wehave that wRw. We know by definition that M, w � ⇤A iff 8x 2 W with wRx, M, x � A. Thus,we have M, w � A and hence M, w � ⇤A ! A as needed. Since M and w were arbitrary, we havethat ⇤A ! A is valid in F .

Theorem 3.3. The 4 axiom is valid in the class of transitive frames.

Proof. Let M := hW,R,�i be an arbitrary model based on a transitive frame F and w 2 W anarbitrary world. Suppose M, w � ⇤A. Let x 2 W be arbitrary such that wRx and y 2 W suchthat xRy. Since R is transitive, we have wRy. Thus, M, y � A because M, w � ⇤A. y 2 Wwas arbitrary such that xRy so M, x � ⇤A, and x 2 W was arbitrary with wRx so we alsoget M, w � ⇤⇤A. Hence, M, w � ⇤A ! ⇤⇤A. Since M and w were arbitrary, we have that⇤A ! ⇤⇤A is valid in F .

Theorem 3.4. The B axiom is valid in the class of symmetric frames.

Proof. Let M := hW,R,�i be an arbitrary model based on a symmetric frame F and w 2 W anarbitrary world. Suppose M, w � A and let x 2 W be arbitrary with wRx. Since R is symmetric,we immediately have xRw. Then M, w � A and xRw, so M, x � ⌃A. But as x 2 W with wRx wasarbitrary, we have that M, w � ⇤⌃A and hence M, w � A ! ⇤⌃A. Since M and w were arbitrary,we have that A ! ⇤⌃A is valid in F .

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3.3 Axiomatic Proof System

The proof system we shall restrict ourselves to for propositional modal logic will be the axiomaticsystem. The version for modal logic is very similar to that for PA and so we shall move swiftly.The main setup is essentially the same: we have axioms and rules of inference. We shall takeany tautologies in L⇤ and any instances of ⇤(A ! B) ! (⇤A ! ⇤B), for A,B 2 FORML⇤ ,

as our basic axioms. The rules of inference will be: modus ponens, i.e. A A ! B

Bfor any

A,B 2 FORML⇤ ; and necessitation, that is from A deduce ⇤A, i.e. A

⇤Afor any A 2 FORML⇤ .

Definition. An (axiomatic) proof or derivation D for A 2 FORML⇤ is a finite sequence of formulasin L⇤, such that the last entry of the sequence is A and each entry of D is either a tautology or anaxiom, or is a logical consequence of a rule of inference.

This gives us a general definition of a proof in modal logic, but it may be the case that we haveaccess to different axioms. For example, if we are in the class of transitive frames, then we havejust seen that formulas of the form ⇤A ! ⇤⇤A are valid and so there is no reason why we cannotassume them to be part of our axiom set.

Definition. We now define several modal logic systems:

(i) K is the modal logic system with all tautologies and all instances of the K axiom as axioms,and modus ponens and necessitation as the only rules of inference.

(ii) T is K but with all instances of the T axiom included in its axiom set.

(iii) K4 is K but with all instances of the 4 axiom included in its axiom set.

(iv) B is K but with all instances of the T and B axioms included in its axiom set.

(v) S4 is K but with all instances of the T and 4 axioms included in its axiom set.

If a modal logic system L has all tautologies and all instances of the K axiom in its set oftheorems, and it is closed under modus ponens and necessitation, then we call L a normal modallogic. Therefore, we can view K as the smallest normal modal logic system, and T, K4, B, S4

(respectively) as the smallest reflexive, transitive, reflexive and symmetric, reflexive and transitive(respectively) normal modal logics. In particular, if we prove any result about K then the resultwill hold for any normal modal logic extending K.

Definition. Let L, M be modal logics. We say L extends M or L is an extension of M, and writeM ✓ L, if every theorem of M is a theorem of L.

Hence, we have the following diagram of normal extensions:

K ✓ T ✓ B✓ ✓

K4 ✓ S4

Notation. For any modal logic system L, we will write L ` A to mean there is a proof of A in theaxiomatic proof system for L, and in which we case we say A is a theorem of L.

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Lemma 3.5. We can introduce a new rule of inference: A ! B

⇤A ! ⇤B, the derived rule of regularity.

Proof. Suppose the line A ! B appears in some proof D. Then we have the following:A ! B⇤(A ! B) Necessitation⇤(A ! B) ! (⇤A ! ⇤B) K axiom⇤A ! ⇤B Modus ponens.Thus, whenever we use this rule we could just replace an application of it with the lines above.

However, to help the the readability of future proofs, we shall allow the derived rule of regularity tobe used as a rule of inference.

For the following results, let A1, . . . , An, A,B 2 FORML⇤ be arbitrary modal formulas.

Corollary 3.6. If K ` A $ B, then K ` ⇤A $ ⇤B.

Proof. K ` A $ B ) K ` A ! B and K ` B ! A. Then by (the derived rule of) regularity,K ` ⇤A ! ⇤B and K ` ⇤B ! ⇤A.

So K ` (⇤A ! ⇤B) ^ (⇤B ! ⇤A) ) K ` ⇤A $ ⇤B.

Lemma 3.7. [2, pp. 6-7]K ` ⇤(A1 ^ · · · Ân) $ (⇤A1 ^ · · · ^⇤An)

Proof. We proceed by induction on n. If n = 1 then we just have K ` ⇤A1 $ ⇤A1, which istrivially true. Now suppose n > 2 and assume the result is true for all m < n. Since

K ` (A1 ^ · · · Ân) $ (A1 ^ (A2 ^ · · · Ân)),

by Corollary 3.6 we have

K ` ⇤(A1 ^ · · · Ân) $ ⇤(A1 ^ (A2 ^ · · · Ân)).

Claim. K ` ⇤(A1 ^ (A2 ^ · · · Ân)) $ (⇤A1 ^⇤(A2 ^ · · · Ân))Proof of Claim.

K ` (A1 ^ (A2 ^ · · · Ân)) ! A1 and K ` (A1 ^ (A2 ^ · · · Ân)) ! (A2 ^ · · · Ân),

so by regularity we also have

K ` ⇤(A1 ^ (A2 ^ · · · Ân)) ! ⇤A1 and K ` ⇤(A1 ^ (A2 ^ · · · Ân)) ! ⇤(A2 ^ · · · Ân).

Thus, K ` ⇤(A1 ^ (A2 ^ · · · Ân)) ! (⇤A1 ^⇤(A2 ^ · · · Ân)).Conversely, we have

K ` A1 ! [(A2 ^ · · · Ân) ! (A1 ^ (A2 ^ · · · Ân))],

and so by regularity we get

K ` ⇤A1 ! ⇤[(A2 ^ · · · Ân) ! (A1 ^ (A2 ^ · · · Ân))].

But note

K ` ⇤[(A2 ^ · · · Ân) ! (A1 ^ (A2 ^ · · · Ân))] ! [⇤(A2 ^ · · · Ân) ! ⇤(A1 ^ (A2 ^ · · · Ân))]

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as it is just an instance of the K axiom. Thus,

K ` ⇤A1 ! [⇤(A2 ^ · · · Ân) ! ⇤(A1 ^ (A2 ^ · · · Ân))],

and hence K ` (⇤A1 ^⇤(A2 ^ · · · Ân)) ! ⇤(A1 ^ (A2 ^ · · · Ân)). End of Claim.

By the Claim, K ` ⇤(A1 ^ · · ·Ân) $ (⇤A1 ^⇤(A2 ^ · · ·Ân)). By the induction hypothesis,K ` ⇤(A2^ · · ·Ân) $ (⇤A2^ · · ·^⇤An) and so K ` ⇤(A1^ · · ·Ân) $ (⇤A1^⇤A2^ · · ·^⇤An)as required.

Corollary 3.8. [2, p. 7] If K ` A1 ^ · · · Ân ! B, then K ` ⇤A1 ^ · · · ^⇤An ! ⇤B.

Proof. By regularity, K ` A1 ^ · · · ^ An ! B ) K ` ⇤(A1 ^ · · · ^ An) ! ⇤B. But by Lemma 3.7above we know K ` ⇤(A1 ^ · · · Ân) $ (⇤A1 ^ · · · ^⇤An), so K ` ⇤A1 ^ · · · ^⇤An ! ⇤B.

Theorem 3.9. [2, p. 7] (Modal Substitution) Suppose S 2 FORML⇤ , p is a sentence letterappearing in S and A,B 2 SENTL⇤ . If K ` A $ B, then K ` S[A/p] $ S[B/p].

Proof. Suppose K ` A $ B. We proceed by induction on the form of S.If S ⌘ ? or S ⌘ q where q is some sentence letter not equal to p, then clearly

S[A/p] = S[B/p] = ? or q respectively. Thus, the results holds because K ` ? $ ? and K ` q $ qtrivially. However, if S ⌘ p then we have S[A/p] = A and S[B/p] = B, so S[A/p] $ S[B/p] isidentically A $ B and hence we have that K ` S[A/p] $ S[B/p].

Suppose S ⌘ Q ^R for some Q,R 2 FORML⇤ , and K ` Q[A/p] $ Q[B/p] andK ` R[A/p] $ R[B/p]. Then clearly K ` (Q[A/p] ^ R[A/p]) $ (Q[B/p] ^ R[B/p]), which is justK ` (Q ^R)[A/p] $ (Q ^R)[B/p] i.e. K ` S[A/p] $ S[B/p].

Now suppose S ⌘ ¬Q for some Q 2 FORML⇤ , and K ` Q[A/p] $ Q[B/p]. Then since(Q[A/p] $ Q[B/p]) $ (¬Q[A/p] $ ¬Q[B/p]) is a tautology, we have K ` ¬Q[A/p] $ ¬Q[B/p],that is K ` S[A/p] $ S[B/p].

Lastly, assume S ⌘ ⇤R for some R 2 FORML⇤ and K ` R[A/p] $ R[B/p]. Then by Corollary3.6, K ` ⇤(R[A/p]) $ ⇤(R[B/p]) ) K ` (⇤R)[A/p] $ (⇤R)[B/p], i.e. K ` S[A/p] $ S[B/p].

Proposition 3.10. [8, p. 74] The rules of inference modus ponens and necessitation are sound; thatis, they produce valid formulas from valid formulas.

Proof. Modus Ponens: Let M := hW,R,�i be an arbitrary model, and A,B 2 FORML⇤ arbitraryformulas such that A and A ! B are both valid in M. Now let w 2 W be arbitrary, then M, w � Aand M, w � A ! B since A,A ! B are both valid in M. M, w � A ! B ) either A is not trueat w or B is true at w, but since M, w � A we must also have M, w � B, i.e. B is also true at w.Therefore, B is also valid in M because w 2 W was an arbitrary world.

Necessitation: Let M := hW,R,�i be an arbitrary model, and A 2 FORML⇤ an arbitraryformula such that A is valid in M. Now let w, x 2 W be arbitrary worlds with wRx. Since A isvalid in M, we have M, x � A. Since x 2 W was arbitrarily chosen such that wRx, we deduceM, w � ⇤A. As w 2 W was also arbitrary, we have ⇤A is valid in M.

Recall from the end of §3.2 that axioms K , T , 4 , B were valid in all, reflexive, transitive,symmetric frames, respectively. Thus, using this and Proposition 3.10 above we have the followingsoundness theorems:

35


Theorem 3.11. (Axiomatic Soundness) For any formula A 2 FORML⇤ ,

(i) K ` A ) A is valid in all frames

(ii) T ` A ) A is valid in all reflexive frames

(iii) K4 ` A ) A is valid in all transitive frames

(iv) B ` A ) A is valid in all symmetric frames

(v) S4 ` A ) A is valid in all frames that are both reflexive and transitive.

Proof. We will show (v) as an example, the other cases are even easier because they involve onlyone frame property.

Suppose A 2 FORML⇤ and S4 ` A. We need to prove A is valid in a frame that is bothreflexive and transitive. To this end, suppose F := hW,Ri is a reflexive and transitive frame. LetM := hW,R,�i be an arbitrary model based on F . Recall that S4 is K but with all instances ofthe T and 4 axioms included in its axiom set and that all instances of the T axiom and 4 axiomare valid in reflexive and transitive frames, respectively. Moreover, all axioms of K are valid inall frames, so all axioms of S4 are valid in F . Therefore, A is valid in frame F because the rulesof inference of S4, namely modus ponens and necessitation, are sound. Since F was an arbitraryreflexive and transitive frame, we have that A is valid in all frames that are both reflexive andtransitive as required.

In light of the results above, we will sometimes abbreviate A is valid in all, reflexive, transitive,symmetric, reflexive and transitive frames, respectively, to A is K-, T-, K4-, B-, S4-valid.

3.4 Completeness for K,T,K4,B and S4

In this section we will prove the semantical completeness for K and S4 in full detail, leaving thedetails of the other proofs to the reader. For semantical completeness, we construct a model for ourmodal logic L 2 {K,T,K4,B,S4}, called the canonical model for L. Then using this model weshow that if a formula A 2 FORML⇤ is L-valid, then A has a proof in the axiomatic proof systemfor L, but we actually prove the contrapositive of this claim. Throughout this section L can be anyof K, T, K4, B, or S4.

Definition. A finite set {A1, . . . , An} of formulas is called L-consistent iff we do not haveL ` (A1^ · · ·^An) ! ?, or equivalently L ` ¬(A1^ · · ·^An). An infinite set � is called L-consistentjust in case every finite subset of � is L-consistent.

Definition. A set � of formulas is said to be maximally L-consistent iff � is L-consistent and if� ✓ �, with � L-consistent, then � = �.

Lemma 3.12. Assume � is a maximally L-consistent set of formulas, then for any A,B 2 FORML⇤

(i) � [ {A} is L-consistent ) A 2 �

(ii) L ` A ) A 2 �

(iii) A 2 � and A ! B 2 � ) B 2 �

36


(iv) either A 2 � or ¬A 2 �, but not both

(v) A ^B 2 � , A 2 � and B 2 �.

Proof. (i) Suppose � [ {A} is L-consistent, then have � ✓ � [ {A} where � [ {A} is L-consistent.Since � is maximally L-consistent, we must have � = � [ {A}, i.e. A 2 �.

(ii) L ` A ) � [ {A} is L-consistent ) A 2 � by (i).(iii) Suppose A,A ! B 2 �, then note that � [ {B} is L-consistent, and so B 2 � by (i).(iv) Suppose A /2 �. We will show that we must have ¬A 2 �.If � [ {A} were L-consistent then we would have A 2 � by (i) , so we must have that � [ {A}

is L-inconsistent. That is, for some C1, . . . , Cn 2 � ✓ FORML⇤ ,

L ` (C1 ^ · · · ^ Cn ^A) ! ? ) L ` (C1 ^ · · · ^ Cn) ! (A ! ?)) L ` (C1 ^ · · · ^ Cn) ! ¬A) [(C1 ^ · · · ^ Cn) ! ¬A] 2 �

by (ii). But C1, . . . , Cn 2 � ) (C1 ^ · · · ^ Cn) 2 �, so by (iii) we have ¬A 2 �. The proof of¬A /2 � ) A 2 � is similar. If both A and ¬A were in �, then we would have {A,¬A} ✓ � andsince L ` (A ^ ¬A) ! ? we would have that � were L-inconsistent, contrary to assumption. Thus,at most one of A or ¬A is a member of �.

