provability with finitely many variables

The Bulletin of Symbolic Logic

Volume 8, Number 3, Sept. 2002

PROVABILITY WITH FINITELY MANY VARIABLES

ROBIN HIRSCH†, IAN HODKINSON∗, AND ROGER D. MADDUX

Abstract. For every finite n ≥ 4 there is a logically valid sentence ϕn with the following

properties: ϕn contains only 3 variables (each of which occurs many times); ϕn contains

exactly one nonlogical binary relation symbol (no function symbols, no constants, and no

equality symbol); ϕn has a proof in first-order logic with equality that contains exactly n

variables, but no proof containing only n − 1 variables. This result was first proved using the

machinery of algebraic logic developed in several research monographs and papers. Here we

replicate the result and its proof entirely within the realm of (elementary) first-order binary

predicate logic with equality. We need the usual syntax, axioms, and rules of inference to

show that ϕn has a proof with only n variables. To show that ϕn has no proof with only

n − 1 variables we use alternative semantics in place of the usual, standard, set-theoretical

semantics of first-order logic.

§1. Introduction. Questions about what can be expressed and proved withonly finitely many variables began with Peirce’s [24] calculus of binary re-lations, as developed by Schroder [25]. They let letters a, b, c, . . . denotebinary relations on some fixed underlying setU (called the “domain”, “uni-verse of discourse”, or “Denkbereich”). New relations are made from theseaccording to operations chosen by Peirce: the union a + b, the intersectiona · b, the complement a, the relative product a ;b, the relative sum a † b,and the converse a. There are four distinguished relations: 1 is the universalrelation on U (namely U × U ), 0 is the empty relation, 1’ is the identityrelation on U , and 0’ is the diversity relation on U , consisting of all pairsof distinct elements of U . Peirce and Schroder write, for example, aij todenote the statement that element i is related to element j by relation a. Theunion, intersection, complement, relative product, sum, converse, and thefour distinguished relations can then be defined as follows:

(a + b)ij iff aij or bij ,

(a · b)ij iff aij and bij ,

Received May 3, 2000; revised May 8, 2002.†Research partially supported by UK EPSRC grants GR/K54946, GR/L82441,

GR/L85961.∗Research partially supported by UK EPSRC grants GR/K54946, GR/L85978.

c© 2002, Association for Symbolic Logic

1079-8986/02/0803-0002/$4.20

348

PROVABILITY WITH FINITELY MANY VARIABLES 349

aij iff not aij ,

(a ;b)ij iff for some k ∈ U, aik and bkj ,

(a † b)ij iff for every k ∈ U, aik or bkj ,

(a)ij iff aji ,

it is never the case that 0ij ,

it is always the case that 1ij ,

1’ij iff i = j,

0’ij iff i 6= j.

Elementary statements about relations and the domain can be made byequations between relations. Here are some examples. The domain has atleast two elements iff 1 = 1;0’;1. The domain has at least three elementsiff 1 = 1;(0’ · (0’;0’));1. If the domain is not empty, then the relation a isnonempty iff 1 = 1;a ;1. The relation a is transitive and contains the iden-tity relation on the domain iff a = a † a. Every equation in the calculus ofrelations can be translated into an elementary statement about the domainand the binary relations used in the equation. For example, the associativelaw for relative multiplication, namely

a ;(b ;c) = (a ;b);c,(1)

is equivalent to the following sentence, which uses only three variables de-noting elements of the domain (namely, i , j, and k):

∀i∀j(∃k(aik ∧ ∃i(bki ∧ cij))⇔ ∃k(∃j(aij ∧ bjk) ∧ ckj)).(2)

It is straightforward to prove that every equation in the calculus of relationsis equivalent to a 3-sentence, that is, a first-order sentence that uses onlythree variables. The converse also holds: every 3-sentence is equivalent toan equation in the calculus of binary relations [28, §3.9], [4, p. 204]. Evenif a sentence uses more than three variables, it may still be expressible asan equation if it can be equivalently rewritten in a way that uses only threevariables. An example for which this is true but far from obvious is the“pairing axiom”, ∀i∀j∃k∀h(ahk ⇔ 1’hi ∨ 1’hj). Although it contains fourvariables in the usual formulation, it is equivalent to a 3-sentence, and so toan equation in the calculus of relations [28, p. 137].Schroder seems to have been the first to consider the problem of whetherevery sentence can be expressed by an equation [25, p. 551]. This problemwas solved by Alwin Korselt. In a letter to Lowenheim [12], Korselt provedthat no equation is equivalent to the following sentence, which asserts thatthe domain has at least four elements:

∃h∃i∃j∃k(0’hi ∧ 0’hj ∧ 0’hk ∧ 0’ij ∧ 0’ik ∧ 0’jk).(3)

350 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D. MADDUX

Since every 3-sentence can be translated into an equation, it follows that (3)is a 4-sentence that is not equivalent to any 3-sentence. Korselt’s theoremwas considerably generalized by Tarski [26, p. 89], [28, p. 61].Tarski proposed several equations as axioms for the calculus of relations,including the six equations listed below. These equations, together withthe associative law (1) and any finite set of equational axioms for Booleanalgebras, axiomatize the class of relation algebras.

(a + b);c = a ;c + b ;c,(4)

a ;1’ = a,(5)

˘a = a,(6)

(a + b)˘= a + b,(7)

(a ;b)˘= b ; a,(8)

a ;a ;b + b = b.(9)

A 3-sentence equivalent to the last equation is

∀i∀j(∃k(aki ∧ ∀i(aki ∨ bij))⇒ bij).(10)

This sentence is logically valid, and has a proof in any complete systemof first-order logic. A proof of (10), in a natural deductive style, goes asfollows. Let i and j be arbitrary elements of the domain. Assume there issome element k such that aki and for every i , either aki or bij . The latterassumption, applied to the arbitrary i we are considering, implies that eitheraki or bij . Butaki , so bij . Wehave so far proved that∃k(aki∧∀i(aki∨bij))⇒

bij . Since we proved this for arbitrary i and j, we may generalize andconclude that (10) holds. Note that we only needed the three variablesthat actually occur in (10). We suspect that, for practically any textbookaxiomatization of first order logic (with or without equality), (10) will stillbe provable if the axiom set is restricted to those formulæ that do not containany variables outside a fixed group of three variables. Principles involvingequality aren’t needed to prove (10), but they are used in the proof of thefollowing 3-sentence, which is equivalent to (5):

∀i∀j(∃k(aik ∧ 1’kj)⇔ aij)).(11)

To prove this, let i and j be arbitrary elements of the domain. For one direc-tion, assume there is some k such that aik and 1’kj . By an axiom assertingthat equality is a congruence relation (that is, an equivalence relation onthe domain with the property that aik and 1’kj imply aij , and akj and 1’kiimply aij), this gives us aij . For the converse, assume aij . Assuming thateverything is equal to something, we conclude there is some k such that 1’jk .Since equality is a congruence relation, it follows that aik and 1’kj , hence


∃k(aik ∧ 1’kj). So far we have proved ∃k(aik ∧ 1’kj) ⇔ aij . Generalizing,we obtain (11).Similar 3-variable proofs can be given for the 3-sentences obtained from(4), (6), (7), and (8). On the other hand, while it is easy to prove (2)using four variables, the difficulties encountered when trying to do it withonly three led Tarski to explicitly include (2) in the axiomatization of the3-variable logic L+3 defined in [28], where the problem of axiomatizing n-variable logic for n ≥ 3 is discussed at length. Let Ls+3 be obtained from L+3by deleting (2) [28, p. 89]. Tarski established that (2) is not provable in Ls+3by considering an algebra, constructed by J. C. C. McKinsey, that satisfies(4)–(9): i.e., all of the equational axioms for relation algebras except theassociative law for relative multiplication (1). See [28, p. 68]; Tarski’s proofwas never published, but another proof was given by Henkin [9].Four variables are sufficient to prove not only the 3-sentences obtainedfrom the equational axioms (1) & (4)–(9) for relation algebras, but alsothe 3-variable translations of all equations derivable from these axioms. Itfollows that if an equation is true in every relation algebra then its 3-variabletranslation has a 4-variable proof. It turns out that the converse is alsotrue: if the 3-variable translation of a given equation has a 4-variable proof,then the equation holds in every relation algebra [14], [16], [17], [28, pp. 92–93], [4, pp. 206–207]. Therefore, if an equation E fails in some relationalgebra and its equivalent 3-sentence T is logically valid, then T is provablebut not with fewer than five variables. Here is an example of an equationwhose equivalent 3-sentence is logically valid and provable with five variablesbut not four [28, pp. 54, 93], [4, p. 198]:

1;((a † a) + [(a ;a) + 1’ + (a + a ;a + a) · a] · a + a ; a) ;1 = 1.(12)

Both this equation and its equivalent 3-sentence use only a single binaryrelation symbol, because they arise from the analysis of a 16-element non-representable relation algebra that is generated by a single element [18].Suppose a 3-sentence T is logically valid, hence provable using somenumber of variables in some textbook axiomatization of first order logicwith equality. Let n be the least number of variables that occur in any proofof T . Then T requires n variables to prove, in the sense that no smallernumber of variables will suffice; in particular, T has no proof that uses onlyn − 1 variables. It has been known for some time that there are 3-sentencesthat require arbitrarily large numbers of variables to prove, that is, for everyinteger n, there is a 3-sentence that requires more than n variables to prove.Monk [21, Th 1.11] proved an algebraic formulation of this result in thecourse of showing that the class of n-dimensional cylindric algebras is notfinitely axiomatizable whenever n ≥ 3. His proof involves the constructionof cylindric algebras that were modified into nonrepresentable polyadic al-gebras by Johnson [11]. Monk [22] outlined the connections between the

https://www.researchgate.net/publication/38323308_Representations_of_integral_relation_algebras?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/242377431_A_Formalization_of_Set_Theory_Without_Variables?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/266002505_Nonfinitizability_of_classes_of_representable_polyadic_algebras?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==


algebra and logic, and applied one of Johnson’s results to show that no finiteset of axiom schemata for first-order logic with equality and only finitelymany variables is both sound and complete. Monk [21, §4] made a conjec-ture about cylindric algebras. In logical terms, Monk’s conjecture was thatfor every n ≥ 3 there is a 3-sentence that requires n variables to prove. Boththe algebraic and logical versions of the conjecture are confirmed in [10]. Itis perhaps an interesting historical fact that the algebras constructed alreadyby Monk [21] can be used to do so. These algebras cannot, however, beused to obtain similar results for languages with a fixed, finite number ofbinary relation symbols. The reason is that the number of generators ofone of Monk’s algebras depends on the size of the algebra; larger algebrasrequire more generators. The same feature afflicts both the nonrepresentablerelation algebras that were constructed from projective geometries by Lyn-don [13] and used by Monk [19] to show that the class of representablerelation algebras is not finitely axiomatizable, and also the corresponding3-dimensional cylindric algebras that were used by Monk [20] to show thatthe class of 3-dimensional cylindric algebras is not finitely axiomatizable.The first-order language L used in [28] has an equality symbol and exactlyone binary relation symbol, the appropriate choice for the development ofset theory, in which the sole nonlogical predicate is the membership relation.To prove for this language that there are 3-sentences that require arbitrarilylarge numbers of variables to prove, it was necessary to modify Monk’salgebras into ones that are generated by a single element [17, §4]. Thisintroduces complexity into the construction; the automorphism group goesfrom large to trivial. Similarly modified versions of Monk’s algebras areused in [10] to show that for every n ≥ 3 there is a 3-sentence in L thatrequires n variables to prove, thus solving a problem posed by Tarski andGivant [28, p. 93], [4, p. 208], [23, p. 735].Here we replicate this result and its proof entirely within the realm offirst-order logic. The structure of the modified algebras shows up in the3-sentences. Our proof proceeds as follows. In §2, we recall a finite Ramseytheoremwhose formalization in first-order logic will provide an upper boundon the number of variables required to prove our 3-sentences. The n-variablefragments of first-order logic, and their proof theory, are defined formallyin §3 and §4. Some proof-theoretic lemmas are given in §5. The main workbegins in §6, where the ‘Ramsey 3-sentence’ (for each n ≥ 4) is specifiedand shown (in §7) to be provable with n variables. In §8, we show that it isnot provable with fewer variables by showing that it is not valid in a certainunorthodox semantics for (n − 1)-variable logic that validates theoremsprovable with n − 1 variables. Finally, in §9, we obtain a similar sentencewritten with only one binary relation symbol.We believe most of this paper can be read as if it were a chapter in a (fairlyterse and sophisticated) text on first-order logic with equality.

