PROVABILITY WITH FINITELY MANY VARIABLES. Abstract. For every nite n 4 there is a logically valid sentence ' n with the

Transcription

1 PROVABILITY WITH FINITELY MANY VARIABLES ROBIN HIRSCH y, IAN HODKINSON, AND ROGER D MADDUX Abstrct. For every nite n 4 there is logiclly vlid sentence ' n with the following properties: ' n contins only 3 vribles (ech of which occurs mny times); ' n contins exctly one nonlogicl binry reltion symbol (no function symbols, no constnts, nd no equlity symbol); ' n hs proof in rst-order logic with equlity tht contins exctly n vribles, but no proof contining only n? 1 vribles. This result ws rst proved using the mchinery of lgebric logic developed in severl reserch monogrphs nd ppers. Here we replicte the result nd its proof entirely within the relm of (elementry) rst-order binry predicte logic with equlity. We need the usul syntx, xioms, nd rules of inference to show tht ' n hs proof with only n vribles. To show tht ' n hs no proof with only n? 1 vribles we use lterntive semntics in plce of the usul, stndrd, set-theoreticl semntics of rst-order logic. x1. Introduction. Questions bout wht cn be expressed nd proved with only nitely mny vribles begn with Peirce's [24] clculus of binry reltions, s developed by Schroder [25]. They let letters, b, c, : : : denote binry reltions on some xed underlying set U (clled the \domin", \universe of discourse", or \Denkbereich"). New reltions re mde from these ccording to opertions chosen by Peirce: the union + b, the intersection b, the complement, the reltive product ;b, the reltive sum y b, nd the converse. There re four distinguished reltions: 1 is the universl reltion on U (nmely U U), 0 is the empty reltion, 1' is the identity reltion on U, nd 0' is the diversity reltion on U, consisting of ll pirs of distinct elements of U. Peirce nd Schroder write, for exmple, ij to denote the sttement tht element i is relted to element j by reltion. The union, intersection, complement, reltive product, sum, converse, nd the four distinguished reltions cn then be dened s follows: y Reserch prtilly supported by UK EPSRC grnts GR/K54946, GR/L82441, GR/L Reserch prtilly supported by UK EPSRC grnts GR/K54946, GR/L

2 2 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX ( + b) ij i ij or b ij ; ( b) ij i ij nd b ij ; ij i not ij ; (;b) ij i for some k 2 U; ik nd b kj ; ( y b) ij i for every k 2 U; ik or b kj ; () ij i ji ; it is never the cse tht 0 ij ; it is lwys the cse tht 1 ij ; 1' ij i i = j; 0' ij i i 6= j: Elementry sttements bout reltions nd the domin cn be mde by equtions between reltions. Here re some exmples. The domin hs t lest two elements i 1 = 1;0';1. The domin hs t lest three elements i 1 = 1;(0' (0';0'));1. If the domin is not empty, then the reltion is nonempty i 1 = 1;;1. The reltion is trnsitive nd contins the identity reltion on the domin i = y. Every eqution in the clculus of reltions cn be trnslted into n elementry sttement bout the domin nd the binry reltions used in the eqution. For exmple, the ssocitive lw for reltive multipliction, nmely (1) ;(b;c) = (;b);c; is equivlent to the following sentence, which uses only three vribles denoting elements of the domin (nmely, i, j, nd k): (2) 8 i 8 j (9 k ( ik 9 i (b ki c ij )), 9 k (9 j ( ij b jk ) c kj )): It is strightforwrd to prove tht every eqution in the clculus of reltions is equivlent to 3-sentence, tht is, rst-order sentence tht uses only three vribles. The converse lso holds: every 3-sentence is equivlent to n eqution in the clculus of binry reltions [28, x3.9], [4, p 204]. Even if sentence uses more thn three vribles, it my still be expressible s n eqution if it cn be equivlently rewritten in wy tht uses only three vribles. An exmple for which this is true but fr from obvious is the \piring xiom", 8 i 8 j 9 k 8 h ( hk, 1' hi 1' hj ). Although it contins four vribles in the usul formultion, it is equivlent to 3-sentence, nd so to n eqution in the clculus of reltions [28, p 137]. Schroder seems to hve been the rst to consider the problem of whether every sentence cn be expressed by n eqution [25, p 551]. This problem ws solved by Alwin Korselt. In letter to Lowenheim [12], Korselt proved tht no eqution is equivlent to the following sentence, which