(v) Suppose A ^ B 2 �. Note (A ^ B) ! A and (A ^ B) ! B are both tautologies, soL ` (A ^B) ! A and L ` (A ^B) ! B. Thus, by (ii) we have (A ^B) ! A, (A ^B) ! B 2 �, soby (iii) we have A 2 � and B 2 �.

Conversely, suppose A 2 � and B 2 �. Note that A ! (B ! A ^ B) is a tautology, soL ` A ! (B ! A ^ B) and so A ! (B ! A ^ B) 2 �. Thus, by (iii) B ! A ^ B 2 � and by (iii)again A ^B 2 �.

Proposition 3.13. [6] FORML⇤ is countable.

Proof. Define�n := {A 2 FORML⇤ | rank(A) 6 n, and i 6 n for all pi appearing in A} where

rank(A) =

8>>>><

>>>>:

0 if A atomicmax(rank(B), rank(C)) + 1 if A ⌘ B ^ C

rank(B) + 1 if A ⌘ ¬Brank(B) + 1 if A ⌘ ⇤B.

The key fact is that the sets �i are all finite, but we will be able to cover all forms of formulas withthis method. Then 8A 2 FORML⇤ , A 2 �R where R = max(rank(A),max{i | pi appears in A}).

Therefore, we can now list all formulas in a single list: A1, A2, . . ..

Proposition 3.14. [8, p. 76] (Lindenbaum’s Lemma) If � is L-consistent, then it can beextended to a maximally L-consistent set.

37


Proof. Suppose � is L-consistent. Let A1, A2, . . . be our countable list of modal formulas. Thendefine the following sets as follows:

�0 := �

�n+1 :=

(�n [ {An+1} if �n [ {An+1} is L-consistent�n else

�⇤ :=S�i

i2N

Claim. �⇤ is a maximally L-consistent extension of �.Proof of Claim. Clearly, �⇤ is a extension of � because � = �0 ✓

Si2N

�i = �⇤. Now suppose

�⇤ is not maximal, then there exists some � ) �⇤. That is, there exists some formula B 2 �\�⇤.Since B 2 FORML⇤ , B ⌘ Ak for some k 2 N. Then note that �k [ {Ak} is L-consistent as�k ✓ �⇤ ✓ � which is L-consistent and Ak ⌘ B 2 �. Then by construction of the �i, we have that�k+1 := �k [ {Ak} ✓ �⇤ so Ak 2 �⇤. Therefore, �⇤ is maximal. It remains to show that �⇤ isconsistent.

Assume for contradiction that �⇤ is not L-consistent. Then there exists from finite subset ⌃ ✓ �⇤

such that ⌃ is not L-consistent. Since ⌃ is finite, we can find the maximum k 2 N such that for allA 2 ⌃ we have A 2 �k. Thus, we have ⌃ ✓ �k and since ⌃ is not L-consistent we cannot have that�k is L-consistent. However, 8i 2 N �i is L-consistent by construction because � is L-consistent. Thus, �⇤ is L-consistent and hence �⇤ is a maximally L-consistent extension of �. End of Claim.

Therefore, any L-consistent set can be extended to a maximally L-consistent set.

Lemma 3.15. [8, p. 77] {¬⇤B,⇤A1,⇤A2, . . .} is L-consistent ) {¬B,A1, A2, . . .} is L-consistent .

Proof. Suppose {¬⇤B,⇤A1,⇤A2, . . .} is L-consistent and, for contradiction, that {¬B,A1, A2, . . .}is L-inconsistent. Then there exists some finite subset of {¬B,A1, A2, . . .} that is not L-consistent.We can assume this subset has the form {¬B,A1, . . . , Ak} for some k 2 N, because any extension ofan L-inconsistent set will be L-inconsistent as well. That is, L ` (¬B ^ A1 ^ · · · ^ Ak) ! ?. Thenwe can do the following all within the axiom system of L:

L ` (¬B Â1 ^ · · · Âk) ! ? AssumptionL ` (A1 ^ · · · Âk) ! (¬B ! ?) Propositional calculusL ` (A1 ^ · · · Âk) ! B ¬C $ (C ! ?)L ` ⇤(A1 ^ · · · Âk) ! ⇤B RegularityL ` (⇤A1 ^ · · · ^⇤Ak) ! ⇤B Lemma 3.7L ` ¬⇤B ! ¬(⇤A1 ^ · · · ^⇤Ak) ContrapositiveL ` ¬⇤B ! ((⇤A1 ^ · · · ^⇤Ak) ! ?) ¬C $ (C ! ?)L ` (¬⇤B ^⇤A1 ^ · · · ^⇤Ak) ! ? Propositional calculus.

Thus, {¬⇤B,⇤A1, . . .⇤Ak} is L-inconsistent, and hence {¬⇤B,⇤A1,⇤A2, . . .} has a finite L-inconsistent subset so is, by definition, L-inconsistent itself.

Therefore, {¬B,A1, A2, . . .} must have been L-consistent.

Definition. The canonical model for L is the structureM⇤

L

:= hWL

, RL

,�L

i where

38


(i) WL

:= {� | � is maximally L-consistent}

(ii) 8�,� 2 WL

, we have �RL

� , 8A 2 FORML⇤ , if ⇤A 2 � then A 2 �

(iii) for all propositional letters pi and for all � 2 WL

, we haveM

L

,� �L

pi , pi 2 �.

From the definition of the canonical model for L, it is easy to see that the canonical model isunique thus justifying our use of ‘the’ in the definition above: any canonical model of L will havethe set of all maximally L-consistent sets of formulas as its set of possible worlds W

L

, and thereis only one such set by extensionality so any two canonical models will in fact be equal, becauseR

L

and �L

are defined the same for all canonical models. Observe also that WL

6= ;, because ; isL-consistent and hence is contained in some maximally L-consistent set, ⌃ say, whence ⌃ 2 W

L

andW

L

6= ;.

Proposition 3.16. [8, pp. 78-9] (Truth Lemma) 8A 2 FORML⇤ and 8� 2 WL

, we have that

M⇤L

,� �L

A , A 2 �.

Proof. Recall that WL

is the set of all maximally L-consistent sets and is non-empty, so let � 2 WL

be an arbitrary maximally L-consistent set. We will prove this claim by induction on the form ofA 2 FORML⇤ .

If A ⌘ ? then note that since � is L-consistent, we have that ? /2 � and M⇤L

,� 1L

? , ? /2 �,i.e. M⇤

L

,� �L

? , ? 2 �, by definition of �L

.If A ⌘ pi for some i 2 N, then again by definition of �

L

, we have M⇤L

,� �L

pi , pi 2 �.Suppose A ⌘ B ^ C for some B,C 2 FORML⇤ . Then

M⇤L

,� �L

A , M⇤L

,� �L

B ^ C, M⇤

L

,� �L

B and M⇤L

,� �L

C (definition of truth in a model), B 2 � and C 2 � by the induction hypothesis, B ^ C 2 � by Lemma 3.12, A 2 �.

Now suppose A ⌘ ¬B for some B 2 FORML⇤ . Then

M⇤L

,� �L

A , M⇤L

,� �L

¬B, M⇤

L

,� 1L

B (definition of truth in a model), B /2 � by the induction hypothesis, ¬B 2 � by Lemma 3.12, A 2 �.

Finally, suppose A ⌘ ⇤B for some B 2 FORML⇤ . We will show the ()) direction by contrapos-itive, so suppose A /2 �. That is, ⇤B /2 � and since � is maximally L-consistent we have ¬⇤B 2 �.Set �0 := {⇤C | ⇤C 2 �} [ {¬⇤B} = {¬⇤B,⇤C1,⇤C2, . . .}. Since �0 ✓ � and � is L-consistent,we must have that �0 is L-consistent. Thus, by Lemma 3.15 we obtain that �1 := {¬B,C1, C2, . . .}is also L-consistent. So by Lindenbaum’s Lemma (Proposition 3.14), we have that �1 is contained insome maximally L-consistent set, � say. Since � is maximally L-consistent, we know that � 2 W

L

.¬B 2 �1 ✓ � ) ¬B 2 �, and since � is maximally L-consistent we know B /2 � from Lemma 3.12.

39


So by the induction hypothesis, M⇤L

,� 1L

B. Note that whenever ⇤Ci 2 �, we have Ci 2 �1 ✓ �,i.e. Ci 2 �, so �R

L

� holds. Therefore, there exists a world � 2 WL

with �RL

� but such thatM⇤

L

,� 1L

B. Hence, by definition of �L

we have M⇤L

,� 1L

⇤B.Conversely, suppose ⇤B 2 �. Let � 2 W

L

be an arbitrary world such that �RL

�. Thus, fromthe definition of R

L

we know B 2 �. Then, by the induction hypothesis, M⇤L

,� �L

B, and since� was arbitrary we have that M⇤

L

,� �L

⇤B, i.e. M⇤L

,� �L

A, from the definition of �L

.

Proposition 3.17. The canonical model M⇤S4

:= hWS4

, RS4

,�S4

i is both reflexive and transitive.

Proof. Let � 2 WL

be arbitrary. We need to show RL

is both reflexive and transitive. Since � is amaximally S4-consistent set, and S4 ` ⇤A ! A and S4 ` ⇤A ! ⇤⇤A for any A 2 FORML⇤ , wehave that ⇤A ! A, ⇤A ! ⇤⇤A 2 � for any A 2 FORML⇤ by Lemma 3.12. Assume ⇤A 2 � forsome A 2 FORML⇤ . We want to show that A 2 � because then we will have �RL� by definitionof R

L

. Since ⇤A,⇤A ! A 2 �, by Lemma 3.12 we know that A 2 � so �RL

� and RL

is reflexiveas � and A were arbitrary.

Now suppose �,⌃ 2 WL

such that �RL

� and �RL

⌃. We want to show �RL

⌃ and that willgive us that R

L

is transitive. Suppose ⇤A 2 � for some A 2 FORML⇤ , then it suffices to showA 2 ⌃ because then by definition of R

L

we will have �RL

⌃. Since ⇤A,⇤A ! ⇤⇤A 2 �, by Lemma3.12 we know ⇤⇤A 2 �. Thus, ⇤A 2 � because �R

L

�. But then A 2 ⌃ because �RL

⌃ and weare done.

Theorem 3.18. (Completeness for S4) If A is S4-valid, then S4 ` A.

Proof. We prove the contraposition: S4 0 A ) A is not S4-valid. So, suppose S4 0 A. As shownin Proposition 3.17, the canonical model M⇤

S4

of S4 is both reflexive and transitive. In order toshow A is not S4-valid, it suffices to show that A is not valid in some reflexive and transitive model,namely M⇤

S4

, and hence we will have proved completeness for S4.Consider {¬A}. If {¬A} were not S4-consistent, then we would have S4 ` ¬A ! ?, i.e. S4 ` A,

but this is contrary to our assumption. So {¬A} is S4-consistent, and thus, by Lindenbaum’s Lemma,can be extended to a maximally S4-consistent set, � say. Since � is maximally S4-consistent, weknow � 2 W

L

by how we defined WL

. ¬A 2 {¬A} ✓ � so ¬A 2 �, which implies A /2 � by Lemma3.12. So by the Truth Lemma, M⇤

S4

,� 1S4

A and hence A is not valid in M⇤S4

. Therefore, A is notS4-valid because there exists a model in which A is not valid, and this proves the contraposition.

Theorem 3.19. (Completeness for K, T, K4 and B)

(i) If A is K-valid, then K ` A.

(ii) If A is T-valid, then T ` A.

(iii) If A is K4-valid, then K4 ` A.

(iv) If A is B-valid, then B ` A.

We omit the proof for Theorem 3.19, but the proofs for K, T, K4 are very similar to, andeasier than, the proof for S4: for K we do not need to show the canonical model of K satisfiesany other properties, then we show the contrapositive in precisely the same way as in the proof ofTheorem 3.18 but with M⇤

K

replacing M⇤S4

; for T and K4 we simply show the canonical models

40


M⇤T

and M⇤K4

are reflexive and transitive, respectively, and then again the proof is just like thatfor S4; and for B we need to show M⇤

B

is symmetric, and we do this in a similar way to showingM⇤

S4

is reflexive (or transitive), then follow the proof of Theorem 3.18 again.

3.5 GL

Now that we have explored the normal modal logics K, T, K4, B and S4, we shall switch ourattention to GL. This modal logic deserves some special focus, and we shall now be entirely fixatedon it. We need to lay down the foundation for Solovay’s Arithmetical Completeness of GL, and inorder to do so we need to traverse the same path we did for the other logics. There is little worthin continuing to discuss GL if we cannot show that it itself is semantically sound and complete.

Definition. GL is the normal modal logic system K but with all instances of G: ⇤(⇤A ! A) ! ⇤Aincluded in its axiom set.

We now make two very important observations concerning GL:

Theorem 3.20. [2, p. 11] For all A 2 FORML⇤ ,

(i) GL 0 ⇤A ! A

(ii) GL ` ⇤A ! ⇤⇤A.

Proof. (i) Suppose GL ` ⇤A ! A for all A 2 FORML⇤ , then in particular GL ` ⇤? ! ?. Thenby necessitation GL ` ⇤(⇤? ! ?). Since GL ` ⇤(⇤A ! A) ! ⇤A for any modal formula, wehave that GL ` ⇤(⇤? ! ?) ! ⇤?. Thus, by modus ponens GL ! ⇤?, and again by modusponens GL ` ?. with the consistency of GL. Therefore, GL 0 ⇤A ! A.

(ii) We give an axiomatic proof:GL ` A ! (⇤⇤A ^⇤A ! ⇤A Â) TautologyGL ` A ! (⇤(⇤A Â) ! ⇤A Â) Lemma 3.7GL ` ⇤A ! ⇤(⇤(⇤A Â) ! ⇤A Â) RegularityGL ` ⇤(⇤(⇤A Â) ! ⇤A Â) ! ⇤(⇤A Â) G axiomGL ` ⇤A ! ⇤(⇤A Â) Propositional calculusGL ` ⇤A ! (⇤⇤A ^⇤A) Lemma 3.7GL ` ⇤A ! ⇤⇤A Propositional calculus.

Corollary 3.21. There is no consistent normal modal logic system that is an extension of both T

and GL.

Proof. T ` ⇤A ! A but GL 0 ⇤A ! A by Theorem 3.20.

Corollary 3.22. GL ◆ K4

Proof. The theorems of K4 are provable from the axioms of K and all instances of ⇤A ! ⇤⇤A,but GL can prove all these axioms and formulas. Therefore, any theorem of K4 is a theorem ofGL, and GL is an extension of K4.

Proposition 3.23. [2, p. 12] GL ` ⇤(⇤A ! A) $ ⇤A $ ⇤(⇤A Â)

41


Proof. GL ` ⇤(⇤A ! A) ! ⇤A because ⇤(⇤A ! A) ! ⇤A is just an axiom form of GL.GL ` A ! (⇤A ! A) TautologyGL ` ⇤A ! ⇤(⇤A ! A) RegularityThus, GL ` ⇤(⇤A ! A) $ ⇤A.Now we will show GL ` ⇤A $ ⇤(⇤A Â):GL ` ⇤A ! ⇤⇤A Theorem 3.20GL ` ⇤(⇤A Â) $ ⇤⇤A ^⇤A Lemma 3.7GL ` ⇤A ! (⇤⇤A ! (⇤⇤A ^⇤A)) TautologyGL ` ⇤A ! (⇤⇤A ^⇤A) Propositional calculusGL ` ⇤A ! ⇤(⇤A Â) Propositional calculusGL ` ⇤⇤A ^⇤A ! ⇤A TautologyGL ` ⇤(⇤A Â) ! ⇤A Propositional calculus.Thus, GL ` ⇤A $ ⇤(⇤A Â).