https://www.researchgate.net/publication/38322976_On_Representable_Relation_Algebras?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/220315228_Relation_Algebra_Reducts_of_Cylindric_Algebras_and_An_Application_to_Proof_Theory?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/220315228_Relation_Algebra_Reducts_of_Cylindric_Algebras_and_An_Application_to_Proof_Theory?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/242377431_A_Formalization_of_Set_Theory_Without_Variables?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==

https://www.researchgate.net/publication/38322822_Relation_algebras_and_projective_geometries?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==


§2. A Ramsey theorem. Define ñ : ù → ù by ñ(0) = 1 and ñ(k + 1) =1+ (k+1) · ñ(k) for every k < ù. This function gives an upper bound for acertain Ramsey number [7], [6, p. 6], [8, Cor 3], [2, p. 440–443]. Its first fivevalues are, respectively, 1, 2, 5, 16, and 65. For any set X , let [X ]2 be the setof 2-element subsets of X :

[X ]2 = {{x, y} : x, y ∈ X, x 6= y}.

For 1 ≤ k < ù, a k-coloring of X is a partition of [X ]2 into k pieces (called‘colors’) so that no ‘monochromatic triangle’ appears, that is, there is no3-element subset of X with all three of its 2-element subsets in the samepiece of the partition.

Proposition 2.1. Let k ≥ 1. If a set X has more than ñ(k) elements thenX has no k-coloring.

Proof. Here is the standard proof. First case: k = 1. No 1-coloring canexist if X has more than ñ(1) = 2 elements, because every 3-element subsetis a monochromatic triangle for every 1-coloring. Next, we assume the resultholds for a particular k ≥ 1 and prove it for k+1. Assume X has a (k + 1)-coloring. Choose such a coloring. Fix one element x ∈ X , and define anequivalence relation∼ on X \ {x} by letting y ∼ y ′ if and only if {x, y} and{x, y ′} have the same color in the chosen coloring. For distinct y ∼ y ′ inX \{x}, if {x, y} and {x, y ′} belong to color c, say, then {y, y ′} cannot alsobelong to c, else {x, y, y ′} ismonochromatic. Hencewe can partitionX \{x}into at most k + 1 parts, given by the ∼-classes, and within each part onlyk colors are used. Thus, each of the parts has a k-coloring and inductivelyhas size not exceeding ñ(k). So |X | ≤ 1 + (k + 1) · ñ(k) = ñ(k + 1). ⊣

Partitions of [X ]2 can be treated in first-order logic by means of symmet-ric binary relations on X . In a first-order language with binary relationsymbols and equality, it is possible with 2-sentences to assert that the re-lations (denoted by the relation symbols) are symmetric, pairwise disjoint,and their union is the diversity relation (that holds between any two distinctelements). That monochromatic triangles do not occur can be expressed by3-sentences, such as ∀i∀j∀k¬

(aij ∧ aik ∧ akj

). Thus, for any k ≥ 1, there is

a single 3-sentence that is satisfiable in X just in case some set of k binaryrelations on X forms a partition of the diversity relation on X into disjointsymmetric binary relations with no monochromatic triangles—in short, thatthere is a k-coloring ofX . By Proposition 2.1, such a sentence has no modelof cardinality greater than ñ(k). By considering more complicated types ofpartitions, we can further assert with only 3 variables that there are morethan ñ(k) elements. A 3-sentence obtained this way, asserting the existenceof a k-coloring on a set with more than ñ(k) elements, has no model, so itsnegation is logically valid and provable. It is easy to see that it can be provedwith 1 + ñ(k) variables (Remark 6.2 below gives a corresponding semantic


argument). The aim of the next few sections is to get this number down, byformalizing the inductive proof of Proposition 2.1.

§3. Relational languages. Let L be a first order language with equalitysymbol =, any set of binary relation symbols, no function symbols, noconstants, and variables vi for i < ù. For every n ≤ ù let

Varn = {vi : i < n}.

Let Rel be the set of relation symbols of L. (The equality symbol = is not inRel.) Let atomicn be the set of atomic formulæ of L written with variablesfrom Varn. Thus

atomicn = {vi=vj , viRvj : R ∈ Rel, i, j < n}.

The only propositional connectives inL are¬ and⇒, and the only quantifieris ∀. The other connectives and the existential quantifier are introducedas abbreviations in the usual way. We will need to use conjunctions anddisjunctions of formulæ indexed by various finite index sets. The order inwhich the formulæ appear is assumed to be determined by some standardordering of the index set. Parentheses are restored by association to the left.For example,

∨

i≤k

ϕi = (· · · ((ϕ0 ∨ ϕ1) ∨ ϕ2) ∨ · · · ∨ ϕk−1) ∨ ϕk .

At one point (Lemma 7.4) we encounter a disjunction over a possibly emptyindex set, in which case we let

∨

i∈∅ ϕi = ⊥, where ⊥ = ¬∀v0(v0=v0).Similarly,

∧

i∈∅ ϕi = ⊤, where ⊤ = ∀v0(v0=v0).For an L-formula ϕ, let free(ϕ) be the set of variables that occur free in ϕ,defined for any R ∈ Rel, i, j < n, and formulæ ϕ,ø, by:

free(vi=vj) = {vi , vj},

free(viRvj) = {vi , vj},

free(ϕ ⇒ ø) = free(ϕ) ∪ free(ø),

free(¬ϕ) = free(ϕ),

free(∀viϕ) = free(ϕ) \ {vi}.

A formula ϕ is a sentence if free(ϕ) = ∅. For every n ≤ ù let Fmn be the setof formulæ of L that contain only variables in Varn, i.e.,

Fmn =⋂

{X : atomicn ⊆ X, if ϕ ∈ X, ø ∈ X, and x ∈ Varn,

then ¬ϕ ∈ X, ϕ ⇒ ø ∈ X, and ∀xϕ ∈ X}.

We say that ϕ is an n-formula if ϕ ∈ Fmn, and an n-sentence if ϕ ∈ Fmn andfree(ϕ) = ∅.


§4. Axiomatization. We now develop the proof theory of L. For ev-ery ϕ ∈ Fmù and all i, j < ù let Sijϕ be the result of interchanging viand vj , that is, simultaneously replacing every occurrence of vi in ϕ withvj and every occurrence of vj in ϕ with vi . For example, if R ∈ Rel

then S01(∀v0∃v1v0Rv1) = ∀v1∃v0v1Rv0 and S02(∀v0∃v1v0Rv1) = ∀v2∃v1v2Rv1.Note that if i, j < n and ϕ ∈ Fmn, then Sijϕ ∈ Fmn as well. A tautologyis a formula which is mapped to 1 (truth) by the canonical extension to allformulæ of any map from atomic and universally quantified formulæ to thetwo-element set {1, 0}. For every n ≤ ù, let Axn be the set of all formulæ ofthe following types.

Name Axiom Restrictions

(A1) ϕ ϕ ∈ Fmn and ϕ is a tautology

(A2) ∀vi (ϕ ⇒ ø)⇒ (ϕ ⇒ ∀viø) ϕ,ø ∈ Fmn, i < n, and vi /∈ free(ϕ)

(A3) ∀viϕ ⇒ ϕ ϕ ∈ Fmn and i < n

(A4) ∃vi (vi=vj) i, j < n and i 6= j

(A5) (vi=vj)⇒ (ϕ ⇒ ø) ϕ,ø ∈ atomicn, i, j < n, and ø is

obtained from ϕ by replacing some

occurrence of vi in ϕ by vj

(A6) (vi=vj)⇒ (ϕ ⇒ Sijϕ) ϕ ∈ Fmn and i, j < n

The rules of inference are modus ponens (infer ø from ϕ and ϕ ⇒ ø) andn-generalization (infer ∀xϕ from ϕ, for any x ∈ Varn). These rules can bedisplayed in a more graphical style as follows.

ϕ ϕ ⇒ øø

ϕ∀xϕ

A formula ϕ ∈ Fmù is n-provable, in symbols ⊢n ϕ, if ϕ belongs to everyset of formulæ that contains Axn and is closed under modus ponens and n-generalization. As Fmn is such a set, if ⊢n ϕ then ϕ ∈ Fmn. By an n-proofof ϕ we mean a finite sequence of formulæ in Fmn whose last element is ϕ,having the property that every formula in the sequence is either a memberof Axn or follows from one or two previous formulæ by n-generalizationor modus ponens. Evidently, ϕ is n-provable iff there exists an n-proof ofϕ. Two formulæ ϕ,ø ∈ Fmn are n-equivalent, in symbols ϕ ≡n ø, iff⊢n ϕ ⇔ ø.The axiom set (A1)–(A6) is based on the axiom set Σ2 of Tarski [27]. Σ2 isdefined by eight axiom schemata, (B1) through (B8). Tarski’s axioms (B1)–(B3) are tautologies from which every other tautology can be obtained bymodus ponens. Axiom (A1) is our replacement for (B1)–(B3). Axiom (A2)


substitutes for the following two axioms (see Lemmas 5.5 and 5.4 below):

(B4) ∀x(ϕ ⇒ ø)⇒ (∀xϕ ⇒ ∀xø),

(B6) ϕ ⇒ ∀xϕ for x /∈ free(ϕ).

Tarski’s (B5), (B7), and (B8) coincide with our (A3), (A4), and (A5), respec-tively. Axiom (A6) can be derived from (A1)–(A5) when n = ù. However,proofs of (A6) (and related schemata) usually involve the introduction ofvariables that do not occur in ϕ. This may not be possible when n is fi-nite since every variable in the language can then occur in a single formula.Therefore, (A6) has been explicitly included in Axn.Axioms (A1)–(A6) are sound with respect to the usual semantics: allformulæ in Axn are logically valid (true in all models under all assignmentsto their free variables). The rules of inference preserve logical validity. Itfollows that if ⊢n ϕ then ϕ is logically valid. To prove the converse whenn = ù, it suffices to observe that Σ2 is complete [27, Th 5], and that everyformula in Σ2 isù-provable. On the other hand, as was proved byMonk [22],completeness fails when n < ù.We say that a formula requires n variables to prove if it can be proved withn variables, but cannot be proved with only n − 1 variables, i.e., ⊢n ϕ and0n−1 ϕ. Evidently, any tautology in Fmn \ Fmn−1 requires n variables toprove. A more interesting challenge is to find 3-formulæ in Fm3 that requiren variables to prove.

§5. Provability. The results on provability in this sectionwill be used later.Here are three lemmata that hold for every n ≤ ù. Their proofs, which weomit, use only axioms (A1)–(A3) and the two inference rules. None of theequality axioms are involved.

Lemma 5.1. If ϕ ∈ Axn then ⊢n ϕ.

Lemma 5.2. If ⊢n ϕ and i < n then ⊢n ∀viϕ.

Lemma 5.3.