3 PROVABILITY WITH FINITELY MANY VARIABLES 3 sserts tht the domin hs t lest four elements: (3) 9 h 9 i 9 j 9 k (0' hi 0' hj 0' hk 0' ij 0' ik 0' jk ): Since every 3-sentence cn be trnslted into n eqution, it follows tht (3) is 4-sentence tht is not equivlent to ny 3-sentence. Korselt's theorem ws considerbly generlized by Trski [26, p 89], [28, p 61]. Trski proposed severl equtions s xioms for the clculus of reltions, including the six equtions listed below. These equtions, together with the ssocitive lw (1) nd ny nite set of equtionl xioms for Boolen lgebrs, xiomtize the clss of reltion lgebrs. (4) (5) (6) (7) (8) (9) ( + b);c = ;c + b;c; ;1' = ; = ; ( + b)= + b; (;b)= b;; ;;b + b = b: A 3-sentence equivlent to the lst eqution is (10) 8 i 8 j (9 k ( ki 8 i ( ki b ij )) ) b ij ): This sentence is logiclly vlid, nd hs proof in ny complete system of rst-order logic. A proof of (10), in nturl deductive style, goes s follows. Let i nd j be rbitrry elements of the domin. Assume there is some element k such tht ki nd for every i, either ki or b ij. The ltter ssumption, pplied to the rbitrry i we re considering, implies tht either ki or b ij. But ki, so b ij. We hve so fr proved tht 9 k ( ki 8 i ( ki b ij )) ) b ij. Since we proved this for rbitrry i nd j, we my generlize nd conclude tht (10) holds. Note tht we only needed the three vribles tht ctully occur in (10). We suspect tht, for prcticlly ny textbook xiomtiztion of rst order logic (with or without equlity), (10) will still be provble if the xiom set is restricted to those formul tht do not contin ny vribles outside xed group of three vribles. Principles involving equlity ren't needed to prove (10), but they re used in the proof of the following 3-sentence, which is equivlent to (5): (11) 8 i 8 j (9 k ( ik 1' kj ), ij )): To prove this, let i nd j be rbitrry elements of the domin. For one direction, ssume there is some k such tht ik nd 1' kj. By n xiom sserting tht equlity is congruence reltion (tht is, n equivlence reltion on the domin with the property tht ik nd 1' kj imply ij, nd kj nd 1' ki imply ij ), this gives us ij. For the converse, ssume ij. Assuming tht everything is equl to something, we conclude there is some