Lemma 3.24. [2, p. 12] GL ` ⇤? $ ⇤⌃p

Proof. ()) GL ` ? ! ⌃p TautologyGL ` ⇤? ! ⇤⌃p Regularity.(() GL ` ⌃p ! ⌃> TautologyRecall: ⌃ ⌘ ¬⇤¬, soGL ` ⌃> ! ¬⇤¬>> ⌘ ¬? and ¬A $ (A ! ?), soGL ` ⌃> ! (⇤? ! ?)GL ` ⌃p ! (⇤? ! ?) Propositional calculusGL ` ⇤⌃p ! ⇤(⇤? ! ?) RegularityGL ` ⇤(⇤? ! ?) ! ⇤? G axiomGL ` ⇤⌃p ! ⇤? Propositional calculus.Hence, GL ` ⇤? $ ⇤⌃p.

These are just some elementary facts that we can prove about GL, but the next result is veryinteresting. Once we have shown that the theorems of GL are theorems of PA, the next followingstates it is provable in PA that for a sentence ': [' , ' is unprovable] , [' , PA is consistent ][2, p. 14]. This is telling us precisely that PA cannot prove its own consistency. This is thefascinating aspect to mixing formal arithmetic with modal logic; from what seems like two disjointareas of mathematics, we can use one to prove a very powerful theorem about the other.

Theorem 3.25. [2, pp. 13-14] GL ` ⇤(p $ ¬⇤p) $ ⇤(p $ ¬⇤?)

Proof. ())

(1) GL ` ⇤(p $ ¬⇤p) ! ⇤(p ! ¬⇤p) Tautology

(2) GL ` ⇤(p ! ¬⇤p) ! ⇤⇤(p ! ¬⇤p) 4 axiom

(3) GL ` ⇤(p $ ¬⇤p) ! ⇤⇤(p ! ¬⇤p) Propositional calculus (1), (2)

(4) GL ` ⇤⇤(p ! ¬⇤p) ! ⇤(⇤p ! ⇤¬⇤p) K axiom

42


(5) GL ` ⇤(p $ ¬⇤p) ! ⇤(⇤p ! ⇤¬⇤p) Propositional calculus (3), (4)

(6) GL ` ⇤p ! ⇤⇤p 4 axiom

(7) GL ` (⇤⇤p ^⇤¬⇤p) $ ⇤(⇤p ^ ¬⇤p) Lemma 3.7

(8) GL ` (⇤p ^ ¬⇤p) $ ? Tautology

(9) GL ` ⇤(⇤p ^ ¬⇤p) $ ⇤? Regularity

(10) GL ` (⇤⇤p ^⇤¬⇤p) $ ⇤? Propositional calculus (7), (9)

(11) GL ` ⇤(p $ ¬⇤p) ! ⇤(⇤p ! ⇤?) Propositional calculus (5), (6), (10)

(12) GL ` ? ! p Tautology

(13) GL ` ⇤? ! ⇤p Regularity

(14) GL ` ⇤(⇤? ! ⇤p) Necessitation

(15) GL ` ⇤(p $ ¬⇤p) ! ⇤(⇤p $ ⇤?) Propositional calculus (11), (14)

(16) GL ` (⇤p $ ⇤?) $ (¬⇤p $ ¬⇤?) Tautology

(17) GL ` ⇤(p $ ¬⇤p) ! ⇤(¬⇤p $ ¬⇤?) Propositional calculus (15), (16)

(18) GL ` (⇤(p $ ¬⇤p) ^⇤(¬⇤p $ ¬⇤?)) ! ⇤(p $ ¬⇤?) Tautology

(19) GL ` ⇤(p $ ¬⇤p) ! ⇤(p $ ¬⇤?) Propositional calculus (17), (18).

(()

(1) GL ` ⇤(p $ ¬⇤?) ! (⇤p $ ⇤¬⇤?) Regularity

(2) GL ` (⇤p $ ⇤¬⇤?) $ (¬⇤p $ ¬⇤¬⇤?) Tautology

(3) GL ` ⇤(p $ ¬⇤?) ! (¬⇤p $ ¬⇤¬⇤?) Propositional calculus (1), (2)

(4) GL ` ⇤? $ ⇤¬⇤? Lemma 3.24 (¬? in for p)

(5) GL ` (⇤? $ ⇤¬⇤?) $ (¬⇤? $ ¬⇤¬⇤?) Tautology

(6) GL ` ¬⇤? $ ¬⇤¬⇤? Modus ponens (4), (5)

(7) GL ` (¬⇤p $ ¬⇤¬⇤?) $ (¬⇤? $ ¬⇤p) Deduction from (6)

(8) GL ` ⇤(p $ ¬⇤?) ! (¬⇤? $ ¬⇤p) Propositional calculus (3), (7)

(9) GL ` (¬⇤? $ ¬⇤p) ! ⇤(¬⇤? $ ¬⇤p) Necessitation

(10) GL ` ⇤(p $ ¬⇤?) ! ⇤(¬⇤? $ ¬⇤p) Propositional calculus (8), (9)

(11) GL ` (⇤(p $ ¬⇤?) ^⇤(¬⇤? $ ¬⇤p)) ! ⇤(p $ ¬⇤p) Tautology

43


(12) GL ` ⇤(p $ ¬⇤?) ! ⇤(p $ ¬⇤p) Deduction from (10), (11)

(13) GL ` ⇤(p $ ¬⇤p) ! ⇤⇤(p $ ¬⇤p) 4 axiom

(14) GL ` ⇤(p $ ¬⇤?) ! ⇤⇤(p $ ¬⇤p) Propositional calculus (12), (13)

(15) GL ` ⇤⇤(p $ ¬⇤p) ! ⇤(⇤p $ ⇤¬⇤p) K axiom twice

(16) GL ` ⇤(p $ ¬⇤?) ! ⇤(⇤p $ ⇤¬⇤p) Propositional calculus (14), (15)

(17) GL ` ⇤? $ ⇤¬⇤p Lemma 3.24 (¬p in for p)

(18) GL ` ⇤(p $ ¬⇤?) ! ⇤(⇤p $ ⇤?) Propositional calculus (16), (17)

(19) GL ` (⇤p $ ⇤?) $ (¬⇤p $ ¬⇤?) Tautology

(20) GL ` ⇤(p $ ¬⇤?) ! ⇤(¬⇤p $ ¬⇤?) Propositional calculus (18), (19)

(21) GL ` (⇤(p $ ¬⇤?) ^⇤(¬⇤p $ ¬⇤?)) ! ⇤(p $ ¬⇤p) Tautology

(22) GL ` ⇤(p $ ¬⇤?) ! ⇤(p $ ¬⇤p) Deduction from (20), (21).

Hence, GL ` ⇤(p $ ¬⇤p) $ ⇤(p $ ¬⇤?).

We turn to the soundness of GL: we need to work out in which kinds of frames the theoremsof GL are valid. As we saw in §3.3, if A 2 FORML⇤ is a theorem of K4 then A is valid inall transitive frames (Theorem 3.11). But we have additional axioms in GL which might give usadditional theorems, so we need to figure out how that affects in which frames the theorems of GL

are valid. However, one can ask if K4 ` ⇤(⇤A ! A) ! ⇤A for all A 2 FORML⇤ , i.e. is the Gaxiom K4-valid. If this was the case then we would have that K4 and GL have precisely the sametheorems. Alas, the answer to the question is a definitive no, because we are able to construct acounter-model that is transitive but in which ⇤(⇤A ! A) ! ⇤A fails (see Figure 2 below).

1 � ⇤(⇤A ! A)

2 � ¬A,⇤A ! A

3 � ¬A,⇤A ! A

Figure 2: Counter-model where iRj iff i < j

Definition. Let F := hW,Ri be a frame. Then we call F wellfounded iff 8X ✓ W , if X 6= ; thenthere exists w 2 X such that for no x 2 X do we have xRw. w is seen to be an R-least element [2,p75], no world can access w.

We call F converse wellfounded iff 8X ✓ W , if X 6= ; then there exists w 2 X such that for nox 2 X do we have wRx. w is seen to be an R-greatest element [2, p75], w cannot access any otherworld.

Lemma 3.26. If a frame F := hW,Ri is converse wellfounded, then F is also irreflexive.

44


Proof. Suppose F is converse wellfounded. If W = ; then we have many vacuous truths, including8w 2 W ¬(wRw) and so F is irreflexive. Therefore, suppose W 6= ;. Now assume for contradictionthat F is reflexive. Then for an arbitrary w 2 W 6= ; we have wRw. Then note X := {w} is anon-empty set. However, 8x 2 X wRx, i.e. w can get to every world in X, a non-empty set. Thiscontradicts our assumption that F is converse wellfounded. So F must have been irreflexive.

Theorem 3.27. [2, p. 75] (Induction on the Converse of R) Let P be a property of worlds.Let F := hW,Ri be an arbitrary converse wellfounded frame. Then to show that every world in Whas property P, it suffices to prove that for each w 2 W : if x has P for every x 2 W with wRx,then w has P.

Proof. Suppose F := hW,Ri is an arbitrary converse wellfounded frame. Assume that 8w 2 W whas property P whenever P holds for every x 2 W with wRx. Set

X := {w 2 W | w does not have P}.

Since F is converse wellfounded, if we can show that X has no R-greatest element, then we willhave that X = ; and can deduce that 8w 2 W w has P.

Assume for contradiction that X 6= ;, and let x0 2 X be an arbitrary world. Since x0 2 X, weknow that P does not hold of x0 and therefore there exists some world x1 2 W such that x0Rx1 andthat x1 does not have P either. Thus, x1 2 X. By similar reasoning there must exist x2 2 X withx1Rx2, etc. So we will end up with a chain x0Rx1Rx2R . . .. Every world in X can access anotherworld, so X has no R-greatest element. with the converse wellfoundedness of F , and hence X = ;so property P holds for every world in W .

Theorem 3.28. [2, pp. 75-6] (Axiomatic Soundness of GL) For any A 2 FORML⇤ , if GL ` Athen A is valid in all frames that are both transitive and converse wellfounded.

Proof. Suppose A 2 FORML⇤ and GL ` A. Now let F := hW,Ri be an arbitrary transitive andconverse wellfounded frame, and suppose M := hW,R,�i is an arbitrary model based on F . Weneed to show A is valid in M. Recall that GL is K but with all instances of the G axiom. Since wehave already shown that K ` A ) A is valid in all frames, and hence in all transitive and conversewellfounded frames, we need only show that the G axiom is valid in M. However, we shall showthat the logically equivalent contrapositive ¬⇤B ! ¬⇤(⇤B ! B) is valid in M.

Let w 2 W be arbitrary world and suppose M, w � ¬⇤B, that is M, w 1 ⇤B. SetX := {x 2 W | wRx and M, x 1 B}. M, w 1 ⇤B implies that there exists y 2 W such thatwRy and M, y 1 B. Thus, y 2 X and so X 6= ;. Since F is converse wellfounded, M is amodel based on F and X is non-empty, we must have that there exists z 2 X such that for nox 2 X do we have zRx. z 2 X ) wRz and M, z 1 B. Note that any arbitrary world a 2 Wsuch that zRa, we must have a /2 X by the maximality of z. But since wRz and zRa, we knowwRa because F is transitive. Thus, M, a � B as otherwise a 2 X which we know is impossible.Since a 2 W with zRa was arbitrary and M, a � B, we get that M, z � ⇤B. As M, z 1 B,we can logically deduce M, z 1 ⇤B ! B. Hence, M, w 1 ⇤(⇤B ! B) because wRz, which isM, w � ¬⇤(⇤B ! B). Therefore, M, w � ¬⇤B ! ¬⇤(⇤B ! B) and since w 2 W was arbitrarywe have that ¬⇤B ! ¬⇤(⇤B ! B) is valid in M, which gives that ¬⇤B ! ¬⇤(⇤B ! B) is validin F . Hence, ⇤(⇤B ! B) ! ⇤B is valid in F .

45


Proposition 3.29. [2, p. 76] Let F := hW,Ri be a transitive frame. Suppose in addition that F isfinite; that is, suppose W is finite. Then

F is converse wellfounded , F is irreflexive.

Proof. ()) Lemma 3.26 told us that any frame which is converse wellfounded is also irreflexive, andthis certainly holds for a finite, transitive, converse wellfounded frame.

(() Suppose F is a finite, transitive and irreflexive frame. Assume for contradiction that F isnot converse wellfounded. Then there exists a non-empty subset X ✓ W such that for every x 2 Xthere exists some wx 2 X such that xRwx, i.e. there is no R-greatest element of X.

Let x1Rx2R · · ·Rxn be an accessibility chain of worlds in W . Assume for some 1 6 i � j 6 nthat xi = xj . Since F is transitive, we will have xiRxj from xiRxi+1R · · ·Rxj�1Rxj . But thenxiRxi, which contradicts the irreflexivity of F . Thus, whenever 1 6 i � j 6 n we know xi 6= xj .

We will show by induction that for each n 2 N there exists an accessibility chain x1Rx2R · · ·Rxnwith xi 2 X 81 6 i 6 n and all distinct. X 6= ;, so let x1 2 X be arbitrary. Since x1 2 X we know9x2 2 X such that x1Rx2 and x1 6= x2. So our claim holds for n = 1. Now suppose, as the inductionhypothesis, that there is a chain of length k. That is, we have x1Rx2R · · ·Rxk where each xi is inX and all are distinct. Note that we cannot have xkRxi for any i 6 k by our comment above. If@y 2 X such that xkRy then X has an R-greatest element, which contradicts our assumption onX. Therefore, 9xk+1 2 X such that xkRxk+1 and xk+1 6= xi 8i 6 k and hence we get the chainx1Rx2R · · ·RxkRxk+1 of distinct worlds in X.

Thus, for each n 2 N X has at least n distinct elements. Since X ✓ W , for each n 2 N W hasat least n elements. Hence, W is infinite. with the finiteness of W . Hence, there cannot exist anysuch set X giving us that F is in fact converse wellfounded.

Corollary 3.30. For any A 2 FORMLA , if GL ` A then A is valid in all finite, transitive andirreflexive frames.

Proof. Follows immediately from the axiomatic soundness of GL and Proposition 3.29 above.

A little more work for the semantical completeness of GL has to be done compared to that forthe other normal modal logics. Recall that for the semantical completeness for S4, we constructedthe canonical model for S4 and showed that it was reflexive and transitive. It happens that thecanonical model M⇤

GL

:= hWGL

, RGL

,�GL

i for GL (defined in the same way as for the other modallogics) is transitive, but it is not converse wellfounded [2, p. 90]. However, to prove this we needthe Arithmetical Soundness of GL, and we leave that for §4. Therefore, we cannot give the sameformat of proof for the completeness of GL as we did for S4.

Our proof will be, in a sense, a proof schema: we prove that for any sentence A for which GL 0 A,there exists a finite, transitive, converse wellfounded model in which A is not valid. We remark herethat the model, and in particular its size, will depend on A, but it will always be finite. In fact, wecan find a bound on its size dependent on A.

Lemma 3.31. There exists a modal sentence which GL cannot prove.