1. If ⊢n ϕ and ⊢n ϕ ⇒ ø, then ⊢n ø. In particular, if ⊢n ϕ and ϕ ⇒ ø ∈Axn, then ⊢n ø.

2. If ⊢n ϕ0, . . . ,⊢n ϕk and (∧

i≤k ϕi)⇒ ø is a tautology in Fmn, then⊢n ø.

Lemma 5.3 implies that ≡n is an equivalence relation on Fmn, and thatϕ ≡n ø iff ⊢n ϕ ⇒ ø and ⊢n ø ⇒ ϕ.

Lemma 5.4. If ϕ ∈ Fmn, i < n, and vi /∈ free(ϕ), then ⊢n ϕ ⇒ ∀viϕ and⊢n ∃viϕ ⇒ ϕ.

Proof. By (A1) and Lemma 5.1, ⊢n ϕ ⇒ ϕ. By Lemma 5.2, ⊢n ∀vi (ϕ ⇒ϕ). By (A2) and Lemma 5.1, ⊢n ∀vi (ϕ ⇒ ϕ) ⇒ (ϕ ⇒ ∀viϕ). So byLemma5.3, ⊢n ϕ ⇒ ∀viϕ. As vi /∈ free(¬ϕ), this also gives⊢n ¬ϕ ⇒ ∀vi¬ϕ.By Lemma 5.3, we obtain ⊢n ¬∀vi¬ϕ ⇒ ϕ, i.e., ⊢n ∃viϕ ⇒ ϕ. ⊣


Lemma 5.5. If ϕ,ø ∈ Fmn and i < n then

⊢n ∀vi (ϕ ⇒ ø)⇒ (∀viϕ ⇒ ∀viø) .

Proof. By (A3) and Lemma 5.1,

⊢n ∀vi (ϕ ⇒ ø)⇒ (ϕ ⇒ ø) and ⊢n ∀viϕ ⇒ ϕ.

So by Lemma 5.3,

⊢n ∀vi (ϕ ⇒ ø)⇒ (∀viϕ ⇒ ø).

By Lemma 5.2,

⊢n ∀vi (∀vi (ϕ ⇒ ø)⇒ (∀viϕ ⇒ ø)).

Since vi is not free in ∀vi (ϕ ⇒ ø), we can use (A2) and Lemma 5.3 to get

⊢n ∀vi (ϕ ⇒ ø)⇒ ∀vi (∀viϕ ⇒ ø).

By (A2) and Lemma 5.1, ⊢n ∀vi (∀viϕ ⇒ ø) ⇒ (∀viϕ ⇒ ∀viø). ByLemma 5.3, we obtain

⊢n ∀vi (ϕ ⇒ ø)⇒ (∀viϕ ⇒ ∀viø),

proving the lemma. ⊣

Lemma 5.6. Assume that ϕ,ϕ′, ø ∈ Fmn and that ø′ is the result of replac-

ing an occurrence of ϕ in ø by ϕ′. If ϕ ≡n ϕ′, then ø ≡n ø

′. Hence, if ⊢n øthen ⊢n ø

′.

Proof. By induction onø. Ifø isϕ, the result is clear. Inductively assumethe result for ø, so that ø ≡n ø

′, i.e., ⊢n ø ⇔ ø′. For ÷ ∈ Fmn, considerø ⇒ ÷. By (A1) and Lemma 5.1, ⊢n (ø ⇔ ø′)⇒ ((ø ⇒ ÷)⇔ (ø′ ⇒ ÷)).By Lemma 5.3, ⊢n (ø ⇒ ÷) ⇔ (ø′ ⇒ ÷), so that ø ⇒ ÷ ≡n ø

′ ⇒ ÷.The cases ÷ ⇒ ø and ¬ø are similar. Finally, consider ∀viø. We have⊢n ø ⇒ ø′. Lemma 5.2 gives ⊢n ∀vi (ø ⇒ ø′), and Lemmas 5.5 and 5.3give ⊢n ∀viø ⇒ ∀viø

′. Similarly, ⊢n ∀viø′ ⇒ ∀viø, giving ∀viø ≡n ∀viø

′

via Lemma 5.3, as required. ⊣

Lemma 5.7. If ϕ,ø ∈ Fmn and i < n then

⊢n ∀vi (ϕ ⇒ ø)⇒ (∃viϕ ⇒ ∃viø) .

Proof. Observe that Lemmas 5.3, 5.5, and 5.6 give

⊢n ∀vi (ϕ ⇒ ø)⇒ ∀vi (¬ø ⇒ ¬ϕ),⊢n ∀vi (¬ø ⇒ ¬ϕ)⇒ (∀vi¬ø ⇒ ∀vi¬ϕ),⊢n (∀vi¬ø ⇒ ∀vi¬ϕ)⇒ (¬∀vi¬ϕ ⇒ ¬∀vi¬ø).

The lemma now follows by Lemma 5.3. ⊣

We will find the following ‘monotonicity’ lemma handy.

Lemma 5.8. Assume ϕi , øi , ÷ ∈ Fmn and ⊢n ϕi ⇒ øi for each i ∈ I . Then⊢n

∧

i∈I ϕi ⇒∧

i∈I øi and ⊢n∨

i∈I ϕi ⇒∨

i∈I øi . Furthermore, for eachi ∈ I, j < n, ⊢n ∃vjϕi ⇒ ∃vjøi , and if ⊢n ÷ ⇒ ϕi then ⊢n ÷ ⇒ øi .


Proof. All parts but the third follow immediately from Lemma 5.3. Forthe third part, if ⊢n ϕi ⇒ øi then Lemma 5.2 gives ⊢n ∀vj (ϕi ⇒ øi), andLemmas 5.7 and 5.3 give ⊢n ∃vjϕi ⇒ ∃vjøi . ⊣

Lemma 5.9. Let i, j < n and ϕ,ø ∈ Fmn. Then

∃vi (ϕ ∨ ø)≡n ∃viϕ ∨ ∃viø,

∃vj (ϕ ∧ ø)≡n ϕ ∧ ∃vjø if vj /∈ free(ϕ),

∀vi (ϕ ∧ ø)≡n ∀viϕ ∧ ∀viø,

∀vj (ϕ ∨ ø)≡n ϕ ∨ ∀vjø if vj /∈ free(ϕ).

Proof. By Lemmas 5.8 and 5.3, ⊢n (∃viϕ ∨ ∃viø) ⇒ ∃vi (ϕ ∨ ø). Forthe converse, (A1) and Lemma 5.1 give ⊢n ¬ϕ ⇒ (¬ø ⇒ ¬(¬ϕ ⇒ ø)).Lemma 5.2 yields⊢n ∀vi (¬ϕ ⇒ (¬ø ⇒ ¬(¬ϕ ⇒ ø))), and two applicationsof Lemmas 5.5 and 5.3 give us

⊢n ∀vi¬ϕ ⇒ (∀vi¬ø ⇒ ∀vi¬(¬ϕ ⇒ ø)).

As (p ⇒ (q ⇒ r))⇒ (¬r ⇒ (¬¬p ⇒ ¬q)) is a tautology, Lemma 5.3 nowyields ⊢n ¬∀vi¬(¬ϕ ⇒ ø)⇒ (¬¬∀vi¬ϕ ⇒ ¬∀vi¬ø), which, abbreviated, is

⊢n ∃vi (ϕ ∨ ø)⇒ (∃viϕ ∨ ∃viø).

For the second part, (A3) and Lemma 5.1 give ⊢n ∀vj¬ø ⇒ ¬ø. Anapplication of Lemma 5.3 yields ⊢n (ϕ ⇒ ∀vj¬ø) ⇒ (ϕ ⇒ ¬ø), and byLemma 5.2,

⊢n ∀vj ((ϕ ⇒ ∀vj¬ø)⇒ (ϕ ⇒ ¬ø)).

As vj is not free in ϕ ⇒ ∀vj¬ø, (A2) and Lemmas 5.1 and 5.3 give us

⊢n (ϕ ⇒ ∀vj¬ø)⇒ ∀vj (ϕ ⇒ ¬ø).

(A2) and Lemma 5.1 immediately give the converse of this, so

(ϕ ⇒ ∀vj¬ø) ≡n ∀vj (ϕ ⇒ ¬ø).

Lemmas 5.3 and 5.6 now yield

¬(ϕ ⇒ ¬¬∀vj¬ø) ≡n ¬∀vj¬¬(ϕ ⇒ ¬ø),

that is, ϕ ∧∃vjø ≡n ∃vj (ϕ ∧ø), as required. The last two statements followfrom the first two by Lemmas 5.3 and 5.6. ⊣

Now we make use of the equality axioms. Note that we never use (A5).

Lemma 5.10. If i, j < n, ϕ ∈ Fmn, and vi , vj /∈ free(ϕ), then ϕ ≡n Sijϕ.

Proof. If i = j then the result is obvious, so assume i 6= j. By (A6) andLemma 5.1,

⊢n (vi=vj)⇒ (ϕ ⇒ Sijϕ).

By Lemma 5.2,

⊢n ∀vi(

(vi=vj)⇒ (ϕ ⇒ Sijϕ))

.


By Lemmas 5.7 and 5.3,

⊢n ∃vi (vi=vj)⇒ ∃vi (ϕ ⇒ Sijϕ).

By (A4) and Lemmas 5.1 and 5.3,

⊢n ∃vi (ϕ ⇒ Sijϕ).

Clearly, vi /∈ free(ϕ ⇒ Sijϕ), so by Lemmas 5.4 and 5.3,

⊢n ϕ ⇒ Sijϕ.

Similarly, ⊢n ¬ϕ ⇒ ¬Sijϕ, since ¬Sijϕ = Sij(¬ϕ) and vi , vj /∈ free(¬ϕ).So, by Lemma 5.3, ⊢n Sijϕ ⇒ ϕ, whence ϕ ≡n Sijϕ. ⊣

§6. A special language. Now we put restrictions on n and L. For theremainder of the paper, fix n ≥ 4 and â > 2ñ(n − 3) (here, ñ is as in §2),and let L be the first-order language with the following set of binary relationsymbols:

Rel = {Id,E,P1, . . . ,Pn−3,Q0, . . . ,Qâ−1}.

The smallest choices for n and â are n = 4 and â = 1 + 2ñ(1) = 5, andRel = {Id,E,P1,Q0, . . . ,Q4}. It can be checked that â − 1 ≥ n − 3 for alln ≥ 4. Partition Rel3 = Rel × Rel × Rel into two parts, as follows. A triple〈R,S, T 〉 ∈ Rel3 is said to be forbidden if one of the following conditionsholds:

(F1) R = Id and S 6= T, or S = Id and R 6= T, or T = Id and R 6= S,

(F2) {R,S, T} ⊆ {E,Q0},

(F3) {R,S, T} ⊆ {Qi : i < â},

(F4) {R,S, T} ⊆ {Pi ,Qi} for some i with 1 ≤ i ≤ n − 3,

(F5) {R,S, T} = {E,Qi ,Qj} for some i, j < â with |i − j| > 1.

Notice that any permutation of a forbidden triple is also forbidden. Anytriple that is not forbidden is said to be mandatory. Let Mand be the setof mandatory triples, and let Forb be the set of forbidden triples. ThenRel3 = Mand ∪ Forb and ∅ = Mand ∩ Forb. Now we define some special3-sentences that make assertions about the binary relations denoted by therelation symbols of L. They say, among other things, that the relations aresymmetric and form a partition of the universal relation, and that Id is acongruence relation (an identity for relative multiplication from either side).

Any two objects are related by some relation:

α1 = ∀v0∀v1∨

R∈Rel

v0Rv1.


Forbidden triples do not occur:

α2 =∧

〈R,S,T 〉∈Forb

∀v0∀v1∀v2¬(

v0Rv1 ∧ v0Sv2 ∧ v2Tv1)

.