4 4 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX k such tht 1' jk. Since equlity is congruence reltion, it follows tht ik nd 1' kj, hence 9 k ( ik 1' kj ). So fr we hve proved 9 k ( ik 1' kj ), ij. Generlizing, we obtin (11). Similr 3-vrible proofs cn be given for the 3-sentences obtined from (4), (6), (7), nd (8). On the other hnd, while it is esy to prove (2) using four vribles, the diculties encountered when trying to do it with only three led Trski to explicitly include (2) in the xiomtiztion of the 3-vrible logic L + 3 dened in [28], where the problem of xiomtizing n-vrible logic for n 3 is discussed t length. Let Ls + 3 be obtined from L + 3 by deleting (2) [28, p 89]. Trski estblished tht (2) is not provble in Ls + 3 by considering n lgebr, constructed by J. C. C. McKinsey, tht stises (4){(9): i.e. ll of the equtionl xioms for reltion lgebrs except the ssocitive lw for reltive multipliction (1). See [28, p 68]; Trski's proof ws never published, but nother proof ws given by Henkin [9]. Four vribles re sucient to prove not only the 3-sentences obtined from the equtionl xioms (1) & (4){(9) for reltion lgebrs, but lso the 3-vrible trnsltions of ll equtions derivble from these xioms. It follows tht if n eqution is true in every reltion lgebr then its 3-vrible trnsltion hs 4-vrible proof. It turns out tht the converse is lso true: if the 3-vrible trnsltion of given eqution hs 4-vrible proof, then the eqution holds in every reltion lgebr [14], [16], [17], [28, pp 92{93], [4, pp 206{207]. Therefore, if n eqution E fils in some reltion lgebr nd its equivlent 3-sentence T is logiclly vlid, then T is provble but not with fewer thn ve vribles. Here is n exmple of n eqution whose equivlent 3-sentence is logiclly vlid nd provble with ve vribles but not four [28, pp 54, 93], [4, p 198]: (12) 1;(( y ) + [(;) + 1' + ( + ; + ) ] + ;);1 = 1: Both this eqution nd its equivlent 3-sentence use only single binry reltion symbol, becuse they rise from the nlysis of 16-element nonrepresentble reltion lgebr tht is generted by single element [18]. Suppose 3-sentence T is logiclly vlid, hence provble using some number of vribles in some textbook xiomtiztion of rst order logic with equlity. Let n be the lest number of vribles tht occur in ny proof of T. Then T requires n vribles to prove, in the sense tht no smller number of vribles will suce; in prticulr, T hs no proof tht uses only n? 1 vribles. It hs been known for some time tht there re 3-sentences tht require rbitrrily lrge numbers of vribles to prove, tht is, for every integer n, there is 3-sentence tht requires more thn n vribles to prove. Monk [21, Th 1.11] proved n lgebric formultion of this result in the course of showing tht the clss of n-dimensionl cylindric lgebrs is not nitely xiomtizble whenever n 3. His proof

5 PROVABILITY WITH FINITELY MANY VARIABLES 5 involves the construction of cylindric lgebrs tht were modied into nonrepresentble polydic lgebrs by Johnson [11]. Monk [22] outlined the connections between the lgebr nd logic, nd pplied one of Johnson's results to show tht no nite set of xiom schemt for rst-order logic with equlity nd only nitely mny vribles is both sound nd complete. Monk [21, x4] mde conjecture bout cylindric lgebrs. In logicl terms, Monk's conjecture ws tht for every n 3 there is 3- sentence tht requires n vribles to prove. Both the lgebric nd logicl versions of the conjecture re conrmed in [10]. It is perhps n interesting historicl fct tht the lgebrs constructed lredy by Monk [21] cn be used to do so. These lgebrs cnnot, however, be used to obtin similr results for lnguges with xed, nite number of binry reltion symbols. The reson is tht the number of genertors of one of Monk's lgebrs depends on the size of the lgebr; lrger lgebrs require more genertors. The sme feture icts both the nonrepresentble reltion lgebrs tht were constructed from projective geometries by Lyndon [13] nd used by Monk [19] to show tht the clss of representble reltion lgebrs is not nitely xiomtizble, nd lso the corresponding 3-dimensionl cylindric lgebrs tht were used by Monk [20] to show tht the clss of 3-dimensionl cylindric lgebrs is not nitely xiomtizble. The rst-order lnguge L used in [28] hs n equlity symbol nd exctly one binry reltion symbol, the pproprite choice for the development of set theory, in which the sole nonlogicl predicte is the membership reltion. To prove for this lnguge tht there re 3-sentences tht require rbitrrily lrge numbers of vribles to prove, it ws necessry to modify Monk's lgebrs into ones tht re generted by single element [17, x4]. This introduces complexity into the construction; the utomorphism group goes from lrge to trivil. Similrly modied versions of Monk's lgebrs re used in [10] to show tht for every n 3 there is 3-sentence in L tht requires n vribles to prove, thus solving problem posed by Trski nd Givnt [28, p 93], [4, p 208], [23, p 735]. Here we replicte this result nd its proof entirely within the relm of rst-order logic. The structure of the modied lgebrs shows up in the 3-sentences. Our proof proceeds s follows. In x2, we recll nite Rmsey theorem whose formliztion in rst-order logic will provide n upper bound on the number of vribles required to prove our 3-sentences. The n-vrible frgments of rst-order logic, nd their proof theory, re dened formlly in x3 nd x4. Some proof-theoretic lemms re given in x5. The min work begins in x6, where the `Rmsey 3-sentence' (for ech n 4) is specied nd shown (in x7) to be provble with n vribles. In x8, we show tht it is not provble with fewer vribles by showing tht it is not vlid in certin unorthodox semntics for (n? 1)-vrible logic