Proof. If GL could prove every modal sentence, then in particular GL ` p ^ ¬p for any sentenceletter and so GL ` ?. with the consistency of GL. Therefore, there must exist some sentencewhich is not derivable in GL.

46


Remark. By Lemma 3.31, we can pick D 2 SENTL⇤ such that GL 0 D. We will base ourcompleteness of GL on this candidate sentence, and as a result need to refer back to it for theremainder of this section. Moreover, our next definition is dependent on this D.

Definition. If A is a subsentence of D then we will call A or its negation ¬A a D-subformula13.

Definition. We call a set � of D-subformulas GL-consistent iff we do not haveGL ` (A1 ^ · · · Ân) ! ? for any A1, . . . , An 2 �.

Note that this definition is identical to the definition for L-consistent as given in §3.4 for anarbitrary normal modal logic system L, but it is only defined for D-subformulas. It is only D-subformulas that we shall be concerned with until the end of this section. D 2 SENTL⇤ and sois built up from a finite number of other modal sentences, namely all the subsentences of D. Sincethere are only a finite number of them, we can let the list be D1, . . . , Dk. Therefore, the complete listof D-subformulas is D1,¬D1, . . . , Dk,¬Dk and we shall define DFORMS := {D1,¬D1, . . . , Dk,¬Dk}to be the finite set of all D-subformulas.

Definition. [2, p. 79] Let � be a set of D-subformulas, then we call � maximally GL-consistent(with respect to D) iff � is GL-consistent and for every D-subformula A: either A 2 � or ¬A 2 �.

Since we shall only be needing this definition for the rest of this section, we will abbreviateto maximally consistent. The proceeding result is similar to Lemma 3.12, and demonstrates thesimilarity of our previous definition of maximally consistent with the new one.

Lemma 3.32. Let � be a maximally consistent set of D-subformulas. Then for any D-subformulasA,B

(i) either A 2 � or ¬A 2 �, but not both

(ii) for any A1, . . . , An 2 �,GL ` (A1, . . . , An) ! B ) B 2 �

(iii) A ^B 2 � , A 2 � and B 2 �

(iv) A,A ! B 2 � ) B 2 �

(v) � [ {A} is GL-consistent ) A 2 �

(vi) GL ` A ) A 2 �.

Proof. (i) [2, p. 79] � is maximally consistent and A is a subsentence of D so either A 2 � or¬A 2 �. Now suppose both A and ¬A are in �. Note GL ` (A^¬A) ! ?. Therefore, by definitionof GL-consistent, � is GL-inconsistent. Hence, we cannot have both A and ¬A in �.

(ii) [2, p. 79] If B /2 � then by (i), ¬B 2 � and hence {A1, . . . , An,¬B} ✓ �. But noteGL ` (A1 ^ · · · Ân ^ ¬B) ! ? because GL ` A1 ^ · · · Ân ! B. Thus, � is GL-inconsistent. Hence, B 2 �.

(iii) ()) Suppose A^B 2 � and for contradiction that A /2 �. Since � is maximally consistent,we have from (i) that ¬A 2 �. So {A ^B,¬A} ✓ �, but note GL ` (A ^B) ^ ¬A ! ?. Thus, � is

13This is what Boolos refers to and defines as a ‘formula’ [2, pp. 78].

47


GL-inconsistent, contradicting our assumption that � is maximally consistent. Thus, we must haveA 2 �. An identical argument with B replacing A shows B 2 �.

(() Suppose A,B 2 � and that A ^B /2 �. Then ¬(A ^B) 2 �, and so {A,B,¬(A ^B)} ✓ �.However, GL ` (A ^B) ^ ¬(A ^B) ! ? so � is GL-inconsistent. Thus, A ^B 2 �.

Hence, A ^B 2 � , A,B 2 �.(iv) Suppose A,A ! B 2 � and for contradiction that B /2 �. Then ¬B 2 � by (i).

GL ` (A ! B) $ (¬A _B),GL ` (¬A _B) ^ ¬B ! ¬A and GL ` (¬A _B) ^ ¬B Â ! ?,

so GL ` (A ! B)^¬BÂ ! ?. But since {A ! B,¬B,A} ✓ � we have that � is GL-inconsistent. Thus, B 2 �.

(v) Suppose �[{A} is GL-consistent, then we have � ✓ �[{A} where �[{A} is GL-consistent.If A /2 � then by (i) we know ¬A 2 �. Thus, A,¬A 2 � [ {A} and so by (iii) A ^ ¬A 2 � [ {A}.But GL ` A ^ ¬A ! ?, so � [ {A} is GL-inconsistent. Hence, we must have had A 2 �.

(vi) L ` A ) � [ {A} is GL-consistent ) A 2 � by (v).

Proposition 3.33. [2, p. 79] Every GL-consistent set is contained in some maximally consistentset.

Proof. Let � be a GL-consistent set of D-subformulas; that is, � ✓ DFORMS and so is a finite set.Thus, suppose � = {A1, . . . An} where Ai 2 DFORMS for each i 6 n.

Claim. For any sentences p, q we have the following: p $ [(p ^ q) _ (p ^ ¬q)].Proof of Claim. We have the tautologies q_¬q and p ! p, so we get p ! [p^ (q_¬q)] whereby

we get p ! [(p^ q)_ (p^¬q)]. Clearly, [(p^ q)_ (p^¬q)] ! p because in either disjunct we triviallyobtain p. End of Claim.

We show by induction on n that (A1^· · ·Ân) is logically equivalent to a disjunction E1_· · ·_Em,where each disjunct is a conjunction of D-subformulas such that for each i 6 k either Di or ¬Di

appears and all elements of � appear. Firstly, suppose n = 1, from the Claim we knowA1 $ (A1 ^D1) _ (A1 ^ ¬D1). In turn,

(A1 ^D1) _ (A1 ^ ¬D1) $ [(A1 ^D1 ^D2) _ (A1 ^D1 ^ ¬D2)_(A1 ^ ¬D1 ^D2) _ (A1 ^ ¬D1 ^ ¬D2)].

Since there are only finitely many Di, we can continue in this way and eventually stop, resulting in adisjunction as desired. Now suppose that the result holds for any conjunct of length < l = n. Thenby the induction hypothesis we know, A1^ · · ·Âl�1Âl is equivalent to (E1_ · · ·_Em)Ân, suchthat in each Ei either Dj or ¬Dj (for each j 6 k) appears and all of A1, . . . , Al�1 appear. Thentruth-functionally we have

(E1 _ · · · _ Em) Ân $ [(E1 Ân) _ · · · _ (Em Ân)].

Notice that in each disjunct either Dj or ¬Dj (for each j 6 k) appears, because either Dj or ¬Dj

appears in Ei, and all elements of � appear because A1, . . . , Al�1 appeared by induction and nowAl appears. Thus, by induction this happens for any n. Thus, A1 ^ · · · ^ An is logically equivalentto E1 _ · · · _ Em for some m 2 N.

48


If GL ` ¬Ei for each i 6 m, then we would have GL ` ¬E1 ^ · · · ^ ¬Em, whenceGL ` ¬(E1 _ · · · _ Em), i.e. GL ` (A1 ^ · · · ^ An) ! ?. That is, � is GL-inconsistent. Hence,for some l 6 m GL ` El. Then the set of all conjuncts of El will be a maximally consistent setcontaining all members of �.

Recall that we chose D with GL 0 D. If {¬D} were GL-inconsistent then GL ` ¬D ! ?, i.e.GL ` ¬¬D, i.e. GL ` D, which is contrary to our choice of D. Thus, {¬D} is a GL-consistentset and is hence contained in some maximally consistent set of D-subformulas. Suppose �¬D is amaximally consistent set which has {¬D} as a subset.

Now define WD := {� ✓ DFORMS | � is maximally consistent}. Since �¬D is maximally consistentset of D-subformulas, we have �¬D 2 WD and so WD 6= ;. Since we supposed D has k subsentencesand no GL-consistent will contain both a D-subformula and its negation, we know combinatoriallythat there are at most 2k consistent sets of D-subformulas. Hence, there are at most 2k maximallyconsistent sets of D-subformulas and thus |WD| 6 2k, i.e. WD is finite. We now have our set ofworlds for the model we will construct and in which we will show D is invalid.

Definition. Let MD := hWD, RD,�Di be the structure such that

(i) WD is as above

(ii) 8�,� 2 WD, we have �RD� , (a) 8⇤A 2 DFORMS, if ⇤A 2 � then A,⇤A 2 �; and(b) there exists ⇤B 2 � such that ¬⇤B 2 �

(iii) for all propositional letters pi and for all � 2 WD, we haveMD,� �D pi , pi appears in D and pi 2 �.

Hence, by our calculations above MD is a model of size at most 2k, and is thus a finite model.We still need to show that RD is transitive and converse wellfounded to be able to use MD in ourproof of completeness for GL.

Proposition 3.34. [2, p. 82]

(i) RD is a transitive relation.

(ii) RD is an irreflexive relation.

Proof. (i) Let �,�,⌃ 2 WD be arbitrary with �RD� and �RD⌃. We need to show �RD⌃. Supposewe have a D-subformula ⇤A 2 �. Since �RD�, we have A,⇤A 2 � and since �RD⌃ we must haveA,⇤A 2 ⌃. It remains to show that there is a formula ⇤B 2 ⌃ such that ¬⇤B 2 �.

Since �RD�, we know there exists ⇤B 2 � such that ¬⇤B 2 �. Note that ⇤B 2 ⌃ because�RD⌃, so ⇤B works and �RD⌃. Thus, RD is transitive.

(ii) Suppose there exists maximally consistent set � such that �RD�. Then, by definition ofRD, we know there exists ⇤A 2 � such that ¬⇤A 2 �. By Lemma 3.32, ⇤A ^ ¬⇤A 2 � butGL ` (⇤A^¬⇤A) ! ?, i.e. � is GL-inconsistent. Hence, for all � 2 WD we do not have �RD�,which is precisely saying that RD is irreflexive.

Proposition 3.34 establishes that hWD, RDi is a finite, transitive and irreflexive frame. Therefore,hWD, RDi is also converse wellfounded by Proposition 3.29. Thus, all that is left to show is that Dis not valid in the frame hWD, RDi and for this we need the following two results.

49


Lemma 3.35. [2, p. 82] RD is defined in such a way that for any subsentence ⇤A of D and forevery � 2 WD, we have

⇤A 2 � , 8� 2 WD, �RD� ) A 2 �.

Proof. Let � 2 WD be an arbitrary maximally consistent set and ⇤A an arbitrary subsentence ofD.

()) Suppose ⇤A 2 � and take an arbitrary � 2 WD such that �RD�, then in particular weknow A 2 � from the definition of RD. Since � was arbitrarily chosen, this holds for any such� 2 WD and we are done.

(() We show this direction by contraposition. Thus, suppose ⇤A /2 �. We need to show that9� 2 WD with �RD� but such that A /2 �.

Define ⌃ := {¬A,⇤A} [ {B,⇤B | ⇤B 2 �} 6= ;.

Claim. ⌃ is GL-consistent.Proof of Claim. Assume for contradiction that ⌃ is GL-inconsistent. Then for some finite subset

X ✓ ⌃, we have GL `VX ! ? where

VX is the conjunction of all elements of X. Since X is a

finite subset of ⌃ = {¬A,⇤A} [ {B,⇤B | ⇤B 2 �}, X will contain, say, l subsentences B1, . . . , Bl

and m subsentences ⇤Bl+1, . . . ,⇤Bl+m of D. Then we can extend X to the set

Y := {¬A,⇤A,B1, . . . , Bl+m,⇤B1, . . . ,⇤Bl+m}

because Y will still be GL-inconsistent as it contains a finite GL-inconsistent set, namely X. Letn := l +m, then from the GL-inconsistency of Y :

GL ` (¬A ^⇤A ^B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) ! ? AssumptionGL ` ¬(¬A ^⇤A ^B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) Logical equivalenceGL ` (⇤A ^B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) ! A Propositional calculusGL ` (B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) ! (⇤A ! A) Propositional calculusGL ` ⇤(B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) ! ⇤(⇤A ! A) RegularityGL ` ⇤(B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) $ (⇤B1 ^⇤⇤B1 ^ · · · ^⇤Bn ^⇤⇤Bn) Lemma 3.7GL ` ⇤Bi ! ⇤⇤Bi for each i 6 n 4 axiomGL ` ⇤(B1 ^⇤B1 ^ · · · ^Bn ^⇤Bn) $ (⇤B1 ^ · · · ^⇤Bn) Propositional calculusGL ` (⇤B1 ^ · · · ^⇤Bn) ! ⇤(⇤A ! A) Propositional calculusGL ` ⇤(⇤A ! A) ! ⇤A G axiomGL ` (⇤B1 ^ · · · ^⇤Bn) ! ⇤A Propositional calculus.From the way we found the ⇤Bi, we know that for each i 6 n ⇤Bi 2 � and by Lemma 3.32 we

have [(⇤B1 ^ · · · ^ ⇤Bn) ! ⇤A] 2 �. Thus, by Lemma 3.32 again, we deduce ⇤A 2 �. Hence,our assumption that ⌃ was GL-inconsistent is false, i.e. ⌃ is truly GL-consistent. End of Claim.

Hence, Proposition 3.33 gives us that ⌃ is contained in some maximally consistent set, �0

say. We claim that this �0 is the desired � 2 WD with �RD� but such that A /2 �. �0 isa maximally consistent set and so certainly �0 2 WD. Note that 8⇤B 2 � B,⇤B 2 �0 since⌃ = {¬A,⇤A} [ {B,⇤B | ⇤B 2 �} ✓ �0. Furthermore, ⇤A 2 �0, but since we supposed⇤A /2 � we know from Lemma 3.32 that ¬⇤A 2 �. Therefore, �RD�0 by definition of RD. Finally,¬A 2 ⌃ ✓ �0, so ¬A 2 �0 which means we must have A /2 �0 by Lemma 3.32 and �0 is as weclaimed.

Thus, we have shown ⇤A /2 � ) 9� 2 WD with �RD� but such that A /2 �.

50


Proposition 3.36. [2, p. 79] For any subsentence A of D and for all � 2 WD, we have

MD,� �D A , A 2 �.

Proof. Let A be an arbitrary subsentence of D and � 2 WD an arbitrary maximally consistent setof D-subformulas. We prove this by induction on the form of A.

First, suppose A ⌘ ?. Then GL ` ¬?, and since � is maximally consistent it is GL-consistent,so ? /2 �. Further, by definition of truth in a model, MD,� 1D ? always. Thus, MD,� �D A ,A 2 �.

Assume A ⌘ pi for some propositional letter appearing in D. Then by our definition of �D, weknow MD,� �D pi , pi 2 � i.e. MD,� �D A , A 2 �.

If A ⌘ B ^ C for some subsentences B,C of D, then

MD,� �D A , MD,� �D B ^ C, MD,� �D B and MD,� �D C (definition of truth in a model), B 2 � and C 2 � by the induction hypothesis, B ^ C 2 � by Lemma 3.32, A 2 �.

Now suppose A ⌘ ¬B for some subsentence B of D. Then

MD,� �D A , MD,� �D ¬B, MD,� 1D B (definition of truth in a model), B /2 � by the induction hypothesis, ¬B 2 � by Lemma 3.32, A 2 �.