Mandatory triples must occur wherever possible:

α3 =∧

〈R,S,T 〉∈Mand

∀v0∀v1

(

v0Rv1 ⇒ ∃v2 (v0Sv2 ∧ v2Tv1))

.

Now let

α = α1 ∧ α2 ∧ α3.

Of course, α depends implicitly on n. α2 can be expressed in a form similarto α3. By Lemmas 5.1, 5.3, 5.6, and 5.9,

α2 ≡3∧

〈R,S,T 〉∈Forb

∀v0∀v1

(

v0Rv1 ⇒ ¬∃v2 (v0Sv2 ∧ v2Tv1))

.

Using only Lemmas 5.1, 5.3, 5.6, and 5.9, we can derive useful but longeralternative formulations of these sentences.

α1 ≡3∧

S,T∈Rel

∀v0∀v1

(( ∨

〈R,S,T 〉∈Forb

v0Rv1)

∨( ∨

〈R,S,T 〉∈Mand

v0Rv1))

α2 ≡3∧

S,T∈Rel

∀v0∀v1

(( ∨

〈R,S,T 〉∈Forb

v0Rv1)

⇒ ¬∃v2 (v0Sv2 ∧ v2Tv1))

α3 ≡3∧

S,T∈Rel

∀v0∀v1

(( ∨

〈R,S,T 〉∈Mand

v0Rv1)

⇒ ∃v2 (v0Sv2 ∧ v2Tv1))

α ≡3 α1 ∧∧

S,T∈Rel

∀v0∀v1

(

¬∃v2 (v0Sv2 ∧ v2Tv1)⇔∨

〈R,S,T 〉∈Forb

v0Rv1)

α ≡3 α1 ∧∧

S,T∈Rel

∀v0∀v1

(

∃v2 (v0Sv2 ∧ v2Tv1)⇔∨

〈R,S,T 〉∈Mand

v0Rv1)

.

(13)

These conclusions follow only from the assumption that Mand and Forb

partition Rel3; their actual structure has not mattered, but it does in the nextlemma. Let

α4 = ∀v0∃v1v0Idv1.

Lemma 6.1. ⊢3 α1 ∧ α3 ⇒ α4.

Proof. Let R ∈ Rel. Since 〈R, Id, R〉 ∈ Mand, we apply Lemma 5.3 toget ⊢3 α3 ⇒ ∀v0∀v1[v0Rv1 ⇒ ∃v2(v0Idv2 ∧ v2Rv1)]. By axiom (A3) andLemma 5.3, ⊢3 α3 ⇒ [v0Rv1 ⇒ ∃v2(v0Idv2 ∧ v2Rv1)]. As ⊢3 v0Idv2 ∧


v2Rv1 ⇒ v0Idv2, Lemmas 5.8 and 5.3 give ⊢3 α3 ⇒ [v0Rv1 ⇒ ∃v2v0Idv2].By Lemma 5.3, we obtain ⊢3 α3 ⇒

∧

R∈Rel[v0Rv1 ⇒ ∃v2v0Idv2].Now by (A3) and Lemma 5.3 again, ⊢3 α1 ⇒

∨

R∈Rel v0Rv1. Hence byLemma 5.3, ⊢3 (α1∧α3)⇒ ∃v2v0Idv2. Lemma 5.10 shows that ∃v2v0Idv2 ≡3∃v1v0Idv1. So by Lemma 5.6, ⊢3 (α1 ∧ α3) ⇒ ∃v1v0Idv1. By Lemma 5.2,⊢3 ∀v0((α1∧α3)⇒ ∃v1v0Idv1), and by (A2) and Lemma 5.3, ⊢3 (α1∧α3)⇒∀v0∃v1v0Idv1, as required. ⊣

Remark 6.2. It is not hard to show that the 3-sentence α is unsatisfiable.Suppose, to the contrary, that α had a modelM, and pick an element a ofM. By Lemma 6.1,M |= aIdb for some b. Let Σ = {i < â : i is even}.AsM |= α3, the fact that 〈Id,Qi ,Qi〉 is a mandatory triple for all i ∈ ΣforcesM to contain points ci (i ∈ Σ) such thatM |= aQici ∧ ciQib. Leti < j in Σ. By α1,M |= ciRcj for some R ∈ Rel. Choose such an R. SoM |= aQjcj ∧ aQici ∧ ciRcj . From α2 and the definition of Forb, we seethat R 6= Id since Qi 6= Qj , R 6= E since |i − j| > 1, and R 6= Qk for everyk < â since

⟨Qi ,Qj ,Qk

⟩is forbidden. Hence,M |= ciPlcj for some l with

1 ≤ l ≤ n−3. But, by the choice of â , |Σ| > ñ(n−3), so by Proposition 2.1,there is no (n−3)-coloring of Σ. There must therefore be i < j < k in Σ suchthatM |= ciPlck ∧ ciPlcj ∧ cjPlck for some l . However, this contradictsα2, since 〈Pl ,Pl ,Pl 〉 is forbidden. Hence α has no model.This argument can be easily formalized to give a proof of ¬α using 2+ |Σ|variables.

Remark 6.3. We can translate ¬α into an equivalent equation in the cal-culus of relations in the way described in the introduction. (Of course, thepreceding remark shows that 0 = 0 is also equivalent to ¬α.) α1 statesthat

∑

R∈RelR = 1, or equivalently that∏

R∈RelR = 0. α2 states thatR · (S ;T ) = 0 whenever 〈R,S, T 〉 ∈ Forb. α3 states that R · S ;T = 0whenever 〈R,S, T 〉 ∈ Mand. For all R,S, T ∈ Rel, let

ô(R,S, T ) =

{

R · S ;T if 〈R,S, T 〉 ∈ Forb,

R · S ;T if 〈R,S, T 〉 ∈ Mand.

So α states that (∏

R∈RelR) + (∑

R,S,T∈Rel ô(R,S, T )) = 0. The desiredequation equivalent to ¬α is therefore

1;

(( ∏

R∈Rel

R)

+( ∑

R,S,T∈Rel

ô(R,S, T )))

;1 = 1.(14)

From (F1)–(F5) it can be shown that for every R ∈ Rel there is a termtR(x) in the calculus of relations such that if ø is the elementary 3-variabletranslation of tR(E) = R (cf. the introduction) then⊢3 α ⇒ ø. For example,tQ0 = x ;x ·x. Terms like this are used in §9 to eliminate all relation symbolsother than E from α.


Remark 6.4. If we were only interested in 3-sentences that require n vari-ables to prove, without regard for the number of binary relation symbolsused in such 3-sentences, then we could simplify the definition of Forb andMand, although these changes have almost no effect on the length of ¬α orthe corresponding equation (14). First, delete E from the language. Con-ditions (F2) and (F5) in the definition of forbidden triples fall away. Ifα′ is the resulting 3-sentence, then ¬α′ still requires n variables to prove.Next, drop Qi from (F4) and let α

′′ be the resulting 3-sentence. Again ¬α′′

requires n variables to prove. At this point the definition of forbidden andmandatory triplesmatches relation algebras that arise fromMonk’s cylindricalgebras. These algebras can also be obtained by the process of splitting oneatom [1, Ex 6], starting with the algebras called Ek({2, 3}) in [15, p. 510].These algebras are symmetric integral relation algebras whose forbiddentriples are exactly the ones of the form 〈a, a, a〉, for every nonidentity atoma. Sentences similar to ¬α′′ and the corresponding equations (14) are givenby Gordeev [5], where it is shown inter alia that for every n ≥ 4 there isa 3-sentence that can be proved in a proof theory with equality and withn variables, but not with n − 2. If we wish to construct the α that arisesfrom the algebras Ek({2, 3}), we may achieve the same effect by deleting therelation symbols Qi and condition (F3), leaving only Id, the symbols Pj ,(F1), and the simplified (F4). It is not known, however, except for small n,whether ¬α has no model in this case [3], [23, Prob. 5].

We will show in the next section, by formalizing the proof of Proposi-tion 2.1, that the 3-sentence¬α defined according to (F1)–(F5) is n-provable(Theorem 7.5). Then, in §8, we show that α is valid in certain (n − 1)-dimensional, non-standard models; it follows (Theorem 8.9) that ¬α is not(n − 1)-provable. In §9 we alter ¬α by replacing all atomic formulæ otherthan those containing E with formulæ containing only E. The resulting3-sentences (for n = 4, 5, . . . ) still require n variables to prove, and the cor-responding equations (14) can be used to continue the sequence begun with(1) and (12).

§7. ¬α is n-provable. To help the induction in the proof that ⊢n ¬α,we introduce a structure called a spectrum. A 0-spectrum is a subset of{Qi : i < â , i even}. If 1 ≤ m ≤ n − 3, we say that Σ is an m-spectrum withcolor sequenceì ifì is a one-to-one functionì : {1, . . . , m} → {1, . . . , n−3},and

Σ ⊆ {Qi : i < â, i even} × {Pì(1)} × · · · × {Pì(m)}

={⟨

Qi ,Pì(1), . . . ,Pì(m)

⟩

: i < â, i even}

.

Distinct sequences S, S ′ in a spectrum have distinct first elements (S0 6= S′0),

but they agree on all other entries, if any (Si = S′i for 1 ≤ i ≤ m). For each

https://www.researchgate.net/publication/221134740_Combinatorial_Principles_Relevant_to_Finite_Variable_Logic?el=1_x_8&enrichId=rgreq-26f873ba-fdbb-4d0f-9643-330b98639037&enrichSource=Y292ZXJQYWdlOzIyMDM2NjI3NDtBUzoxMjY1NzYyMjIwMTk1ODZAMTQwNzE4OTUzMTY1NA==


��

@@

@@

@@

@@

@@

t t t

t∃vm+1

v0 v1. . . . . . . . . vm

S0 = Qi(some even i < â)

S1 = Pì(1) . . . Sm = Pì(m)

Figure 1. The conjunct of αΣ corresponding to S ∈ Σ.

m-spectrum Σ we also define a formula

αΣ =∧

S∈Σ

∃vm+1∧

i<m+1

viSivm+1,

saying that for every sequence S ∈ Σ there is vm+1 such that v0, . . . , vm andvm+1 ‘realize’ S, as shown in Figure 1.The following two propositions show that ¬α is n-provable.

Proposition 7.1. Let Σ be any 0-spectrum. Then ⊢n α3 ∧ α4 ⇒ αΣ.

Proof. Let Qi ∈ Σ. By (A1) and Lemma 5.1,

⊢n α3 ⇒ ∀v0∀v1(v0Idv1 ⇒ ∃v2(v0Qiv2 ∧ v2Qiv1)).

By (A3) and Lemmas 5.1 and 5.3,

⊢n α4 ⇒ ∃v1(v0Idv1),⊢n α3 ⇒ ∀v1(v0Idv1 ⇒ ∃v2(v0Qiv2 ∧ v2Qiv1)).


⊢n α3 ⇒ (∃v1(v0Idv1)⇒ ∃v1∃v2(v0Qiv2 ∧ v2Qiv1)),⊢n α3 ∧ α4 ⇒ ∃v1∃v2(v0Qiv2 ∧ v2Qiv1).

As ⊢n v0Qiv2 ∧ v2Qiv1 ⇒ v0Qiv2, by Lemmas 5.3 and 5.8 we get

⊢n α3 ∧ α4 ⇒ ∃v1∃v2v0Qiv2.

By Lemma 5.4, ⊢n ∃v1∃v2v0Qiv2 ⇒ ∃v2v0Qiv2, so by Lemma 5.3,

⊢n α3 ∧ α4 ⇒ ∃v2v0Qiv2.

But v1, v2 are not free in ∃v2v0Qiv2, so Lemma 5.10 gives ⊢n ∃v2v0Qiv2 ⇒∃v1v0Qiv1. By Lemma 5.3,

⊢n α3 ∧ α4 ⇒ ∃v1v0Qiv1.