6 6 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX tht vlidtes theorems provble with n? 1 vribles. Finlly, in x9, we obtin similr sentence written with only one binry reltion symbol. We believe most of this pper cn be red s if it were chpter in (firly terse nd sophisticted) text on rst-order logic with equlity. x2. A Rmsey theorem. Dene :!!! by (0) = 1 nd (k + 1) = 1 + (k + 1) (k) for every k <!. This function gives n upper bound for certin Rmsey number [7], [6, p 6], [8, Cor 3], [2, p 440{443]. Its rst ve vlues re, respectively, 1, 2, 5, 16, nd 65. For ny set X, let [X] 2 be the set of 2-element subsets of X: [X] 2 = ffx; yg : x; y 2 X; x 6= yg: For 1 k <!, k-coloring of X is prtition of [X] 2 into k pieces (clled `colors') so tht no `monochromtic tringle' ppers, tht is, there is no 3-element subset of X with ll three of its 2-element subsets in the sme piece of the prtition. Proposition 2.1. Let k 1. If set X hs more thn (k) elements then X hs no k-coloring. Proof. Here is the stndrd proof. First cse: k = 1. No 1-coloring cn exist if X hs more thn (1) = 2 elements, becuse every 3-element subset is monochromtic tringle for every 1-coloring. Next, we ssume the result holds for prticulr k 1 nd prove it for k + 1. Assume X hs (k + 1)-coloring. Choose such coloring. Fix one element x 2 X, nd dene n equivlence reltion on X n fxg by letting y y 0 if nd only if fx; yg nd fx; y 0 g hve the sme color in the chosen coloring. For distinct y y 0 in X n fxg, if fx; yg nd fx; y 0 g belong to color c, sy, then fy; y 0 g cnnot lso belong to c, else fx; y; y 0 g is monochromtic. Hence we cn prtition X n fxg into t most k + 1 prts, given by the -clsses, nd within ech prt only k colors re used. Thus, ech of the prts hs k-coloring nd inductively hs size not exceeding (k). So jxj 1 + (k + 1) (k) = (k + 1). Prtitions of [X] 2 cn be treted in rst-order logic by mens of symmetric binry reltions on X. In rst-order lnguge with binry reltion symbols nd equlity, it is possible with 2-sentences to ssert tht the reltions (denoted by the reltion symbols) re symmetric, pirwise disjoint, nd their union is the diversity reltion (tht holds between ny two distinct elements). Tht monochromtic tringles do not occur cn be expressed by 3-sentences, such s 8 i 8 j 8 k : ( ij ik kj ). Thus, for ny k 1, there is single 3-sentence tht is stisble in X just in cse some set of k binry reltions on X forms prtition of the diversity reltion on X into disjoint symmetric binry reltions with no monochromtic tringles in short, tht there is k-coloring of X. By Proposition 2.1,