Finally, assume A ⌘ ⇤B for some subsentence B of D. Then

MD,� �D A , MD,� �D ⇤B, 8� 2 WD, �RD� ) MD,� �D B (definition of truth in a model), 8� 2 WD, �RD� ) B 2 � by the induction hypothesis, ⇤B 2 � by Lemma 3.35, A 2 �.

Recall that �¬D 2 WD is a maximally consistent set of D-subformulas containing ¬D, so fromLemma 3.32 we know D /2 �¬D. Hence, MD,�¬D 1D D by Proposition 3.36 above, so D is not validin the model MD. Since MD is based on hWD, RDi, a (finite) transitive and converse wellfoundedframe, and D is not valid in MD then D cannot be valid in the class of all transitive and conversewellfounded frames, i.e. is not GL-valid. Therefore, we have established:

Theorem 3.37. (Completeness for GL) For any A 2 FORML⇤ , if A is GL-valid then GL ` A.

Definition. We call a transitive frame F := hW,Ri a tree if we have that 8x, y, z 2 W xRz andyRz ) x = y or xRy or yRx.

51


Proposition 3.38. [2, p. 84] A modal sentence A is valid in all finite, transitive and irreflexiveframes iff it is valid in all finite, transitive and irreflexive frames that are trees.

Proof. Let A an arbitrary modal sentence.()) Clearly, if A is valid in all finite, transitive and irreflexive frames then it is valid in any that

are also trees.(() Now let M := hW,R,�i be an arbitrary finite, transitive and irreflexive frame. We will call

a function f an R-chain if there exists k 2 N such that f : {0, 1, . . . , k} ! W and 8i < k we havef(i)Rf(i+ 1), i.e. we have the chain f(0)Rf(1)R · · ·Rf(k).

Set F := {f | f is an R-chain}. Define a new relation S on F by 8f, g 2 F fSg iff there existm < n 2 N such that f : {0, . . . ,m} ! W , g : {0, . . . , n} ! W and 8i 6 m f(i) = g(i).

Since W is finite (by assumption), there can only be finitely many ordered pairs in R and thusF can only be finite. S is transitive because if fSg and gSh then for some k 2 N we will havef(i) = g(i) = h(i) for all i 6 k, and hence fSh. Since f(i) = f(i) for all i 2 N, we can never havefSf , so S is irreflexive. If we have f, g, h 2 F such that fSh and gSh, then there exists m, k < nsuch that f : {0, . . . ,m} ! W , h : {0, . . . , n} ! W , g : {0, . . . , k} ! W and for all i 6 m f(i) = h(i)and for all j 6 k g(j) = h(j). Either m < k or k < m or m = k, and so either fSg or gSf or g = f ,respectively. So hF , Si is a finite, transitive and irreflexive frame that is a tree.

Now let N := hF , S,�N i where we define �N by N , f �N p , M, f(k) � p for f 2 F wheref : {0, . . . , k} ! W .

Claim. For all f 2 F , f : {0, . . . , k} ! W , we have N , f �N A , M, f(k) � A.Proof of Claim. If A ⌘ p for some propositional letter p, then by definition of �N we immediately

have N , f �N p , M, f(k) � p.Suppose A ⌘ B ^ C, then

N , f �N A , N , f �N B ^ C, N , f �N B and N , f �N C (definition of truth in a model), M, f(k) � B and M, f(k) � C by the induction hypothesis, M, f(k) � B ^ C, M, f(k) � A.

Suppose A ⌘ ¬B, then

N , f �N A , N , f �N ¬B, N , f 1N B (definition of truth in a model), M, f(k) 1 B by the induction hypothesis, M, f(k) � ¬B (definition of truth in a model), M, f(k) � A.

Suppose A ⌘ ⇤B, then

N , f �N A , N , f �N ⇤B, 8g 2 F with fSg N , g �N B (definition of truth in a model), 8g(kg) 2 W, f(k)Rg(kg) ) M, g(kg) � B by the induction hypothesis, M, f(k) � ⇤B (definition of truth in a model), M, f(k) � A.

End of Claim.

52


If A is valid in all finite, transitive and irreflexive frames that are trees, then A is valid in N .Note for each w 2 W we can find f = fw 2 F , f : {0, . . . , k} ! W, such that f(k) = w, soN , f �N A , M, f(k) � A , M, w � A. That is, 8w 2 W we have M, w � A, i.e. A is valid inM.

4 Arithmetical Completeness in Provability Logic

In §2 we saw that formal arithmetic has much to offer, especially because it can cope with and provemuch number theory, and in §3 the focus changed to a seemingly disjoint area of mathematics thathas close connections to the philosophy of language. In this section, we seek to combine these twoareas of mathematics by giving the modal operator ⇤ a meaning in formal arithmetic.

4.1 Arithmetical Soundness of GL

The easy part of our main task is to show that if a modal sentence A is provable in GL, then anytranslation of A into the language of arithmetic is provable in PA.

Definition. An (arithmetical) realization is a function ⇤ : SENTL⇤ ! SENTLA , which we defineinductively as follows:

(i) p⇤i =⇤(pi) for any sentence letter pi

(ii) ?⇤ = ?

(iii) (A ^B)⇤ = (A⇤ ^B⇤) for any A,B 2 SENTL⇤

(iv) (¬A)⇤ = ¬(A⇤) for any A 2 SENTL⇤

(v) (⇤A)⇤ = Prov(pA⇤q), which is just the sentence of PA asserting that A is provable, for anyA 2 SENTL⇤ .

We call A⇤ the translation of the modal sentence A under the realization ⇤.

Remark. Directly from how a realization ⇤ is inductively defined, one can check routinely throughlogical equivalences that for any A,B 2 SENTL⇤

(i) (A _B)⇤ = A⇤ _B⇤

(ii) (A ! B)⇤ = A⇤ ! B⇤.

A specific realization essentially provides us with a bi-lingual dictionary between the language ofmodal logic L⇤ and the language of arithmetic LA; in that it decides to which sentence of arithmeticeach “word” (propositional letter) pi gets sent, and therefore how a string of “words” and “connectingwords” (the logical connectives) - i.e. a sentence A - gets translated. Continuing with this analogy,it becomes clear that if two realizations, ⇤ and # say, give us two identical dictionaries then wehave ⇤ = #. In particular, if we have a sentence A of modal logic and for each sentence letter piappearing in A we have p⇤i = p#i then A⇤ = A#. Moreover, if A contains no propositional letters,e.g. A ⌘ ¬⇤?, then A⇤ = A# for all realizations ⇤ and #.

53


Lemma 4.1. There are infinitely many realizations.

Proof. Suppose there are only finitely many distinct realizations, ⇤1 , . . . ,⇤k say. Pick any proposi-tional letter p, then for each i 6 k p⇤i = 'mi for some sentence 'mi of arithmetic. Then define therealization ⇤ by: for any propositional letter q 6= p, q⇤ = q⇤1 but p⇤ = p⇤1 ^ p⇤2 = 'm1 ^ 'm2 . Since'mi are sentences for all i 6 k, 'm1 ^ 'm2 is still a sentence as we have not freed any variablesappearing in either of 'm1 or 'm2 . Note that since p⇤ 6= p⇤i for all i 6 k, we have that ⇤ is distinctfrom each ⇤i . Thus, ⇤ is a new realization not included in our finite list ⇤1 , . . . ,⇤k . Hence, theremust be infinitely many realizations.

Definition. For any A 2 SENTL⇤ , if PA ` A⇤ for all realizations ⇤, then we call A always provable.

It can be seen that a specific realization gives rise to a specific possible world of formal arithmeticpurely because the translations of sentences have truth values (because they contain no free vari-ables), so an always provable sentence A is one that is provable in each possible world, i.e. PA-valid.Our end goal for the moment is to see that if GL ` A then A is always provable. We will showthis by first introducing another modal logic K4LR, which turns out to have precisely the sametheorems as GL, and then showing that K4LR is arithmetically sound.

Definition. We define the Löb Rule to be the rule of inference in a modal logic axiomatic systemwhich allows one to deduce A from ⇤A ! A for any A 2 FORML⇤ , i.e. for anyA 2 FORML⇤

⇤A ! A

A.

Definition. Define the modal logic system K4LR

14 to be K4 but with the Löb Rule as an additionalrule of inference.

Proposition 4.2. [2, p. 59] For any A 2 SENTL⇤ , GL ` A , K4LR ` A.

Proof. Let A 2 SENTL⇤ be arbitrary.()) For this direction we only need to show that GL is closed under the Löb Rule because we

saw in Theorem 3.20 (ii) that GL ` ⇤A ! ⇤⇤A, the 4 axiom. Therefore, we need to show thatfrom ⇤A ! A we can deduce A in the axiomatic proof system of GL.

(1) GL ` ⇤A ! A Assumption

(2) GL ` ⇤(⇤A ! A) Necessitation (1)

(3) GL ` ⇤(⇤A ! A) ! ⇤A G axiom

(4) GL ` ⇤A Modus ponens (2), (3)

(5) GL ` A Modus ponens (4), (1).

So whenever we know GL ` ⇤A ! A, we also know GL ` A from the steps above, i.e. GL isclosed under the Löb Rule.

(() The converse requires us to show that K4LR can prove any instance of the G axiom⇤(⇤A ! A) ! ⇤A.

14We keep the same name as [2, p. 59].

54


(1) K4LR ` ⇤(⇤(⇤A ! A) ! ⇤A) ! (⇤⇤(⇤A ! A) ! ⇤⇤A) K axiom

(2) K4LR ` ⇤(⇤A ! A) ! ⇤⇤(⇤A ! A) 4 axiom

(3) K4LR ` ⇤(⇤(⇤A ! A) ! ⇤A) ! (⇤(⇤A ! A) ! ⇤⇤A) Propositional calculus (1), (2)

(4) K4LR ` ⇤(⇤A ! A) ! (⇤⇤A ! ⇤A) K axiom

(5) K4LR ` ⇤(⇤(⇤A ! A) ! ⇤A) ! (⇤(⇤A ! A) ! ⇤A) Propositional calculus (3), (4)

(6) K4LR ` ⇤(⇤A ! A) ! ⇤A Löb Rule.

Hence, all formulas of the form ⇤(⇤A ! A) ! ⇤A are derivable in K4LR and we are done.

Theorem 4.3. [2, pp. 52-3, p. 59] For all A 2 SENTL⇤ and every realization ⇤, if K4LR ` Athen PA ` A⇤.

Proof. Through induction on the lines of a proof in K4LR. Let A be an arbitrary modal sentenceand ⇤ an arbitrary realization and suppose K4LR ` A, then A is the conclusion of some axiomaticproof in the modal system K4LR. In axiomatic proofs in K4LR, each line B will be one of thefollowing: a conclusion of modus ponens, necessitation or the Löb Rule, or a tautology, an instanceof the K axiom or an instance of the 4 axiom. Therefore, if we show that our claim holds for eachof these, then we will have our desired result because each line will then be provable in PA under⇤, and in particular the last line of the proof will be A⇤ and provable in PA.

Modus ponens: If B is the conclusion of an application of modus ponens, then there existsC 2 SENTL⇤ such that K4LR ` C and K4LR ` C ! B. Thus, by the induction hypothesis,PA ` C⇤ and PA ` (C ! B)⇤. But recall that (C ! B)⇤ = C⇤ ! B⇤ from our Remark earlier, soPA ` C⇤ ! B⇤. Thus, by modus ponens in PA, we have PA ` B⇤.

Necessitation: If B is the conclusion of an application of necessitation, then there existsC 2 SENTL⇤ such that B ⌘ ⇤C and K4LR ` C. By the induction hypothesis, we know PA ` C⇤.Since Prov(x) is a provability predicate for PA (Proposition 2.33) and PA ` C⇤, we know that(P1) implies PA ` Prov(pC⇤q). From the definition of ⇤, we know B⇤ ⌘ (⇤C)⇤ = Prov(pC⇤q) andso PA ` B⇤.

Löb Rule: If B is the conclusion of an application of the Löb Rule, then there exists C 2 SENTL⇤such that B ⌘ C and K4LR ` ⇤C ! C. From the definition of ⇤:

(⇤C ! C)⇤ = (⇤C)⇤ ! C⇤ = Prov(pC⇤q) ! C⇤

and by the induction hypothesis PA ` (⇤C ! C)⇤, i.e. PA ` Prov(pC⇤q) ! C⇤. Then by Löb’sTheorem (Theorem 2.35), PA ` C⇤ which is just PA ` B⇤.

Tautology: If B is just a tautological truth-functional combination of modal sentences, then underany realization B⇤ will be another tautological combination of sentence of arithmetic. Therefore, ifB appears in the K4LR-proof of A then PA ` B⇤.

K axiom: Suppose B ⌘ ⇤(C ! D) ! (⇤C ! ⇤D) for some C,D 2 SENTL⇤ and K4LR ` B,then

B⇤ = (⇤(C ! D) ! (⇤C ! ⇤D))⇤

= (⇤(C ! D))⇤ ! (⇤C ! ⇤D)⇤

= (⇤(C ! D))⇤ ! ((⇤C)⇤ ! (⇤D)⇤)= Prov(p(C ! D)⇤q) ! (Prov(pC⇤q) ! Prov(pD⇤q))= Prov(pC⇤ ! D⇤q) ! (Prov(pC⇤q) ! Prov(pD⇤q))

55


by how realizations are defined. Note that C⇤, D⇤ are just sentences of arithmetic, so by Proposition2.33 (P2)

PA ` Prov(pC⇤ ! D⇤q) ! (Prov(pC⇤q) ! Prov(pD⇤q)),which is PA ` (⇤(C ! D) ! (⇤C ! ⇤D))⇤, i.e. PA ` B⇤ as required.

4 axiom: Suppose B ⌘ ⇤C ! ⇤⇤C for some C 2 SENTL⇤ and K4LR ` B, then

B⇤ = (⇤C ! ⇤⇤C)⇤

= (⇤C)⇤ ! (⇤⇤C)⇤

= Prov(pC⇤q) ! Prov(p(⇤C)⇤q)= Prov(pC⇤q) ! Prov(pProv(pC⇤q)q)

by the inductive definition of ⇤. Since C⇤ is a sentence of arithmetic, we know by (P3) thatPA ` Prov(pC⇤q) ! Prov(pProv(pC⇤q)q), i.e. PA ` B⇤.

Hence, arithmetical soundness will hold for each line of the proof of A, and thus if K4LR ` Athen PA ` A⇤.

Theorem 4.4. [2, p. 59] (Arithmetical Soundness of GL) For all A 2 SENTL⇤ , if GL ` Athen PA ` A⇤ for every realization ⇤,.

Proof. Suppose GL ` A for an arbitrary A 2 SENTL⇤ , then by Proposition 4.2 K4LR ` A.Whence, by Theorem 4.3, PA ` A⇤ for any realization ⇤.

The arithmetical soundness of GL is easy to prove relative to the arithmetical completeness ofGL due to Solovay which we prove later. However, before beginning our final climb to Solovay’sresult, we will demonstrate the consequences of Theorem 4.4 itself.

Corollary 4.5. [2, p. 61] (Formalised Gödel’s Second Incompleteness Theorem for PA)

Gödel’s Second Incompleteness Theorem for PA is derivable within PA itself.