This holds for each Qi ∈ Σ, so by Lemma 5.3 again we obtain ⊢n α3 ∧α4 ⇒αΣ, as required. ⊣

Proposition 7.2. Suppose that 0 ≤ m ≤ n−3 and Σ is anm-spectrum with|Σ| > ñ(n − 3−m). Then ⊢n α1 ∧ α2 ⇒ ¬αΣ.


��

@@

@@

@@

@@

s s s

svm+1

v0 v1. . . vm

S0 = Qi S1 · · · Sm

AAAAAAAA

��

��

��

��

S ′0 = Qi ′ S ′1 · · · S ′m

s vm+2

Figure 2. The meaning of ø(S, S ′).

Proof. The proof will proceed by downward induction on m, startingwith m = n − 3 and ending at m = 0. Therefore, we will need to constructan (m+1)-spectrum from anm-spectrum. The two preliminary lemmas 7.3and 7.4 allow us to do this. Form,Σ as in the proposition, and S, S ′ ∈ Σ, letø(S, S ′) be the following Fmn-formula, illustrated in Figure 2:

ø(S, S ′) =∧

i<m+1

(viSivm+1 ∧ viS

′ivm+2

).

Lemma 7.3. αΣ ≡n∧

S∈Σ

∃vm+1∧

S′∈Σ\{S}

∃vm+2ø(S, S′).

Proof.( ∧

i∈I

pi)

⇔

(∧

i∈I

(

pi ∧∧

j∈I\{i}

pj))

is a tautology, so by (A1) and

Lemma 5.1,

αΣ ≡n∧

S∈Σ

((

∃vm+1∧

i<m+1

viSivm+1)

∧( ∧

S′∈Σ\{S}

∃vm+1∧

i<m+1

viS′ivm+1

))

.

By Lemma 5.10, we have

∧

S′∈Σ\{S}

∃vm+1∧

i<m+1

viS′ivm+1 ≡n

∧

S′∈Σ\{S}

∃vm+2∧

i<m+1

viS′ivm+2.

So by Lemma 5.6, we obtain

αΣ ≡n∧

S∈Σ

((

∃vm+1∧

i<m+1

viSivm+1)

∧( ∧

S′∈Σ\{S}

∃vm+2∧

i<m+1

viS′ivm+2

))

.


Distribute ∃vm+1 over the conjuncts in which vm+1 does not appear free(Lemmas 5.9 and 5.6):

αΣ ≡n∧

S∈Σ

∃vm+1

(( ∧

i<m+1

viSivm+1)

∧( ∧

S′∈Σ\{S}

∃vm+2∧

i<m+1

viS′ivm+2

))

.

Distribute conjunctions over conjunctions, using (A1) and Lemmas 5.1and 5.3:

αΣ ≡n∧

S∈Σ

∃vm+1

( ∧

S′∈Σ\{S}

(( ∧

i<m+1

viSivm+1)

∧(

∃vm+2∧

i<m+1

viS′ivm+2

)))

.

Now distribute ∃vm+2 over the conjuncts in which vm+2 does not appear free,using Lemmas 5.9 and 5.6:

αΣ ≡n∧

S∈Σ

∃vm+1∧

S′∈Σ\{S}

∃vm+2

(( ∧

i<m+1

viSivm+1)

∧( ∧

i<m+1

viS′ivm+2

))

.

Use the associativity and commutativity of conjunction ((A1), Lemma 5.1,Lemma 5.6). Conclude that

αΣ ≡n∧

S∈Σ

∃vm+1∧

S′∈Σ\{S}

∃vm+2

( ∧

i<m+1

(

viSivm+1 ∧ viS′ivm+2

))

︸︷︷︸

ø(S,S′)

,

as required. ⊣

Let Σ have color sequence ì : {1, . . . , m} → {1, . . . , n − 3}, and let

Tì = {Pj : 1 ≤ j ≤ n − 3, j /∈ rng(ì)},(15)

where rng(ì) is the range of ì. Note that Tì = ∅ if m = n − 3.

Lemma 7.4. For all distinct S, S ′ ∈ Σ,

⊢n α1 ∧ α2 ∧ ø(S, S′)⇒

∨

R∈Tì

∧

i<m+1

(viS

′ivm+2 ∧ vm+1Rvm+2

).

Proof. Let S, S ′ ∈ Σ with S 6= S ′. It follows by Lemma 5.10 that α1 ≡n∀vm+1∀vm+2

∨

R∈Rel vm+1Rvm+2. So by axiom (A3) and Lemma 5.3,

⊢n α1 ∧ ø(S, S′)⇒

( ∨

R∈Rel

vm+1Rvm+2)

∧∧

i<m+1

(viSivm+1 ∧ viS

′ivm+2

).

By Lemma 5.3,

(16) ⊢n α1∧ø(S, S′)⇒

∨

R∈Rel

∧

i<m+1

(viS

′ivm+2 ∧ viSivm+1 ∧ vm+1Rvm+2

).

Now using Lemma 5.10 to change the variables of α2, and then (A3) andLemma 5.3 to remove the outer quantifiers and irrelevant conjuncts, we


obtain

(17) ⊢n α2 ⇒ ¬(viPvm+2 ∧ viQvm+1 ∧ vm+1Rvm+2)

for any i < m + 1, 〈P,Q,R〉 ∈ Forb.

From these last two statements we see that every conjunct in (16) of theform viS

′ivm+2 ∧ viSivm+1 ∧ vm+1Rvm+2 is contradicted by α2 according to

(17) whenever 〈S ′i , Si , R〉 ∈ Forb. Since S0, S′0 ∈ {Qi : i < â, i even} and

S0 6= S ′0, the definition of Forb implies that 〈S0, S′0, R〉 ∈ Forb whenever

R ∈ {Id,E,Qi : i < â}. Also, for 1 ≤ i ≤ m, Si = S′i = Pì(i) so

〈Si , S′i , R〉 ∈ Forb if R = Pì(i). So (15), (17), and Lemma 5.3 give

(18) if R ∈ Rel \ Tì then

⊢n α2 ⇒ ¬∧

i<m+1

(viS


).

Combining (16) and (18) by Lemma 5.3 gives

⊢n α1 ∧ α2 ∧ ø(S, S′)⇒

∨

R∈Tì

∧

i<m+1

(viS


),

and Lemma 5.8 (or 5.3) now yields

⊢n α1 ∧ α2 ∧ ø(S, S′)⇒

∨

R∈Tì

∧

i<m+1

(

viS′ivm+2 ∧ vm+1Rvm+2

)

,

as required. If Tì = ∅, this is ⊢n α1 ∧ α2 ∧ ø(S, S′)⇒ ⊥. ⊣

The proof of Proposition 7.2 now proceeds by induction on n−3−m. Thebase case is n−3−m = 0. In the base case we have n ≥ 4,m = n−3 ≥ 1, ìis surjective, ñ(n− 3−m) = 1, and |Σ| ≥ 2. It follows that there are distinctS, S ′ ∈ Σ giving non-empty conjunctions in Lemma 7.3, and the disjunctionover Tì in Lemma 7.4 is empty, so Lemmas 5.8 and 5.3 give us

⊢n αΣ ⇒ ∃vm+1∃vm+2ø(S, S′),(19)

⊢n α1 ∧ α2 ∧ ø(S, S′)⇒ ⊥.(20)

Hence, in the usual way,

⊢n α1 ∧ α2 ⇒ ¬ø(S, S ′) by (20) and Lemma 5.3,

⊢n ∀vm+2(α1 ∧ α2 ⇒ ¬ø(S, S ′)) by Lemma 5.2,

⊢n α1 ∧ α2 ⇒ ∀vm+2¬ø(S, S′) by (A2) and Lemma 5.3,

⊢n α1 ∧ α2 ⇒ ¬∃vm+2ø(S, S′) by Lemma 5.3,

⊢n α1 ∧ α2 ⇒ ¬∃vm+1∃vm+2ø(S, S′) similarly,

⊢n α1 ∧ α2 ⇒ ¬αΣ by (19) and Lemma 5.3.

This completes the proof of the base case.


Assume that n−3−m > 0. The inductive hypothesis is that ⊢n α1∧α2 ⇒¬αΣwhenever Σ is an (m+1)-spectrumof size larger thanñ(n−3−(m+1)) =ñ(n − 4 − m). We will show from this assumption that ⊢n α1 ∧ α2 ⇒ ¬αΣwhenever Σ is an m-spectrum of size greater than ñ(n − 3−m).Fix S ∈ Σ. By Lemmas 7.3 and 5.3,

⊢n α1 ∧ α2 ∧ αΣ ⇒ α1 ∧ α2 ∧ ∃vm+1∧

S′∈Σ\{S}

∃vm+2ø(S, S′).

Using Lemmas 5.3, 5.6, and 5.9, we obtain

⊢n α1 ∧ α2 ∧ αΣ ⇒ ∃vm+1∧

S′∈Σ\{S}

∃vm+2(α1 ∧ α2 ∧ ø(S, S′)).


⊢n α1 ∧ α2 ∧ αΣ

⇒ ∃vm+1∧

S′∈Σ\{S}

∃vm+2

( ∨

R∈Tì

∧

i<m+1

(viS′ivm+2 ∧ vm+1Rvm+2)

)

.

Distribute the existential quantifiers ∃vm+2 over disjunctions (Lemmas 5.9and 5.6):

⊢n α1 ∧ α2 ∧ αΣ

⇒ ∃vm+1∧

S′∈Σ\{S}

∨

R∈Tì

∃vm+2

( ∧

i<m+1

(

viS′ivm+2 ∧ vm+1Rvm+2

))

.

Then, in the critical step, distribute conjunctions over disjunctions usingLemma 5.6 and (A1) with the tautology

( ∧

i∈I

∨

j∈J

pij)

⇔( ∨

g : I→J

∧

i∈I

pi,g(i))

,

exponentially expanding the length of the formula:

⊢n α1 ∧ α2 ∧ αΣ

⇒ ∃vm+1∨

g : Σ\{S}→Tì

∧

S′∈Σ\{S}

∃vm+2

( ∧

i<m+1

viS′ivm+2 ∧ vm+1g(S

′)vm+2)

.

Finally, distribute∃vm+1 over the resulting disjunctions (Lemmas 5.9 and 5.6):

⊢n α1 ∧ α2 ∧ αΣ

⇒∨

g : Σ\{S}→Tì

∃vm+1∧

S′∈Σ\{S}

∃vm+2

( ∧

i<m+1


′)vm+2)

.

Now each function g : Σ \ {S} → Tì induces a partition of Σ \ {S} whosepieces are the preimages of the symbols in Tì. The number of pieces is at


most |Tì| = n − 3 − m. For each g, let P(g) be one of the largest pieces.Restricting the conjunctions over Σ \ {S} to P(g) we obtain, by Lemma 5.8,

(21) ⊢n α1 ∧ α2 ∧ αΣ

⇒∨

g : Σ\{S}→Tì

∃vm+1∧

S′∈P(g)

∃vm+2

( ∧

i<m+1


′)vm+2)

.