7 PROVABILITY WITH FINITELY MANY VARIABLES 7 such sentence hs no model of crdinlity greter thn (k). By considering more complicted types of prtitions, we cn further ssert with only 3 vribles tht there re more thn (k) elements. A 3-sentence obtined this wy, sserting the existence of k-coloring on set with more thn (k) elements, hs no model, so its negtion is logiclly vlid nd provble. It is esy to see tht it cn be proved with 1+(k) vribles (Remrk 6.2 below gives corresponding semntic rgument). The im of the next few sections is to get this number down, by formlizing the inductive proof of Proposition 2.1. x3. Reltionl lnguges. Let L be rst order lnguge with equlity symbol =, ny set of binry reltion symbols, no function symbols, no constnts, nd vribles v i for i <!. For every n! let Vr n = fv i : i < ng: Let Rel be the set of reltion symbols of L. (The equlity symbol = is not in Rel.) Let tomic n be the set of tomic formul of L written with vribles from Vr n. Thus tomic n = fv i =v j ; v i Rv j : R 2 Rel; i; j < ng: The only propositionl connectives in L re : nd ), nd the only quntier is 8. The other connectives nd the existentil quntier re introduced s bbrevitions in the usul wy. We will need to use conjunctions nd disjunctions of formul indexed by vrious nite index sets. The order in which the formul pper is ssumed to be determined by some stndrd ordering of the index set. Prentheses re restored by ssocition to the left. For exmple, ik ' i = ( ((' 0 ' 1 ) ' 2 ) ' k?1 ) ' k : At one point (Lemm 7.4) we encounter disjunction over possibly empty index set, in which cse we let W i2; ' i =?, where? = :8 v0 (v 0 =v 0 ). Similrly, V i2; ' i = >, where > = 8 v0 (v 0 =v 0 ). For n L-formul ', let free(') be the set of vribles tht occur free in ', dened for ny R 2 Rel; i; j < n, nd formul ';, by: free(v i =v j ) = fv i ; v j g; free(v i Rv j ) = fv i ; v j g; free(' ) ) = free(') [ free( ); free(:') = free('); free(8 vi ') = free(') n fv i g:

8 8 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX A formul ' is sentence if free(') = ;. For every n! let Fm n be the set of formul of L tht contin only vribles in Vr n, i.e., Fm n = T f X : tomic n X; if ' 2 X, 2 X, nd x 2 Vr n ; then :' 2 X; ' ) 2 X; nd 8 x ' 2 X g: We sy tht ' is n n-formul if ' 2 Fm n, nd n n-sentence if ' 2 Fm n nd free(') = ;. x4. Axiomtiztion. We now develop the proof theory of L. For every ' 2 Fm! nd ll i; j <! let S ij ' be the result of interchnging v i nd v j, tht is, simultneously replcing every occurrence of v i in ' with v j nd every occurrence of v j in ' with v i. For exmple, if R 2 Rel then S 01 (8 v0 9 v1 v 0 Rv 1 ) = 8 v1 9 v0 v 1 Rv 0 nd S 02 (8 v0 9 v1 v 0 Rv 1 ) = 8 v2 9 v1 v 2 Rv 1. Note tht if i; j < n nd ' 2 Fm n, then S ij ' 2 Fm n s well. A tutology is formul which is mpped to 1 (truth) by the cnonicl extension to ll formul of ny mp from tomic nd universlly quntied formul to the two-element set f1; 0g. For every n!, let Ax n be the set of ll formul of the following types. Nme Axiom Restrictions (A1) ' ' 2 Fm n nd ' is tutology (A2) 8 vi (' ) ) ) (' ) 8 vi ) '; 2 Fm n ; i < n; nd v i =2 free(') (A3) 8 vi ' ) ' ' 2 Fm n nd i < n (A4) 9 vi (v i =v j ) i; j < n nd i 6= j (A5) (v i =v j ) ) (' ) ) '; 2 tomic n ; i; j < n; nd is obtined from ' by replcing some occurrence of v i in ' by v j (A6) (v i =v j ) ) (' ) S ij ') ' 2 Fm n nd i; j < n The rules of inference re modus ponens (infer from ' nd ' ) ) nd n-generliztion (infer 8 x ' from ', for ny x 2 Vr n ). These rules cn be displyed in more grphicl style s follows. ' ' ) ' 8 x ' A formul ' 2 Fm! is n-provble, in symbols `n ', if ' belongs to every set of formul tht contins Ax n nd is closed under modus ponens nd n-generliztion. As Fm n is such set, if `n ' then ' 2 Fm n. By n n-proof of ' we men nite sequence of formul in Fm n whose lst element is ', hving the property tht every formul in the sequence is either member of Ax n or follows from one or two previous formul by n-generliztion or modus ponens. Evidently, ' is n-provble i there