Proof. GL ` ⇤(⇤? ! ?) ! ⇤? G axiomGL ` (⇤(⇤? ! ?) ! ⇤?) ! (¬⇤? ! ¬⇤(⇤? ! ?)) TautologyGL ` ¬⇤? ! ¬⇤(⇤? ! ?) Modus ponens.Then by the arithmetical soundness of GL, we have that PA ` (¬⇤? ! ¬⇤(⇤? ! ?))⇤. But

that is just PA ` ¬Prov(p?q) ! ¬Prov(pProv(p?q) ! ?q), which is

PA ` ¬Prov(p?q) ! ¬Prov(p¬Prov(p?q)q).

In words, this is says that PA can prove: if arithmetic is consistent then PA cannot prove its ownconsistency ; this is Gödel’s Second Incompleteness Theorem for PA.

Corollary 4.6. [2, p. 61] PA can prove that if the inconsistency of arithmetic is not formallyprovable (in PA), then the consistency of arithmetic is undecidable.

56


Proof. GL ` ⇤? ! ⇤⇤? G axiomGL ` (⇤? ! ⇤⇤?) ! (¬⇤⇤? ! ¬⇤?) TautologyGL ` ¬⇤⇤? ! ¬⇤? Modus ponensGL ` ¬⇤? ! ¬⇤¬⇤? Corollary 4.5GL ` ¬⇤⇤? ! ¬⇤¬⇤? Propositional calculusGL ` ¬⇤⇤? ! (¬⇤¬⇤? ^ ¬⇤⇤?) Propositional calculus.

Thus, by Theorem 4.4 (arithmetical soundness of GL), we have

PA ` ¬Prov(pProv(p?q)q) ! (¬Prov(p¬Prov(p?q)q) ^ ¬Prov(pProv(p?q)q)).

That is, not being able to formally prove the inconsistency of arithmetic implies that, firstly, it isnot formally provable that arithmetic is consistent and, secondly, it is not formally provable thatarithmetic is inconsistent. Hence, the formal unprovability of the inconsistency of arithmetic impliesthat the consistency of arithmetic is undecidable.

Corollary 4.7. [2, p. 62] (Formalised Löb) PA ` Prov(pProv(p'q) ! 'q) ! Prov(p'q) forany ' 2 SENTLA .

Proof. Let ' 2 SENTLA be arbitrary. Note GL ` ⇤(⇤A ! A) ! ⇤A for any A 2 SENTL⇤ . Inparticular, GL ` ⇤(⇤p ! p) ! ⇤p for any sentence letter p and we can find a realization ⇤ suchthat p⇤ = '. Since GL is arithmetically sound, PA ` (⇤(⇤p ! p) ! ⇤p)⇤ but as

(⇤(⇤p ! p) ! ⇤p)⇤ = Prov(pProv(pp⇤q) ! p⇤q) ! Prov(pp⇤q)= Prov(pProv(p'q) ! 'q) ! Prov(p'q)

we have PA ` Prov(pProv(p'q) ! 'q) ! Prov(p'q). This holds for any sentence of arithmeticbecause ' was arbitrary.

Corollary 4.8. [2, p. 62] (Strengthened Löb) For any ', 2 SENTLA ,

if PA ` Prov(p'q) ^ Prov(p q) ! , then PA ` Prov(p'q) ! .

Proof. (1) PA ` Prov(p'q) ^ Prov(p q) ! Assumption

(2) PA ` Prov(p'q) ! (Prov(p q) ! ) Propositional calculus (1)

(3) PA ` Prov(pProv(p'q) ! (Prov(p q) ! )q) (P1) on (2)

(4) PA ` Prov(pProv(p'q)q) ! Prov(pProv(p q) ! q) (P2) on (3)

(5) PA ` Prov(pProv(p q) ! q) ! Prov(p q) Formalised Löb

(6) PA ` Prov(pProv(p'q)q) ! Prov(p q) Propositional calculus (4), (5)

(7) PA ` Prov(p'q) ! Prov(pProv(p'q)q) (P3)

(8) PA ` Prov(p'q) ! Prov(p q) Propositional calculus (6), (7)

(9) PA ` Prov(p'q) ! Propositional calculus (2), (8).

57


As we claimed in our comments after Corollary 3.30, the canonical model for GL,M⇤

GL

:= hWGL

, RGL

,�GL

i, is transitive but not converse wellfounded. The proof of the nexttheorem is due to Giancarlo Meloni. The idea of the proof is to show that M⇤

GL

is not irreflexive,and so cannot be converse wellfounded because we know converse wellfounded implies irreflexive(Lemma 3.26).

Theorem 4.9. [2, pp. 90-1] The canonical model M⇤GL

is not converse wellfounded.

Proof. Let ⇤ be an arbitrary realization and define � := {A | A⇤ is true}. Note that � is non-emptybecause for any A 2 SENTL⇤ that is a tautological combination of sentences, we know A⇤ is trueas it will also be a tautological combination of sentences but just in arithmetic. That is, � containsall modal tautologies, so for example contains p = p for any sentence letter p of modal logic, hence� 6= ;. We will show that � is maximally GL-consistent (as defined in the very beginning of §3.4),so � 2 W

GL

, but that �RGL

� and hence RGL

is not irreflexive.

Claim. � is GL-consistent.Proof of Claim. Assume for contradiction that � is GL-inconsistent, then there exists some

finite subset {A1, . . . , Ak} ✓ � such that GL ` (A1 ^ · · · ^ Ak) ! ?. For each i 6 k, Ai 2 � so byconstruction of � we have A⇤

i is true. By the arithmetical soundness of GL, we knowPA ` (A⇤

1 ^ · · · ^ A⇤k) ! ? and by the soundness of PA we have (A⇤

1 ^ · · · ^ A⇤k) ! ? is true.

Therefore, A⇤j is false for some j 6 k. Hence, � is GL-consistent. End of Claim.

Claim. � is maximally GL-consistent.Proof of Claim. Assume for contradiction that � is not maximal. Then there exists some max-

imally GL-consistent set � with � ( �. Then there exists A0 2 �\�. If A⇤0 were true then A0

would be an element of � by construction of �, which is contrary to assumption on A0, so A⇤0 is

false. Therefore, we know ¬A⇤0 is true and so ¬A0 2 �. But then ¬A0 2 �, so A0 ^¬A0 2 � which

implies � is GL-inconsistent. Hence, � is maximal. End of Claim.

Claim. �RGL

�.Proof of Claim. Recall that 8⌃,� 2 W

GL

, we have ⌃RGL

� , 8A 2 FORML⇤ , if⇤A 2 ⌃ then A 2 �. Let A 2 FORML⇤ be arbitrary such that ⇤A 2 �. We need to show thatA 2 �. If ⇤A 2 � then (⇤A)⇤ is true, and hence Prov(pA⇤q) is true. Therefore, PA ` A⇤ and sincePA is sound we know A⇤ is true, so A 2 �. Hence, �R

GL

�. End of Claim.

Thus, RGL

is not irreflexive and hence cannot be converse wellfounded as explained above. Sothe canonical model M⇤

GL

for GL is not converse wellfounded and cannot be used to prove thesemantical completeness of GL.

Definition. For any ' 2 SENTLA , we call (Prov(p'q) ! ') 2 SENTLA the reflection principlefor '.

So then, is there a sentence of arithmetic that is consistent with PA and implies all reflectionprinciples? If there was, then by Löb’s Theorem we would be able to prove any sentence of arithmetic.However, as great as this looks on first glance, it takes just a little thought to realise the catastrophicconsequence: PA would be inconsistent.

58


Proposition 4.10. There exists no sentence of arithmetic which implies all reflection principles forsentences of arithmetic.

Proof. Assume for contradiction that there exists such a sentence, 0 say. That is for any' 2 SENTLA , we have PA ` 0 ! (Prov(p'q) ! '). Since 0 2 SENTLA , we also know¬ 0 2 SENTLA . Therefore, PA ` 0 ! (Prov(p¬ 0q) ! ¬ 0) and if PA ` 0 thenPA ` Prov(p¬ 0q) ! ¬ 0 by modus ponens. Thus, by Löb’s Theorem PA ` ¬ 0. So PA ` ?and this is a contradiction. Thus, no such 0 can exist.

Lemma 4.11. For any sentence G 2 SENTLA ,

PA ` (G $ ¬Prov(pGq)) ) PA ` ((Prov(pGq) ! G) $ G).

Proof. We will firstly show PA ` (G $ ¬Prov(pGq)) ) PA ` ((Prov(pGq) ! G) ! G) and thenPA ` (G $ ¬Prov(pGq)) ) PA ` (G ! (Prov(pGq) ! G)).

Suppose PA ` G $ ¬Prov(pGq) and PA ` Prov(pGq) ! G. Assume for contradiction thatPA ` ¬G. Note that

PA ` (Prov(pGq) ! G) ! (¬G ! ¬Prov(pGq)),

so PA ` ¬Prov(pGq) by two applications of modus ponens. Then by our first premise, PA ` G soPA ` G ^ ¬G, i.e. PA ` ?. Thus, we must have PA ` G and so PA ` (Prov(pGq) ! G) ! G.

Now suppose PA ` G $ ¬Prov(pGq), PA ` G and PA ` Prov(pGq). Then trivially we havePA ` Prov(pGq) ! G, and so PA ` G ! (Prov(pGq) ! G).

Therefore, PA ` (Prov(pGq) ! G) $ G.

Therefore, if PA is consistent, then by Gödel’s First Theorem we can find a sentence G ofarithmetic such that PA ` G $ ¬Prov(pGq). Then by Lemma 4.11, PA cannot prove the reflectionprinciple for G, because otherwise PA would prove G, and hence Prov(pGq) by (P1), but also¬Prov(pGq) which is a contradiction. We close this subsection with the following remark abouttruth in the standard model of arithmetic.Remark. One can show the soundness of PA by induction on the axiomatic proofs in PA: that is,show that the rules of inference are sound and the axioms of PA are true in that standard model.Thus, we have that every theorem of PA is true in the standard model. Now for any sentence ofarithmetic ', if we show that Prov(p'q) is true then ' will be provable in PA, and hence true itself.So for every sentence of arithmetic ', we know Prov(p'q) ! ' is true - by our results above, weknow Prov(p'q) ! ' is certainly not provable in PA for all ' - and therefore, for any sentence Aof modal logic and any realization ⇤, we know (⇤A ! A)⇤ is true.

4.2 Arithmetical Completeness of GL

Solovay’s result, expressing the Arithmetical Completeness of GL, is the last piece of the puzzlewhich solidifies the relationship between GL and PA. The proof of this result relies on the semanticalcompleteness for GL and then involves, in the words of Artemov and Beklemishev, “embedding”Kripke models into arithmetic [1, pp. 222-7]. We prove the contrapositive and the outline of theproof is thus: if GL 0 A for some modal sentence A, then there is a finite, transitive, conversewellfounded Kripke model in which A is not valid; and then we will define a function h (due to

59


Solovay [19, pp. 296-7] and with the aid of the Diagonal Lemma) that is used to show A⇤ is notprovable in PA, where ⇤ is a specific realization which we will define.

We now state and prove Solovay’s result. The proof15 is long and consists of several claims.

Theorem 4.12. [19, Theorem 4.6, p. 297] (Solovay’s Arithmetical Completeness of GL) Forall A 2 SENTL⇤ , if for every realization ⇤

PA ` A⇤ then GL ` A.

Proof. Let A 2 SENTL⇤ be arbitrary sentence of modal logic and suppose that GL 0 A. Thenby the semantical completeness of GL (Theorem 3.37), there exists a finite, transitive and conversewellfounded Kripke model M0 := hW 0, R0,�0i in which A is not valid. We may assume without anyloss of generality that W 0 = {1, . . . , n}, 1R0i iff 1 < i 6 n and that M0, 1 10 A since A is not validin M0. We should observe that for no 1 6 i 6 n do we have iR01, because otherwise we would haveiR01R0i and by the transitivity of R0 also iR0i but R0 is irreflexive since it is converse wellfounded, sothis is impossible. Therefore, we can view M0 as a tree (we have no cycles by the previous sentence)with root 1. However, note that we put no conditions on how the other worlds (excluding 1) relateto worlds other than 1, so it is not necessarily true that this tree is linear, i.e. it is not necessarilytrue that iR0j iff 1 6 i < j 6 n. For example, here are some very different possibilities for M0,where we suppose iR0j ) i < j and 1R0i for all 1 < i 6 n is true by the transitivity of R0.

1

2

...

n� 1

n

Figure 3: Linear Possibility

1

2 3 . . . n� 1 n

Figure 4: Non-Linear Possibility

Or something less simple.15We largely follow [2, §9, pp. 124-130]. However, the beginning follows [19, p. 296].

60


1

2

3 4

7 9

6

8

10

...

n� 1

n

Figure 5: Second Possibility

Before continuing, we augment M0 by adding in a sub-root, namely 0, and increasing the size ofR0. Set W := W [ {0} and R := R0 [ {h0, ii | 1 6 i 6 n}. Just like hW 0, R0i, hW,Ri is still finite,transitive and converse wellfounded. We define the relation � in M := hW,R,�i as follows: for anypropositional letter p, M, 0 � p , M0, 1 �0 p and 81 6 i 6 n M, i � p , M0, i �0 p.

We now make the definition of h and we follow [19, p. 296]. Define a function h as follows:

h(0) = 0

h(m+ 1) =

(j if h(m) = i ^ iRj ^ Pf(m, p` 6= jq)h(m) = i else

where ` 6= j ⌘ ¬9N8n[n > N ! h(n) = j]. Consider how we determine the value for h(m + 1). Ifm, a natural number, is the Gödel number of a proof in PA then m codes a finite sequence, namelythe sequence of lines of the proof which is finite. Then we pick out the last line, i.e. the last entry ofthe sequence, and check if that is equivalent to ` 6= j for some j such that iRj and where h(m) = i.If so then set h(m + 1) = j and otherwise we stay at i. There is some ‘apparent circularity’ [19,p. 296]: h is defined on ` and yet ` relies on knowing the eventual tendencies of h. One way tobreak out of the circularity is by using the Diagonal Lemma, but another is to use Kleene’s SecondRecursion Theorem16 just at Solovay does in his original proof [19, p. 296].

The following analogy of the behaviour of h is insightful and we think of h(m) = b as the refugeeis in country b at moment m [1, p. 190].

Consider a refugee who can only acquire a visa for a country only if he/she can prove that he/shewill not remain in that country forever. Further, if the refugee is not allowed back into a countryonce having left it, then the refugee will eventually reside somewhere as there are only finitelymany countries in the world. Thus, an honest refugee will remain in his/her native country.17

The author provides his own version (‘The Toy Story’) where h(m) = b should be thought of as thechild has toy b at moment m:

16

Theorem 4.13. [12, Theorem XXVII, §66, p. 352] (Kleene’s Second Recursion Theorem) Let (z, x1, . . . , xn)be total any recursive function. Then a number e can be found which defines (e, x1, . . . , xn) recursively, i.e. such

that

(e,m1, . . . ,mn) =

( (z,m1, . . . ,mn) if (z,m1, . . . ,mn) is definedundefined if (z,m1, . . . ,mn) is undefined.

17This analogy comes from [1, p. 190].

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Consider a child who owns just one toy, cannot hold more than one toy but must always beholding a toy. You have a (finite) box of toys and if the child promises to give you back a toyfrom your box, then you can give the child that toy. However, if you give the child a toy thenhe/she must drop his/her other toy and it will break. Thus, the child will eventually be holdingone toy as the box is finite and a good (honest) child will never ask for any toys.