For each g : Σ\{S} → Tì, let Σ(g) = {〈S ′0, S′1, . . . , S

′m, g(S

′)〉 : S ′ ∈ P(g)}.Since g is constant on P(g), Σ(g) is an (m + 1)-spectrum. So with thisnotation, we can rewrite (21):

⊢n α1 ∧ α2 ∧ αΣ ⇒∨

g : Σ\{S}→Tì

∃vm+1∧

S+∈Σ(g)

∃vm+2

( ∧

i<m+2

viS+i vm+2

)

︸︷︷︸

αΣ(g)

—that is,

⊢n α1 ∧ α2 ∧ αΣ ⇒∨

g : Σ\{S}→Tì

∃vm+1αΣ(g).(22)

Now a partitioned set cannot have more elements than the number of piecestimes the size of a largest piece, so for any g : Σ \ {S} → Tì we have

(n − 3−m)|P(g)| ≥ |Σ| − 1.(23)

Our assumption on the size of Σ is that

|Σ| > ñ(n − 3−m).(24)

Since |Σ(g)| = |P(g)|, the definition of ñ and inequalities (23) and (24) give

1 + (n − 3−m)|Σ(g)| ≥ |Σ| > ñ(n − 3−m)

= 1 + (n − 3−m) · ñ(n − 4−m),

so |Σ(g)| > ñ(n−3−(m+1)).Thus, each Σ(g) is a sufficiently large (m+1)-spectrum that, by the inductive hypothesis, ⊢n α1 ∧ α2 ⇒ ¬αΣ(g). As in thebase case, we deduce from this that

⊢n α1 ∧ α2 ⇒ ¬∃vm+1αΣ(g).

This holds for all g, so by (22) and Lemma 5.3, ⊢n α1 ∧ α2 ⇒ ¬αΣ. Thiscompletes the induction and the proof of the proposition. ⊣

Theorem 7.5. ⊢n ¬α.

Proof. Let Σ be the 0-spectrum {Qi : i < â, i even}. By Proposition 7.1,⊢n α3 ∧ α4 ⇒ αΣ. By Proposition 7.2 since |Σ| > ñ(n − 3), ⊢n α1 ∧ α2 ⇒¬αΣ. Also, by Lemma 6.1, ⊢n α1 ∧ α3 ⇒ α4. So by Lemma 5.3, we have⊢n ¬(α1 ∧ α2 ∧ α3), or ⊢n ¬α, as required. ⊣

Thus the 3-sentence ¬α is n-provable without using (A5). In the nextsection we show that ¬α is not (n − 1)-provable.


§8. Alternative semantics. To show that ¬α is not (n − 1)-provable, wewill use non-classical semantics for which ⊢n−1 is sound and α is valid.Suppose A is an (n − 1)-by-(n − 1) matrix of relation symbols in Rel. Fori, j < n − 1, the symbol in row i and column j is denoted by Aij or Ai,j .Thus,

A =

A00 A01 A02 · · · A0,n−2A10 A11 A12 · · · A1,n−2A20 A21 A22 · · · A2,n−2...

......

...An−2,0 An−2,1 An−2,2 · · · An−2,n−2

.

Think of A as a list of binary relations that hold among n objects, that is, Ais just a special kind of quantifier-free n-type having the same ‘meaning’ asthe conjunction of the atomic formulæ viAijvj :

‘A⇔∧

i,j<n−1

viAijvj ’.

LetM be the set of (n − 1)-by-(n − 1) matricesA of relation symbols in Rel

that satisfy the following conditions:

(M1) Aii = Id for all i < n − 1,(M2)

⟨

Aij , Aik , Akj⟩

∈ Mand for all i, j, k < n − 1.

The matrices inM are called atomic matrices.

Lemma 8.1. If i, j, k < n − 1, A ∈ M, and Aij = Id, then Aik = Akj .Hence, any atomic matrix A ∈ M is symmetric: it satisfies

(M3) Aij = Aji for all i, j < n − 1.

Proof. By (M2),⟨

Aij , Aik , Akj⟩

=⟨

Id, Aik , Akj⟩

∈ Mand. From thedefinition ofMandwe see thatAik = Akj . By (M1),Aii = Id for anyA ∈ Mand i, j < n − 1, so by the foregoing, Aij = Aji , giving (M3). ⊣

For any i, j < n − 1, let [i, j] and [i/j] be the functions mapping n − 1 ton − 1 defined by the following conditions for all k < n − 1:

[i, j]i = j [i/j]i = j[i, j]j = i[i, j]k = k if k 6= i, j [i/j]k = k if k 6= i

[i, j] is a transposition and [i/j] is a replacement. If A is an atomic matrixand g is any function mapping n − 1 into n − 1, then Ag is the matrix ofrelation symbols defined by (Ag)ij = Ag(i),g(j) for all i, j < n − 1. It is easyto check that Ag is an atomic matrix. In particular, A[i, j] and A[i/j] areatomicmatrices. Let k, l < n−1. Two atomicmatricesA,B are said to agreeup to k, written A ∼=k B , if Aij = Bij whenever i, j < n − 1 and i, j 6= k,and they agree up to k, l , writtenA ∼=kl B , ifAij = Bij whenever i, j < n−1and i, j /∈ {k, l}. We will need to know thatM has the following properties.


Lemma 8.2.

(C1) if A,C ∈ M, i, j < n − 1, i 6= j, and A ∼=ij C , then there is someB ∈ M such that A ∼=i B and B ∼=j C ,

(C2) if A ∈ M and i, j < n − 1 then A[i/j] ∈ M,(C3) if A ∈ M and i, j < n − 1 then A[i, j] ∈ M.

Proof. (C2) and (C3) were discussed above. The hard part is (C1). As-sume A,C ∈ M and A ∼=ij C . We wish to find some B ∈ M such thatA ∼=i B and B ∼=j C . The requirements A ∼=i B , B ∼=j C determineBkl whenever {k, l} 6= {i, j}, so we only need to choose Bij = Bji so that⟨Bji , Ajk , Cki

⟩∈ Mand for every k < n − 1, k 6= i, j. It is easily seen that

this is sufficient to ensure that B ∈ M.Let K = {k : i, j 6= k < n − 1}. If Ajk = Id for some k ∈ K then welet Bij = Bji = Cik . Then by Lemma 8.1,

⟨

Bji , Ajl , Cli⟩

= 〈Cki , Ckl , Cli〉 ∈Mand for all l ∈ K . Similarly, if Cik = Id for some k ∈ K then we maylet Bij = Bji = Akj . We may therefore assume that Ajk 6= Id 6= Cik forall k ∈ K . Suppose that for every k ∈ K there is some l ∈ {1, . . . , n − 3}such that {Ajk , Cki} ⊆ {Pl ,Ql}. Then

⟨E, Ajk , Cki

⟩is mandatory whenever

k ∈ K , so we may let Bij = Bji = E. If not, then because |K | = n − 3and there are n − 3 pairs of the form {Pl ,Ql}, 1 ≤ l ≤ n − 3, there must besome l ∈ {1, . . . , n − 3} such that for every k ∈ K , {Ajk , Cki} * {Pl ,Ql}.Then since Ajk , Cki 6= Id,

⟨Pl , Ajk , Cki

⟩is mandatory for every k ∈ K , so

let Bij = Bji = Pl . ⊣

The satisfaction relation |=associated withM is the unique binary relationbetween matrices and (n−1)-formulæ that satisfies the following conditionsfor all A ∈ M, ϕ,ø ∈ Fmn−1, i, j < n − 1, and R ∈ Rel:

A |= vi=vj iff Aij = Id,A |= viRvj iff Aij = R,A |= ¬ϕ iff A 2 ϕ,A |= ϕ ⇒ ø iff either A 2 ϕ or A |= ø or both,A |= ∀viϕ iff B |= ϕ whenever B ∈ M and B ∼=i A.

A formula that isM-satisfied by every atomic matrix is said to beM-valid.

Lemma 8.3. If ϕ ∈ Fmn−1, vi /∈ free(ϕ), and A ∼=i B , then A |= ϕ iffB |= ϕ.

Proof. The proof is by induction onϕ. Suppose vi /∈ free(ϕ) andA ∼=i B .If ϕ is atomic, then ϕ is vj=vk or vjRvk with R ∈ Rel, and i 6= j, k. ThenAjk = Bjk since A ∼=i B , hence A |= ϕ iff B |= ϕ. Assume the lemma holdsfor ø, and let ϕ = ¬ø. Then vi /∈ free(ϕ) = free(ø), so inductively, A |= ϕiff A 2 ø iff B 2 ø iff B |= ϕ. It is equally easy to show that if the lemmaholds for ø and ÷, then it holds for ø ⇒ ÷. Assume the lemma holds for ø,and let ϕ = ∀vjø. We only show that B |= ϕ whenever A |= ϕ and A ∼=i B ,since the converse follows by interchanging A and B . Assume A |= ϕ and


A ∼=i B . To show B |= ϕ, it suffices to assume B ∼=j C and show C |= ø.If j = i , then C |= ø because A ∼=i C , so assume j 6= i . From A ∼=i Band B ∼=j C we conclude that A ∼=ij C . SinceM has property (C1), theremust be some D ∈ M such that A ∼=j D and D ∼=i C . Then D |= ø sinceA |= ∀vjø and A

∼=j D. Now vi /∈ free(ø) since j 6= i , so it follows fromthe inductive hypothesis and D ∼=i C that C |= ø, as desired. ⊣

Lemma 8.4. For all i, j < n − 1, A ∈ M, and ϕ ∈ Fmn−1, we have A |=Sijϕ iff A[i, j] |= ϕ.

Proof. The proof is again by induction. If ϕ = xRy, where x, y ∈ Varn−1and R ∈ Rel, then there are several cases to be distinguished according towhether x or y is either vi , vj , or neither. We will treat only one case,as an example. Suppose x = vi , y = vk , and k 6= i, j. Then, sinceSijϕ = SijviRvk = vjRvk , and A[i, j]ik = Ajk , we have A |= Sijϕ iffA |= vjRvk iff Ajk = R iff A[i, j]ik = R iff A[i, j] |= viRvk iff A[i, j] |= ϕ.(Notice that (C3) of Lemma 8.2, giving A[i, j] ∈ M, is needed here.)Suppose the lemma holds for ø, and let ϕ = ¬ø. Then, since Sijϕ =

Sij¬ø = ¬Sijø, we have A |= Sijϕ iff A 2 Sijø iff A[i, j] 2 ø iff A[i, j] |=¬ø iff A[i, j] |= ϕ. It is equally easy to treat the case ϕ = (ø ⇒ ÷).Finally, suppose the lemma holds for ø, and let ϕ = ∀vkø. We firstassume A |= Sij∀vkø and prove A[i, j] |= ∀vkø. To get this conclusion, weassume B ∼=k A[i, j] and show B |= ø. Let l = [i, j]k. Then k = [i, j]l ,and it is easy to check that B[i, j] ∼=l A. Now A |= Sij∀vkø = ∀vlSijø, soB[i, j] |= Sijø. But then B = B[i, j][i, j] |= ø by the inductive hypothesis.Thus A[i, j] |= ∀vkø whenever A |= Sij∀vkø. Conversely, assume A[i, j] |=∀vkø. To show that A |= Sij∀vkø = ∀vlSijø, we suppose B

∼=l A and showB |= Sijø. From B ∼=l A we get B[i, j] ∼=k A[i, j], so B[i, j] |= ø, andhence B |= Sijø by the inductive hypothesis. ⊣

Lemma 8.5. If A ∈ M and ϕ ∈ Axn−1 then A |= ϕ. Thus, every axiom isM-valid.

Proof. Both the alternative and the usual semantics treat the connectives¬ and⇒ in the same way, so the proof that every tautology is semanticallyvalid also shows that every tautology isM-valid. Hence the lemma holdswhen ϕ is an instance of (A1).Suppose ϕ is an instance ∀vi (ø ⇒ ÷) ⇒ (ø ⇒ ∀vi÷) of (A2), wherevi /∈ free(ø). To show that A |= ϕ, it suffices to assume A |= ∀vi (ø ⇒ ÷),A |= ø, and B ∼=i A, and then show B |= ÷. From the first and third ofthese assumptions we know that B |= ø ⇒ ÷, i.e., either B 2 ø or elseB |= ÷. However, the second and third assumptions imply, by Lemma 8.3and vi /∈ free(ø), that B |= ø. Hence B |= ÷.Suppose ϕ is an instance ∀viø ⇒ ø of (A3). If A |= ∀viø, then A |= øsince A ∼=i A. Hence A |= ϕ.


If ϕ is an instance of (A4), then ϕ is ∃vi (vi=vj) for some distinct i, j <n − 1. Now A |= ∃vi (vi=vj) iff there is some B ∈ M such that B ∼=i A andB |= vi=vj . Let B = A[i/j]. Then B ∈ M sinceM has property (C2).Furthermore, A ∼=i B , for if k, l 6= i then [i/j]k = k and [i/j]l = l soAkl = A[i/j]kl . Finally, B |= vi=vj , since Bij = A[i/j]ij = Ajj = Id.We will treat only one type of instance of (A5), namely (vi=vj) ⇒(viRvk ⇒ vjRvk). Suppose A |= vi=vj and A |= viRvk . Then Aij = Id

and Aik = R. By Lemma 8.1, Ajk = R, hence A |= vjRvk . The other casesare similar.Finally, let ϕ be (vi=vj) ⇒ (ø ⇒ Sijø), an instance of (A6). SupposeA |= vi=vj . ThenAij = Id, and it is easy to prove from this thatA = A[i, j].(For example, if k 6= i, j then by Lemma 8.1, Aik = Ajk = A[i, j]ik .) If alsoA |= ø, then A[i, j] |= ø, so A |= Sijø by Lemma 8.4. Thus A |= ϕ forevery A ∈ M. ⊣

Lemma 8.6. If ϕ,ϕ ⇒ ø ∈ Fmn−1 areM-valid, then so are ø and ∀viϕ forall i < n − 1.

Proof. Let A ∈ M. Then A |= ϕ and A |= ϕ ⇒ ø since ϕ and ϕ ⇒ øareM-valid, and it follows thatA |= ø. We getA |= ∀viϕ because ifA

∼=i Bthen by theM-validity of ϕ, B |= ϕ. ⊣

Lemma 8.7. If ⊢n−1 ϕ then ϕ isM-valid.

Proof. The set ofM-valid formulæ contains Axn−1 by Lemma 8.5, and isclosed under modus ponens and (n − 1)-generalization by Lemma 8.6, so itmust therefore include all (n − 1)-provable formulæ. ⊣

Lemma 8.8. α isM-valid.

Proof. We use the second alternative form (13) of α:

α ≡3 ∀v0∀v1∨

R∈Rel

v0Rv1

∧∧

S,T∈Rel

∀v0∀v1

((

∃v2 (v0Sv2 ∧ v2Tv1))

⇔∨

〈R,S,T 〉∈Mand

v0Rv1)

.

Certainly, ∀v0∀v1∨

R∈Rel v0Rv1 isM-valid. Suppose S,T ∈ Rel. LetA ∈ M.If A |= ∃v2 (v0Sv2 ∧ v2Tv1) then for some B ∈ M with B ∼=2 A, we haveB |= v0Sv2 ∧ v2Tv1. By (M2), B |=

∨

〈R,S,T 〉∈Mand v0Rv1, so by Lemma 8.3,A |=

∨

〈R,S,T 〉∈Mand v0Rv1. Conversely, suppose that R ∈ Rel, 〈R,S, T 〉 ∈

Mand, and A |= v0Rv1. Define ð = [3/0][4/0] · · · [n − 2/0] : (n − 1) →{0, 1, 2}, and define B by

Blm =

Id if ð(l) = ð(m),

R if {ð(l), ð(m)} = {0, 1},

S if {ð(l), ð(m)} = {0, 2},

T if {ð(l), ð(m)} = {2, 1},


where l, m < n − 1. It is easy to check that B ∈ M. Now evidently,

A ∼=3 A[3/0] ∼=4 A[3/0][4/0] ∼=5 · · ·(25)∼=n−2 A[3/0] · · · [n − 2/0] = Að ∼=2 B.

By (C2), all matrices shown are inM. As B02 = S and B21 = T , we haveB |= ∃v2 (v0Sv2 ∧ v2Tv1). But free(∃v2(v0Sv2 ∧ v2Tv1)) = {v0, v1}, so byrepeated use of Lemma 8.3 on (25), we obtain A |= ∃v2(v0Sv2 ∧ v2Tv1), asrequired. ⊣

Theorem 8.9. 0n−1 ¬α.

Proof. By Lemma 8.8, α is M-valid. Hence ¬α is not M-valid. ByLemma 8.7, 0n−1 ¬α. ⊣

We have shown that ¬α is n-provable without (A5), but is not (n − 1)-provable, even using (A5).

§9. Reduction to one binary relation symbol. Finally, we obtain from α a3-sentence, using only the binary relation symbol E and without equality,that requires n variables to prove. For every ϕ ∈ Fmn, define a formula ϕ

ä

next. The definition of ϕä formalizes the fact that the relation symbol E‘generates’ all the others. Below, i, j < n, i 6= j, k is the least element ofn \ {i, j}, 1 ≤ l < â − 2, and 1 ≤ m ≤ n − 3.

(vi=vi)ä = (vi=vi) ,

(vi Idvi)ä = (vi=vi) ,

(viRvi)ä = ¬ (vi=vi) if R ∈ Rel \ {Id},

(vi=vj

)ä =(vi=vj

),

(viEvj

)ä = viEvj ,(viQ0vj

)ä = ¬∃vk(viEvk ∧ vkEvj

)∧ ¬viEvj ,

(

vi Idvj)ä= ∃vk

(

(viQ0vk)ä ∧

(

vkQ0vj)ä

)

∧ ¬∃vk

(

viEvk ∧(

vkQ0vj)ä

)

,

(viQ1vj

)ä = ∃vk

(

viEvk ∧(vkQ0vj

)ä)

∧ ¬∃vk

(

(viQ0vk)ä ∧

(vkQ0vj

)ä)

,

(

viQ2vj)ä= ∃vk

(

viEvk ∧(

vkQ1vj)ä

)

∧ ¬∃vk

(

viEvk ∧(

vkQ0vj)ä

)

∧ ∃vk(viEvk ∧ vkEvj

),

(

viQl+2vj)ä= ∃vk

(

viEvk ∧(

vkQl+1vj)ä

)

∧ ¬∃vk

(

viEvk ∧(

vkQlvj)ä

)

,

(viPmvj

)ä = ∃vk

(

(viQ0vk)ä ∧

(vkQ0vj

)ä)

∧ ¬∃vk

(

(viQmvk)ä ∧

(

vkQmvj)ä

)

,

(ϕ ⇒ ø)ä =

(

ϕä ⇒ øä)

,


(¬ϕ)ä = ¬ϕä ,

(∀viϕ)ä = ∀viϕ

ä .

It is easy to prove by induction that free(ϕ) = free(ϕä) for all ϕ ∈ Fmn.Note that if 3 ≤ l ≤ n and ϕ ∈ Fml then ϕ

ä ∈ Fml . Because no atomicsubformula of α is an equality or of the form viRvi for R ∈ Rel, it followsthat ¬αä is a 3-sentence with only one binary relation symbol in it, namelyE. It does not involve equality.

Lemma 9.1. 0n−1 ¬αä .

Proof. We use the notation and results of the preceding section. First,we show by induction on ϕ that for all ϕ ∈ Fmn−1, ϕ ⇔ ϕä isM-valid.Let ϕ ∈ Fmn−1 be atomic. If it is vi=vi or vi=vj or viEvj there is nothingto prove. Suppose it is viRvi for some R ∈ Rel. For every A ∈ M,A |= viRvi iff Aii = R, so by (M1), A |= viRvi iff R = Id. We haveA |= vi Idvi and A |= (vi Idvi)

ä , and A 2 (viRvi) and A 2 (viRvi)ä wheneverR ∈ Rel \ {Id}. So for all R ∈ Rel, A |= viRvi iff A |= (viRvi)ä , and itfollows that viRvi ⇔ (viRvi)

ä isM-valid.The remaining cases viRvj for i 6= j and R = Q0, Id,Q1, . . . ,P1, . . .are established by investigating instances and consequences of the secondalternative form (13) of α for various choices of S and T , with k = min(n \{i, j}) < n − 1. By Lemmas 8.7 and 8.8, this alternative form isM-valid,and by symmetry ofM (or by Lemmas 5.10 and 8.7), we see that

∨

R∈Rel

viRvj ∧∧

S,T∈Rel

∀vi∀vj

((

∃vk(viSvk ∧ vkTvj

) )

⇔∨

〈R,S,T 〉∈Mand

viRvj)

is alsoM-valid. Consider first E with itself:

∃vk(viEvk ∧ vkEvj

)⇔

(vi Idvj ∨ viQ1vj ∨ · · · ∨ viQâ−1vj

∨ viP1vj ∨ · · · ∨ viPn−3vj)isM-valid.

Since∨

R∈Rel viRvj isM-valid, we getM-validity of

¬∃vk(

viEvk ∧ vkEvj)

⇔(

viEvj ∨ viQ0vj)

(26)

and so of

viQ0vj ⇔ ¬∃vk(viEvk ∧ vkEvj

)∧ ¬viEvj .

By the definition of ä , this says that

viQ0vj ⇔(viQ0vj

)ä isM-valid.(27)

Next consider E with Q0, and Q0 with Q0. We obtainM-validity of

∃vk(

viEvk ∧ vkQ0vj)

⇔(

viQ1vj ∨ viP1vj ∨ · · · ∨ viPn−3vj)

,(28)

and of

∃vk(

viQ0vk ∧ vkQ0vj)

⇔ (vi Idvj ∨ viP1vj ∨ · · · ∨ viPn−3vj).(29)


Hence,

vi Idvj ⇔ ∃vk(viQ0vk ∧ vkQ0vj

)∧ ¬∃vk

(viEvk ∧ vkQ0vj

)

and

viQ1vj ⇔ ∃vk(

viEvk ∧ vkQ0vj)

∧ ¬∃vk(

viQ0vk ∧ vkQ0vj)

areM-valid. Using these last two statements and (27), we getM-validity of

vi Idvj ⇔ ∃vk

(

(viQ0vk)ä ∧

(vkQ0vj

)ä)

∧ ¬∃vk

(

viEvk ∧(vkQ0vj

)ä)

and

viQ1vj ⇔ ∃vk

(

viEvk ∧(

vkQ0vj)ä

)

∧ ¬∃vk

(

(viQ0vk)ä ∧

(

vkQ0vj)ä

)

,

and hence, by the definition of ä ,

vi Idvj ⇔(vi Idvj

)ä and viQ1vj ⇔(viQ1vj

)ä areM-valid.(30)

Next, considering E with Q1, we get theM-validity of

(31) ∃vk(

viEvk ∧ vkQ1vj)

⇔(viEvj ∨ viQ0vj ∨ viQ1vj ∨ viQ2vj ∨ viP1vj ∨ · · · ∨ viPn−3vj

),

which implies, by (26) and (28), theM-validity of

viQ2vj ⇔ ∃vk(

viEvk ∧ vkQ1vj)

∧ ¬∃vk(

viEvk ∧ vkQ0vj)

∧ ∃vk(

viEvk ∧ vkEvj)

,

and, by (27) and (30), theM-validity of

viQ2vj ⇔ ∃vk

(

viEvk ∧(vkQ1vj

)ä)

∧ ¬∃vk

(

viEvk ∧(

vkQ0vj)ä

)

∧ ∃vk(

viEvk ∧ vkEvj)

,

hence viQ2vj ⇔(viQ2vj

)ä isM-valid. The argument for viQ3vj is similar,using E and Q2. We have the followingM-validities:

∃vk(viEvk ∧ vkQ2vj

)

⇔(viEvj ∨ viQ1vj ∨ viQ2vj ∨ viQ3vj ∨ viP1vj ∨ · · · ∨ viPn−3vj

)

by definition ofMand, so by (31),

viQ3vj ⇔ ∃vk(

viEvk ∧ vkQ2vj)