We now show how to find this function h18 using the Diagonal Lemma on the following formula:B(y,m, b) ⌘

9s[FinSeq(s) ^ lh(s) = m+++ 1 ^ s0 = 0 ^ sm = b^

8x < mnV

i=0

"sx = i !

(V

j:iRj

[Pf(x, notlim(y, j)) ! sx+1 = j]

^"(

Vj:iRj

¬Pf(x, notlim(y, j))

)! sx+1 = sx

#)#].

where notlim(k, j) ⌘ p¬9N8a(a > N ! 9c[c = j ^ Fk])q and Fk is the formula with Gödel numberk. B(y,m, b) is just a formula of arithmetic so we can apply the Diagonal Lemma to it in order toobtain formula H(m, b) such that

PA ` H(m, b) $ B(pH(m, b)q,m, b).

Now let the Gödel number of H(m, b) be k so that

notlim(pH(m, b)q, j) ⌘ p¬9N8a(a > N ! H(a, j))q.

For each 0 6 j 6 n set the Solovay sentence for j as

Sj ⌘ 9N8a[a > N ! H(a, j)].

That is, Sj expresses that the limit of h is j, so ¬Sj expresses that the limit of h 6= j. Thus,p¬Sjq = notlim(pH(m, b)q, j) and

PA ` H(m, b) $ 9s[FinSeq(s) ^ lh(s) = m+++ 1 ^ s0 = 0 ^ sm = b^

8x < mnV

i=0

"sx = i !

(V

j:iRj

[Pf(x, p¬Sjq) ! sx+1 = j]

^"(

Vj:iRj

¬Pf(x, p¬Sjq))

! sx+1 = sx

#)#].

H(m, b) is equivalent to a ⌃1 formula and describes our function h above. We will now prove aseries of claims [2, pp. 128-130], [1, pp. 190-192] about formally provable facts about H(m, b), andequivalently h. To help the flow of the proof we sometimes give only informal arguments but whichcan be formalised in PA if need be.

Claim (C1). PA ` 9!bH(m, b).Proof of Claim (C1). Since h(0) = 0 we know PA ` H(0,0), and since there is a unique

sequence h0i we have PA ` 9!bH(0, b). Now suppose that PA ` 9!bH(k, b), say h(k) = i, then18We follow [2, pp. 127-8].

62


there is a sequence s = ha0, . . . , aki such that ak = i. Then h(k + 1) = i or j for some j suchthat iRj. In either case, we augment the sequence s to get t = ha0, . . . , ak, ak+1i where ak = i

unless k is the Gödel number of a proof of ¬Sj for some j with iRj and in that case ak = j. Thus,PA ` 9bH(k + 1, b). The uniqueness follows from the uniqueness of sequences: sequence t is oflength k + 2 and any other sequence v = hb0, . . . , bk+1i satisfying the conditions in the equivalenceof H(k + 1, b) will necessarily have final entry bk+1 = ak+1. Hence, PA ` 9!bH(k + 1, b) and we aredone by induction on m. End of Claim (C1).

Claim (C2). 80 6 i < j 6 n, we have PA ` ¬(Si ^ Sj).Proof of Claim (C2). Suppose for some 0 6 i < j 6 n that PA ` Si ^ Sj , then PA ` Si

and PA ` Sj . This implies there exist N1 and N2 such that PA ` 8a[a > N1 ! H(a, i)] andPA ` 8a[a > N2 ! H(a, j)]. Then setting N := max(N1, N2) we clearly see that PA ` H(N, i) andPA ` H(N, j), but then by (C1) we have a contradiction because i 6= j. End of Claim (C2).

Claim (C3). PA ` H(m, i) ! (Si _W

j:iRj

Sj).

Proof of Claim (C3). We prove this by induction on the converse of R (Theorem 3.27).Fix an 0 6 i 6 n and suppose that 8j with iRj PA ` H(m, j) ! (Sj _

Wk:jRk

Sk). By definition

of h, we know that

PA ` H(m, i) !

0

@8a

2

4a > m ! (H(a, i) __

j:iRj

H(a, j))

3

5

1

A

so by the induction hypothesis

PA ` H(m, i) !

0

@8a[a > m ! H(a, i)] __

j:iRj

0

@Sj __

k:jRk

Sk

1

A

1

A .

But since R is transitive, this is just

PA ` H(m, i) !

0

@8a[a > m ! H(a, i)] __

j:iRj

Sj

1

A .

And since PA ` 8a[a > m ! H(a, i)] ! 9N8a[a > N ! H(a, i)], i.e.PA ` 8a[a > m ! H(a, i)] ! Si,

PA ` H(m, i) !

0

@Si __

j:iRj

Sj

1

A .

End of Claim (C3).

Claim (C4). PA ` S0 _ S1 _ · · · _ Sn.

63


Proof of Claim (C4). Above we saw PA ` H(0,0), so by (C3) PA ` S0 _

Wj:0Rj

Sj

!. But since

1Rj for each 1 < j 6 n, 0R1 and R is transitive we have that

PA ` S0 _ S1 _ · · · _ Sn.

End of Claim (C4).

Claim (C5). If iRj then PA ` Si ! ¬Prov(p¬Sjq).Proof of Claim (C5). For this claim we give an informal argument that may be formalised in

PA. Recall that ` was the number defined to be the limit of h. If ` = i, then there exists N 2 Nsuch that for any a > N h(a) = ` = i. Now suppose j is such that iRj and that PA ` ` 6= j, i.e.that PA can prove that the limit of h is not j, and since PA is sound this is also true. Then thereexists a natural number k which is the Gödel number of a formal proof of ` 6= j. For any theoremthere are infinitely many proofs (we can just keep extending the proof by adding in the theorem asanother line), so we can suppose this k > N . Then h(k + 1) = j as Pf(k, p` 6= jq), but this impliesthat ` 6= i. Hence, PA 0 ` 6= j. End of Claim (C5).

Claim (C6). 8i > 1, we have PA ` Si ! Prov(p¬Siq).Proof of Claim (C6). Let i > 1 be arbitrary. Recall that Si ⌘ 9N8a[a > N ! H(a, i)], so

PA ` Si ! 9mH(m, i), say h(k+1) = i. Then by the definition of h, and since h(0) = 0 and i > n,we have that k is the Gödel number of a proof of ¬Si. So PA ` 9mH(m, i) ! 9yPf(y, p¬Siq), i.e.PA ` 9mH(m, i) ! Prov(p¬Siq). Therefore, PA ` Si ! Prov(p¬Siq). End of Claim (C6).

Claim (C7). 8i > 1, we have PA ` Si ! Prov(p Wj:iRj

Sjq).

Proof of Claim (C7). Suppose i > 1 throughout.

(1) PA ` H(a, i) ! (Si _WiRj

Sj) By (C3)

(2) PA ` 9aH(a, i) ! (Si _WiRj

Sj) Propositional calculus

(3) PA ` Prov(p9aH(a, i) ! (Si _WiRj

Sj)q) (P1) on (2)

(4) PA ` Prov(p9aH(a, i)q) ! Prov(pSi _WiRj

Sjq) (P2) on (3)

(5) PA ` Prov(p¬Siq) ^ Prov(pSi _WiRj

Sjq) ! Prov(pWiRj

Sjq) Tautology

(6) PA ` 9aH(a, i) ! Prov(p9aH(a, i)q) PA ⌃1-complete and H is ⌃1

(7) PA ` Si ! 9aH(a, i) As in proof of (C6)

(8) PA ` Si ! Prov(p9aH(a, i)q) Propositional calculus (6), (7)

(9) PA ` Si ! Prov(pSi _WiRj

Sjq) Propositional calculus (4), (8)

64


(10) PA ` Si ! Prov(p¬Siq) (C6)

(11) PA ` Si ! Prov(p¬Siq) ^ Prov(pSi _WiRj

Sjq) Propositional calculus (9), (10)

(12) PA ` Si ! Prov(pWiRj

Sjq) Propositional calculus (5), (11).

End of Claim (C7).We now specify a realization: define ⇤ by p⇤ =

WM,i�p

Si for each propositional letter p. The next

claim is a result concerning this realization.

Claim (C8). For any subsentence B of A and each 1 6 i 6 n

(i) if M0, i �0 B, then PA ` Si ! B⇤

(ii) if M0, i 10 B, then PA ` Si ! ¬B⇤

Proof of Claim (C8). We prove this claim by induction on the form of B. Let 1 6 i 6 n bearbitrary.

Firstly, suppose B ⌘ p for some sentence letter p. Thus, B⇤ = p⇤ =W

M,i�pSi. If M0, i �0 p then

we know M, i � p and thus PA ` Si !W

M,i�pSi. However, if M0, i 10 p, then M, i 1 p. In every

disjunct of p⇤ =W

M,j�pSj , Si 6= Sj because M, i 1 p but M, j � p. So by (C2) (uniqueness of the limit

of h), PA ` Si ! ¬Sj for each j appearing in p⇤. Hence, PA `V

M,j�p¬Sj , i.e. PA ` ¬

W

M,j�pSj

!,

i.e. PA ` ¬B⇤.Now suppose B ⌘ C ^D for some other subsentences C,D of A and that the conclusion holds

for C and D.M0, i �0 B , M0, i �0 C ^D

, M0, i �0 C and M0, i �0 D (definition of truth in a model)) PA ` Si ! C⇤ and PA ` Si ! D⇤ by the induction hypothesis) PA ` Si ! C⇤ ^D⇤ propositional calculus) PA ` Si ! B⇤.

M0, i 10 B , M0, i 10 C ^D, M0, i 10 C or M0, i 10 D (definition of truth in a model)) PA ` Si ! ¬C⇤ or PA ` Si ! ¬D⇤ by the induction hypothesis) PA ` Si ! ¬C⇤ _ ¬D⇤ propositional calculus) PA ` Si ! ¬(C⇤ ^D⇤) De Morgan’s law) PA ` Si ! ¬B⇤.

Suppose B ⌘ ¬C for some other subsentence C of A and that the conclusion holds for C.

M0, i �0 B , M0, i �0 ¬C, M0, i 10 C (definition of truth in a model)) PA ` Si ! ¬C⇤ by the induction hypothesis) PA ` Si ! B⇤

65


M0, i �0 B , M0, i 10 ¬C, M0, i �0 ¬¬C (definition of truth in a model), M0, i �0 C C $ ¬¬C) PA ` Si ! C⇤ by the induction hypothesis) PA ` Si ! ¬¬C⇤ C $ ¬¬C) PA ` Si ! ¬B⇤.

Finally, suppose B ⌘ ⇤C from subsentence C of A and that the conclusion holds for C. Therefore,B⇤ = (⇤C)⇤ = Prov(pC⇤q).

M0, i �0 B , M0, i �0 ⇤C, 8j with iR0j M0, j �0 C (definition of truth in a model)) 8j with iR0j PA ` Sj ! C⇤ by the induction hypothesis) 8j with iRj PA ` Sj ! C⇤ Since i > 1 ) [iR0j iff iRj]) PA `

Wj:iRj

Sj ! C⇤ propositional calculus

) PA ` Prov

p Wj:iRj

Sj ! C⇤q!

(P1)

) PA ` Prov

p Wj:iRj

Sjq!

! Prov(pC⇤q) (P2)

) PA ` Si ! Prov(pC⇤q) (C7)) PA ` Si ! B⇤

M0, i 10 B , M0, i 10 ⇤C, 9j > 1 with iR0j M0, j 10 C (definition of truth in a model)) 9j > 1 with iRj M0, j 10 C Since i > 1 ) [iR0j iff iRj]) PA ` Sj ! ¬C⇤ by the induction hypothesis) PA ` C⇤ ! ¬Sj propositional calculus) PA ` Prov(pC⇤ ! ¬Sjq) (P1)) PA ` Prov(pC⇤q) ! Prov(p¬Sjq) (P2)) PA ` ¬Prov(p¬Sjq) ! ¬Prov(pC⇤q) propositional calculus) PA ` Si ! ¬Prov(pC⇤q) (C5) since i > 1 and iRj) PA ` Si ! ¬B⇤.

This concludes the proof. End of Claim (C8).

Claim (C9). PA ` S0 ! ¬Prov(pA⇤q).Proof of Claim (C9). Since A is a subsentence of itself and M0, 1 10 A, we know by (C8) that

PA ` S1 ! ¬A⇤. Then by contra-posing, we get PA ` A⇤ ! ¬S1. Applying the derivabilityconditions (P1) and (P2) produce PA ` Prov(pA⇤q) ! Prov(p¬S1q), and contra-posing againgives

PA ` ¬Prov(p¬S1q) ! ¬Prov(pA⇤q).

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By (C5) for 0R1, we know PA ` S0 ! ¬Prov(p¬S1q), so then PA ` S0 ! ¬Prov(pA⇤q), asrequired. End of Claim (C9).

Note that the remainder of the proof cannot be formalised in PA. Since PA is sound, we knowthat every theorem of PA is true in the standard model. By (C6) we know that if i > 1 and Si istrue, then Prov(p¬Siq) is true, hence ¬Si is a provable in PA and ¬Si is true. Therefore, for all1 6 i 6 n Si cannot be true, otherwise we come to the contradiction just discovered. Recall that(C4) states PA ` S0 _ · · · _ Sn, so S0 _ · · · _ Sn is true and because no Si can be true for i > 1 wemust have S0 is true. Then by (C9), ¬Prov(pA⇤q) is true i.e. PA 0 A⇤. This finishes the proof ofSolovay’s Arithmetical Completeness of GL.

In particular,

PA `nV

i=1[Prov(p¬Siq) ! ¬Si] ) PA `

nVi=1

¬Si by Löb’s Theorem

) PA ` (S0 _ · · · _ Sn) ^nV

i=1¬Si by (C4)

) PA ` S0 Propositional calculus) PA ` ¬Prov(pA⇤q) by (C9).

Therefore, the fact that PA 0 A⇤ follows from a finite conjunction of reflection principles, namelythe reflection principles of the Solovay sentences for each 1 6 i 6 n.

However, there seems to be part of this proof which would be nice to improve. The end result,i.e. PA 0 A⇤, relied on this specifically defined realization ⇤, so is there a single realization whichtreats all such modal sentences A with GL 0 A? The answer is yes, and is handled by the nexttheorem due to Artemov, Avron, Boolos, Montagna, and Visser.

Theorem 4.14. [2, pp. 132-4] (Uniform Arithmetical Completeness of GL) There exists arealization ⇤ such that for any modal sentence A, PA ` A⇤ ) GL ` A.

Proof. We give a proof following [2, pp. 132-4]. We prove this in a similar fashion to Theorem 4.12.Firstly, we make the following observation: suppose A is a modal sentence such that GL 0 A, thenif A is a string of k symbols there is a finite, transitive, converse wellfounded model hW,R,�i inwhich A is not valid; in particular, as noted before, |W | = n 6 2k, and |R| 6 2(2

k·2k) = 222k , because

wRx ) 1 6 w, x 6 n 6 2k; and finally | � | 6 22k·k = 2k2

k , so there are no more than 2k · 222k · 2k2k

possibilities for such a hW,R,�i.Now we let Q(x, y) be a ⌃1 pterm for a total function that codes the 5-tuple hWk, Rk,�k, wk, Aki

for each k 2 N, and setting Mk := hWk, Rk,�ki gives that

(i) Wk is a finite set of strictly positive natural numbers;

(ii) Wk = {wk} [ {i | wkRki} and Rk is transitive and irreflexive, and hence converse wellfoundedas Wk is finite;

(iii) if i 2 Wk then for any propositional letter p not in Ak, we have Mk, i 1k p;

(iv) Mk, wk 1 Ak;

(v) if m 6= n, then Wm and Wn are completely disjoint;

67


(vi) for each i 2 N, 1 6 i, we have i 2Sk2N

Wk; and

(vii) if A 2 SENTL⇤ is not a theorem of GL, then there exists k 2 N such that A ⌘ Ak.

Now we set R0 :=Sk2N

Rk [ {h0, ii | i 2 Wk for some k 2 N}.

Claim. R’ is transitive and converse wellfounded.Proof of Claim. Suppose for some w, x, y that wR0x and xR0y. We need to show wR0y. wR0x

implies either w = 0 or w 2 Wk for some k 2 N, and xR0y gives either x = 0 or x 2 Wn for somen 2 N. If w = 0 then x 6= 0 as h0, 0i /2 R0, so x 2 Wk giving y 2 Wk by (v) above, and in particularh0, yi 2 R0 by definition of R0. Now suppose w 6= 0, then w 2 Wk for some k and x 6= 0. By (v), wemust have x, y 2 Wk, so then wRkxRky and since Rk is transitive we have wRky, i.e. wR0y.

Let X be a non-empty subset of R0. We must show that there exists a world in X which canaccess no other world via R0. Since X ✓ R0 and X 6= ;, then there exists some � 2 X. Whenever� = hw, xi 2 X for some k with hw, xi 2 Rk, then since Rk is converse wellfounded, we know thereexists an Rk-greatest element, yk say, but then since R0 is transitive we know that for no z 2 Wk dowe have hyk, zi 2 X. Further, we certainly do not have hyk, 0i 2 X by definition of R0 and by (v)we do not have ykR0a for any a /2 Wk.

Now suppose � = h0, ii 2 X for some i 2 Wk for some k. i cannot access 0, so at best iRkj forsome j 2 Wk. Just as before, we obtain an Rk-greatest element. End of Claim.

Since R0 is converse wellfounded, we know then that it is also irreflexive (recall that this impli-cation does not require finiteness) and hence for no i 2 N do we have hi, 0i 2 R0. Observe also thatif for some i > 1 iR0j, then i 2 Wk for some k and by (v) j 2 Wk, but also Wk finite so iR0j foronly finitely many j 2 Wk. Define �0:=

Sk2N

�k.

Set R(x, y) to be a �1 formula defined from Q(x, y) with the following properties:

(i) PA ` R(0, y) $ y 6= 0

(ii) for each i > 1, PA ` R(i, y) $W

j:iR0jy = j

(iii) PA ` 8x8y8z[R(x, y) ^R(y, z) ! R(x, z)]

(iv) PA ` 8x[8y(R(x, y) ! '(y)) ! '(x)] ! 8x'(x)

(i) and (ii) tell us that R(x, y) arithmetically defines R0 in PA; note that since R(x, y) is �1 weknow iR0j ) PA ` R(i, j) and ¬(iR0j) ) PA ` ¬R(i, j). (iii), (iv) tell us that R(x, y) is transitiveand converse wellfounded, respectively.

Now let ex(x1, x2) be a ⌃1 pterm which arithmetically defines the function Ext(x1, x2) given by

Ext(m, r) =

(j if r = p¬9N8a[a > N ! 9b(b = j ^ Fm)]q0 else

where m = pFmq. Then we can choose ex(x1, x2) and nonlim(x1, ex(x1, x2)) in such a way thatPA ` ex(x1, x2) 6= 0 ! Pf(x2, nonlim(x1, ex(x1, x2))). Using the Diagonal Lemma on a formula

68


similar to that in the proof of Theorem 4.12, we obtain a ⌃1 formula G(r, b) with Gödel number gthat defines the following function

h(0) = 0

h(r + 1) =

(j if h(r) = i ^ iR0j ^ Pf(r, pl 6= jq)h(r) else

where l is the limit of h. Set S(x) := 9N8a[a > N ! G(a, x)], and thus for each j 2 N S(j) assertsthat the limit of this function h is j. We now prove a series of claims just as before, but omittingparts which are similar to (C1)-(C9) above.

Claim (C1’). PA ` S(i) ^ S(j) ! i = j.Proof of Claim (C1’). PA ` 9!bG(r, b) which we can prove by induction as before.

End of Claim (C1’).

Claim (C2’). PA ` G(r, x) ! S(x) _ 9y[R(x, y) ^ S(y)].Proof of Claim (C2’). We prove this by induction on the formalised converse of R0. Let

'(x) ⌘ G(r, x) ! S(x) _ 9y[R(x, y) ^ S(y)].

Suppose 8y[R(x, y) ! '(y)] and G(r, x). We can prove the following formally in PA, but we give aninformal argument. Recall the definition of h above, then from that it is clear that for each d > r,either h(d) = x or h(d) = y for some y such that xR0y. That is,PA ` 8d[d > r ! G(d, x) _ 9y[R(x, y) ^ G(d, y)]]. Since we assume 8y[R(x, y) ! '(y)], we alsohave that

PA ` 8d[d > r ! G(d, x) _ 9y[R(x, y) ^ (S(y) _ 9z[R(y, z) ^ S(z)])]].

This gives PA ` S(x) _ 9y[R(x, y) ^ (S(y) _ 9z[R(y, z) ^ S(z)])], and then we can deducePA ` S(x) _ 9y[R(x, y) ^ S(y)] using (C1’). Therefore, by induction on the converse of R0 we aredone.


Claim (C3’). PA ` 9xS(x).Proof of Claim (C3’). From (C2’), PA ` G(r,0) ! S(0)_9y[R(0, y)^S(y)], and since h(0) = 0

we know PA ` G(0,0). These two provable statements give us PA ` 9xS(x) since h0, ii 2 R0 forall i 2 N.


Claim (C4’). If iR0j, then we have PA ` S(i) ! ¬Prov(p¬S(j)q).Proof of Claim (C4’). This is entirely similar to (C5) and we omit the proof.


Claim (C5’). 8i > 1, we have PA ` S(i) ! Prov(p¬S(i)q).Proof of Claim (C5’). Let i > 1 be arbitrary. Assume G(r, i), then we must have r > 0 because

we know PA ` G(0,0) and i > 0. We argue the following informally but it may be formalisedin PA. Since G(r, i) holds, then h(r) = i > 0 and so r � 1 must have been the Gödel numberof a proof that the limit of h is not i, that is a proof of ¬S(i). Thus, Prov(p¬S(i)q) is true

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and ⌃1, so PA ` Prov(p¬S(i)q). Hence, PA ` S(i) ! Prov(p¬S(i)q) because PA can proveS(i) ⌘ 9N8a[a > N ! G(a, x)] implies 9xG(x, i) implies 9yPf(y, p¬S(i)q). End of Claim (C5’).

Claim (C6’). 8i > 1, we have PA ` S(i) ! Prov(p Wj:iR0j

S(i)q).

Proof of Claim (C6’). Note that PA ` 9rG(r, i) ! G(m, i) for some m, so by (C2’) we have

PA ` 9rG(r, i) ! S(i) _ 9y[R(i, y) ^ S(y)].

Then by (ii) in our definition of R(x, y),

PA ` 9rG(r, i) ! S(i) _ 9y

0

@_

j:iR0j

y = j ^ S(y)

1

A .

Simplifying, we get

PA ` 9rG(r, i) ! S(i) _

0

@_

j:iR0j

S(j)

1

A .

Then an argument similar to that in the proof of (C7) produces our desired result.End of Claim (C6’).

Recall that �0:=Sk2N

�k. Set M0 := hW 0, R0,�0i where W 0 :=Sk2N

Wk[{0}. Consider the relation

{hi, ni | hi, pni 2�0} ✓ N2. Then we can find a �1 formula from Q(x, y) that arithmetically definesthe relation; let this formula be V (x, y). Now define the realization ⇤ by p⇤n = 9x[S(x) ^ V (x,n)]8n 2 N.

Claim (C7’). 8k 2 N and for any subsentence B of Ak and each i 2 Wk

(i) if Mk, i �k B, then PA ` S(i) ! B⇤

(ii) if Mk, i 1k B, then PA ` S(i) ! ¬B⇤.

Proof of Claim (C7’). We prove this by induction on the form of B, a subsentence of Ak. Letk 2 N be arbitrary and let i 2 Wk be some world. We will let Mk := hWk, Rk,�ki as before.

Firstly, suppose B ⌘ pn for some n 2 N, so B⇤ = p⇤n = 9x[S(x) ^ V (x,n)]. If Mk, i �k pn,then M0, i �0 pn and in which case PA ` V (i,n). Trivially, PA ` S(i) ! 9x[x = i ^ S(x)]and PA ` V (i,n) ! 9x[x = i ^ V (x,n)], and thus PA ` S(i) ! 9x[S(x) ^ V (x,n)]. Hence,PA ` S(i) ! B⇤. On the other hand, if Mk, i 1k pn then M0, i 10 pn. So PA ` ¬V (i,n) as V (x, y)is �1, thus by (C1’) PA ` S(i) ! ¬9x[S(x) ^ V (x,n)], i.e. PA ` S(i) ! ¬B⇤.

The case of B ⌘ C ^ D and B ⌘ ¬C are just like that in (C8). Suppose B ⌘ ⇤C for somesubsentence C. Below we use the fact that since i 2 Wk, if iRkj then not only do we have j 2 Wk

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but also iR0j.

Mk, i �k B , Mk, i �k ⇤C, 8j with iRkj Mk, j �k C (definition of truth in a model)) 8j with iRkj PA ` S(j) ! C⇤ by the induction hypothesis) PA `

Wj:iR0j

S(j) ! C⇤ propositional calculus

) PA ` Prov

p Wj:iR0j

S(j) ! C⇤q!

(P1)

) PA ` Prov

p Wj:iR0j

S(j)q!

! Prov(pC⇤q) (P2)

) PA ` S(i) ! Prov(pC⇤q) (C6’)) PA ` S(i) ! B⇤

Mk, i 1k B , Mk, i 1k ⇤C, 9j > 1 with iRkj Mk, j 1k C (definition of truth in a model)) PA ` S(j) ! ¬C⇤ by the induction hypothesis) PA ` C⇤ ! ¬S(j) contra-posing) PA ` Prov(pC⇤ ! ¬S(j)q) (P1)) PA ` Prov(pC⇤q) ! Prov(p¬S(j)q) (P2)) PA ` ¬Prov(p¬S(j)q) ! ¬Prov(pC⇤q) contra-posing) PA ` S(i) ! ¬Prov(pC⇤q) (C4’) since i, j > 1

and iRkj , iR0j) PA ` S(i) ! ¬B⇤.

Thus, by induction on the form of B we are done. End of Claim (C7’).

Suppose GL 0 A, then by our definition of the function which Q(x, y) defines, we must havethat A ⌘ Ak for some k 2 N. So Mk, wk 1k Ak, i.e. Mk, wk 1k A, where wk is the positivenatural number found in the 5-tuple hWk, Rk,�k, wk, Aki associated to k. Then by (C7’), we knowPA ` S(w

k

) ! ¬A⇤ because A is a subsentence of itself and wk 2 Wk. Then contra-posing,applying (P1), (P2) and contra-posing again we obtain: PA ` ¬Prov(p¬S(w

k

)q) ! ¬Prov(pA⇤q).By our construction of R0, we know 0R0wk and so PA ` S(0) ! ¬Prov(p¬S(w

k

)q) by (C4’). Thus,PA ` S(0) ! ¬Prov(pA⇤q), and by the soundness of PA we know S(0) ! ¬Prov(pA⇤q) is in facttrue.

Set i = wk, then (C3’) and (C2’) tell us that S(i) is true for just one i. Suppose i > 1, thenfrom (C5’) we know PA ` S(i) ! Prov(p¬S(i)q), so S(i) ! Prov(p¬S(i)q) is true. Then if S(i)is true we will have that Prov(p¬S(i)q) is true. Therefore, PA ` ¬S(i) and ¬S(i) is true, and wehave come to a contradiction. Hence, we must have i = 0 and so S(0) is true. Thus, ¬Prov(pA⇤q)is true and PA 0 A⇤.

We now have two proofs of the Arithmetical Completeness of GL, but there is something stillsignificant to be discussed. The application of these results is definitely worth mentioning. Aswe saw earlier, the unprovability of the modal sentence A is given to us by not all the reflectionprinciples we can formulate in PA, but by a conjunction of a finite number of them. We discussed

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in §1.1 how PA can be interpreted in ZF set theory, and by choosing an appropriate interpretationthis conjunction of reflection principles can be proved in ZF [2, p. 166]. That is, we will havea ZF set-theoretical derivation for each conjunct, and hence the conjunction. Moreover, sincePA `

nVi=1

[Prov(p¬Siq) ! ¬Si] we have a translation ofnV

i=1[Prov(p¬Siq) ! ¬Si] into the language

of ZF set theory, and together with the provable conjunction we can actually derive (in ZF) that A⇤

is not provable in PA. That is, the assertion ¬ProvPA

(pA⇤q) (under an appropriate interpretationinto the language of ZF) is derivable in ZF.

All this being well, we encounter the same problem as we did for PA for ZF: the formal unprov-ability of A⇤ in ZF is a consequence of a conjunction of reflection principles in the language of ZF,where the conjunction cannot be proved formally in ZF so long as ZF is consistent [2, p. 166].

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References

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[2] Boolos, G., The Logic of Provability, Cambridge University Press 1993

[3] Boolos, G., J. Burgess and R. Jeffrey, Computability and Logic, Cambridge University Press2007, 5th Edition

[4] Dean, W., Logic II: Metatheory [Lecture Notes], University of Warwick Fall 2010

[5] Dean, W., Logic III: Incompleteness and Undecidability [Lecture Notes], University of WarwickSpring 2012

[6] Dean, W., Modal Logic [Lecture Notes], University of Warwick Spring 2011

[7] Epstein, A., Set Theory [Lecture Notes], University of Warwick Fall, 2012

[8] Fitting, M. and R. Mendelsohn, First-Order Modal Logic, Kluwer Academic Publishers 1998

[9] Hájek, P. and P. Pudlák, Metamathematics of First-Order Arithmetic, Springer Verlag 1998

[10] Kaye, R., Models of Peano Arithmetic, Clarendon Press 1991

[11] Kaye, R., The Mathematics of Logic, Cambridge University Press 2008

[12] Kleene, S., Introduction to Metamathematics, D. Van Nostrand Co., Inc. 1952

[13] Lindström, P., Aspects of Incompleteness, Springer Verlag 1997

[14] Mackie, J., Ethics: Inventing Right and Wrong, Penguin Books 1990

[15] Marker, D., Model Theory: An Introduction, Springer Verlag 2002

[16] Smith, P., An Introduction to Gödel’s Theorems, Cambridge University Press 2009, 4th Edition

[17] Smoryński, C., Self-Reference and Modal Logic, Springer Verlag 1985

[18] Smullyan, R., First-Order Logic, Dover Publications, Inc. 1994

[19] Solovay, R., “Provability Interpretations of Modal Logic,” Israel Journal of Mathematics 25

(1976), 287-304

[20] van Dalen, P., Logic and Structure, Springer Verlag 2008, 4th Edition

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shah - solovay's arithmetical completeness for provability logic

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