∧ ¬∃vk(

viEvk ∧ vkQ1vj)

,

so by the result for Q1 and Q2,

viQ3vj ⇔ ∃vk

(

viEvk ∧(

vkQ2vj)ä

)

∧ ¬∃vk

(

viEvk ∧(

vkQ1vj)ä

)

,

whence viQ3vj ⇔(

viQ3vj)äis M-valid. This pattern continues, and

viQlvj ⇔(

viQlvj)äis M-valid whenever 3 < l ≤ â − 1. To complete


the proof, it suffices to continue in a similar way for Pm (1 ≤ m ≤ n − 3),using the definition ofMand and (29) to obtainM-validity of

∃vk(viQmvk ∧ vkQmvj

)⇔ vi Idvj ∨ viEvj ∨

∨

l 6=m

viPlvj ,

viPmvj ⇔ ∃vk(viQ0vk ∧ vkQ0vj

)

∧ ¬∃vk(

viQmvk ∧ vkQmvj)

,

viPmvj ⇔ ∃vk

(

(viQ0vk)ä ∧

(vkQ0vj

)ä)

∧ ¬∃vk

(

(viQmvk)ä ∧

(

vkQmvj)ä

)

,

and viPmvj ⇔ (viPmvj)ä . Thus, ϕ ⇔ ϕä isM-valid for all atomic ϕ ∈

Fmn−1. To show the same for all ϕ ∈ Fmn−1 is now a straightforwardinduction on ϕ, left to the reader.ByLemma8.8,α isM-valid. We have shownα ⇔ αä to beM-valid, soαä

is alsoM-valid. It now follows that0n−1 ¬αä , exactly as in Theorem 8.9. ⊣

After two more lemmata, we can show the main result of this paper, that¬αä requires n variables to prove.

Lemma 9.2. Let ø ∈ Fmn and l, m < n. Then Slm

(

øä)

≡n (Slmø)ä .

Proof. The proof is by induction. First, consider atomic ø, of the formviRvj , for i, j < n and R ∈ Rel ∪ {=}. If R is = or E, or if i = j, the resultholds trivially. For example,

Slm

((viEvj

)ä)

= Slm(viEvj

)= v[l,m](i)Ev[l,m](j)

= (v[l,m](i)Ev[l,m](j))ä = (Slm

(

viEvj)

)ä .

Now assume that i 6= j. Let k be the least element of n \ {i, j}, and let h bethe least element of n \ {[l, m](i), [l, m](j)}. Let i ′ = [l, m](i), and similarlydefine j ′, k′. Then by Lemmas 5.6 and 5.10,

Slm

((viQ0vj

)ä)

= ¬∃vk′(vi ′Evk′ ∧ vk′Evj′

)∧ ¬vi ′Evj′

≡n Shk′(

¬∃vk′(vi ′Evk′ ∧ vk′Evj′

)∧ ¬vi ′Evj′

)

= ¬∃vh(vi ′Evh ∧ vhEvj′

)∧ ¬vi ′Evj′

=(vi ′Q0vj′

)ä =(Slm

(viQ0vj

))ä,

so the lemma holds for Q0. Next we prove the lemma for Id. We useLemma 5.10, the definition of (vi Idvj)

ä , the fact that the lemma holds for

Q0, Lemma 5.6, and the definition of (vi ′ Idvj′)ä , in that order. Note that


h, k′ /∈ {i ′, j ′} = free(Slm(vi Idvj)) = free(Slm((vi Idvj

)ä)).

Slm

((

vi Idvj)ä

)

≡n Shk′Slm

((

vi Idvj)ä

)

= ∃vh

(

Shk′Slm(

(viQ0vk)ä )

∧ Shk′Slm( (vkQ0vj

)ä ))

∧ ¬∃vh

(

Shk′Slm (viEvk) ∧ Shk′Slm( (

vkQ0vj)ä ))

≡n ∃vh

((

Shk′Slm (viQ0vk))ä

∧(

Shk′Slm(

vkQ0vj) )ä

)

∧ ¬∃vh

(

Shk′Slm (viEvk) ∧(Shk′Slm(vkQ0vj)

)ä)

= ∃vh

(

(vi ′Q0vh)ä ∧

(

vhQ0vj′)ä

)

∧ ¬∃vh

(

vi ′Evh ∧(vhQ0vj′

)ä)

= (vi ′ Idvj′)ä

= (Slm(vi Idvj))ä ,

so the lemma holds for Id. Proceed similarly through the remaining relationsymbols to complete the proof for atomic ø. By Lemma 5.6, the same holdsfor arbitrary ø ∈ Fmn, completing the proof of the lemma. ⊣

Lemma 9.3. Suppose that ϕ ∈ Fmn has an n-proof not using axiom (A5).Then ⊢n ϕ

ä .

Proof. Let ð0, . . . , ðl = ϕ be an n-proof of ϕ not using (A5). We showby induction on m = 0, . . . , l that ⊢n ð

äm. Letm ≤ l and assume inductively

that ⊢n ðäs for s < m. If ðm is any instance of (A1)–(A4), it is clear that ð

äm

is also an instance of the same axiom, so that ⊢n ðäm. Assume that ðm is an

instance vi = vj ⇒ (ø ⇒ Sijø) of (A6). The previous lemma shows that

for any ø and i, j < n, (Sijø)ä ≡n Sij(ø

ä). Then ðäm is

vi = vj ⇒ (øä ⇒ (Sijø)

ä),

which by the above and Lemma 5.6 is ≡n-equivalent to

vi = vj ⇒ (øä ⇒ Sij(ø

ä)),

another instance of (A6). So ⊢n ðäm. Finally, if ðm follows from earlier ðs

by a rule of inference, it is clear by definition of ä that ðäm follows from theearlier ðäs by the same rule. This completes the induction, and shows that⊢n ϕ

ä . ⊣

Theorem 9.4. ¬αä requires n variables to prove.

Proof. We proved 0n−1 ¬αä in Lemma 9.1. It follows from the precedinglemma that ⊢n ¬α

ä , because (A5) was never used in the n-proof of ¬α givenin Theorem 7.5. ⊣


REFERENCES

[1] H.Andreka, R. D.Maddux, and I. Nemeti, Splitting in relation algebras,Proceedingsof the American Mathematical Society, vol. 111 (1991), no. 4, pp. 1085–1093.[2] C. Berge, Graphs and hypergraphs, North-Holland Publishing Co., Amsterdam, 1973,

Translated from the French by Edward Minieka, North-Holland Mathematical Library,vol. 6.[3] S. D. Comer, Color schemes forbidding monochrome triangles, Proceedings of the Four-

teenth Southeastern conference on combinatorics, graph theory and computing (Boca Raton,Fla., 1983), vol. 39, 1983, pp. 231–236.[4] S. Givant, Tarski’s development of logic and mathematics based on the calculus of

relations, Algebraic Logic (Budapest, 1988) (H. Andreka, J. Monk, and I. Nemeti, editors),Colloquia Mathematica Societatis Janos Bolyai, vol. 54, North-Holland, Amsterdam, 1991,pp. 189–215.[5] L. Gordeev, Combinatorial principles relevant to finite variable logic, RelMiCS 2000

(J. Desharnais, editor), Departement d’Informatique, Universite Laval, 2000, pp. 95–111.[6] R. L. Graham, Rudiments of Ramsey theory, CBMS, AmericanMathematical Society,

Providence, RI, 1981.[7] R. L. Graham, B. L. Rothschild, and J. H. Spencer, Ramsey theory, Wiley, New

York, 1980.[8] R. E. Greenwood andA.M. Gleason, Combinatorial relations and chromatic graphs,

Canadian Journal of Mathematics, vol. 7 (1955), pp. 1–7.[9] L. Henkin, Internal semantics and algebraic logic, Truth, syntax and modality (Proc.

Conf. Alternative Semantics, Temple Univ., Philadelphia, Pa., 1970), Studies in Logic and theFoundations of Mathematics, vol. 68, North-Holland, Amsterdam, 1973, pp. 111–127.[10] R. Hirsch, I. Hodkinson, and R. D. Maddux, Relation algebra reducts of cylindric

algebras and an application to proof theory, The Journal of Symbolic Logic, vol. 67 (2002),no. 1, pp. 197–213.[11] J. Johnson, Nonfinitizability of classes of representable polyadic algebras, The Journal

of Symbolic Logic, vol. 34 (1969), pp. 344–352.[12] L. Lowenheim, UberMoglichkeiten imRelativkalkul,Mathematische Annalen, vol. 76

(1915), pp. 447–470.[13] R. C. Lyndon, Relation algebras and projective geometries, TheMichigan Mathemat-

ical Journal, vol. 8 (1961), pp. 21–28.[14] R. D. Maddux, Topics in relation algebras, Ph.D. thesis, University of California,

Berkeley, 1978, pp. iii+241.[15] , Some varieties containing relation algebras, Transactions of the American

Mathematical Society, vol. 272 (1982), no. 2, pp. 501–526.[16] , A sequent calculus for relation algebras, Annals of Pure and Applied Logic,

vol. 25 (1983), no. 1, pp. 73–101.[17] , Nonfinite axiomatizability results for cylindric and relation algebras, The

Journal of Symbolic Logic, vol. 54 (1989), no. 3, pp. 951–974.[18] R. N. McKenzie, Representations of integral relation algebras, The Michigan Mathe-

matical Journal, vol. 17 (1970), pp. 279–287.[19] J. D. Monk, On representable relation algebras, The Michigan Mathematical Journal,

vol. 11 (1964), pp. 207–210.[20] , Model-theoretic methods and results in the theory of cylindric algebras, The

Theory of Models (Proc. 1963 Internat. Sympos. Berkeley) (J. Addison, L. Henkin, andA. Tarski, editors), North-Holland, Amsterdam, 1965, pp. 238–250.[21] , Nonfinitizability of classes of representable cylindric algebras, The Journal of

Symbolic Logic, vol. 34 (1969), pp. 331–343.


[22] , Provability with finitely many variables, Proceedings of the American Mathe-matical Society, vol. 27 (1971), pp. 353–358.[23] Participants, Open problems, Algebraic Logic (Budapest, 1988) (H. Andreka,

J. Monk, and I. Nemeti, editors), Colloquia Mathematica Societatis Janos Bolyai, vol. 54,North-Holland, Amsterdam, 1991, pp. 727–746.[24] C. S. Peirce, Note B: the logic of relatives, Studies in Logic by Members of the Johns

Hopkins University (C. S. Peirce, editor), Little, Brown, and Co., Boston, 1883, pp. 187–203.[25] F.W.K. E. Schroder,Vorlesungen uber die Algebra der Logik (exacte Logik), Volume

3, “Algebra und Logik der Relative” Part I, second ed., Chelsea, Bronx, New York, 1966, firstpublished in Leipzig, 1895.[26] A. Tarski, On the calculus of relations, The Journal of Symbolic Logic, vol. 6 (1941),

pp. 73–89.[27] , A simplified formalization of predicate logic with identity, Arch. Math. Logik

Grundlag., vol. 7 (1965), pp. 61–79.[28] A. Tarski and S. Givant, A formalization of set theory without variables, American

Mathematical Society, Providence, RI, 1987.

DEPARTMENT OF COMPUTER SCIENCE

UNIVERSITY COLLEGE

GOWER STREET, LONDONWC1E 6BT, UK

E-mail: [email protected]

DEPARTMENT OF COMPUTING

IMPERIAL COLLEGE

180 QUEEN’S GATE, LONDON SW7 2BZ, UK


DEPARTMENT OFMATHEMATICS

400 CARVER HALL

IOWA STATE UNIVERSITY

AMES, IOWA 50011-2066, USA


provability with finitely many variables

Documents