9 PROVABILITY WITH FINITELY MANY VARIABLES 9 exists n n-proof of '. Two formul '; 2 Fm n re n-equivlent, in symbols ' n, i `n ',. The xiom set (A1){(A6) is bsed on the xiom set 2 of Trski [27]. 2 is dened by eight xiom schemt, (B1) through (B8). Trski's xioms (B1){(B3) re tutologies from which every other tutology cn be obtined by modus ponens. Axiom (A1) is our replcement for (B1){(B3). Axiom (A2) substitutes for the following two xioms (see Lemms 5.5 nd 5.4 below): (B4) 8 x (' ) ) ) (8 x ' ) 8 x ); (B6) ' ) 8 x ' for x =2 free('): Trski's (B5), (B7), nd (B8) coincide with our (A3), (A4), nd (A5), respectively. Axiom (A6) cn be derived from (A1){(A5) when n =!. However, proofs of (A6) (nd relted schemt) usully involve the introduction of vribles tht do not occur in '. This my not be possible when n is nite since every vrible in the lnguge cn then occur in single formul. Therefore, (A6) hs been explicitly included in Ax n. Axioms (A1){(A6) re sound with respect to the usul semntics: ll formul in Ax n re logiclly vlid (true in ll models under ll ssignments to their free vribles). The rules of inference preserve logicl vlidity. It follows tht if `n ' then ' is logiclly vlid. To prove the converse when n =!, it suces to observe tht 2 is complete [27, Th 5], nd tht every formul in 2 is!-provble. On the other hnd, s ws proved by Monk [22], completeness fils when n <!. We sy tht formul requires n vribles to prove if it cn be proved with n vribles, but cnnot be proved with only n? 1 vribles, i.e., `n ' nd 6 `n?1 '. Evidently, ny tutology in Fm n n Fm n?1 requires n vribles to prove. A more interesting chllenge is to nd 3-formul in Fm 3 tht require n vribles to prove. x5. Provbility. The results on provbility in this section will be used lter. Here re three lemmt tht hold for every n!. Their proofs, which we omit, use only xioms (A1){(A3) nd the two inference rules. None of the equlity xioms re involved. Lemm 5.1. If ' 2 Ax n then `n '. Lemm 5.2. If `n ' nd i < n then `n 8 vi '. Lemm If `n ' nd `n ' ), then `n. In prticulr, if `n ' nd ' ) 2 Ax n, then V `n. 2. If `n ' 0 ; : : :; `n ' k nd ( ik ' i) ) is tutology in Fm n, then `n. Lemm 5.3 implies tht n is n equivlence reltion on Fm n, nd tht ' n i `n ' ) nd `n ) '.

10 10 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX Lemm 5.4. If ' 2 Fm n, i < n, nd v i nd `n 9 vi ' ) '. =2 free('), then `n ' ) 8 vi ' Proof. By (A1) nd Lemm 5.1, `n ' ) '. By Lemm 5.2, `n 8 vi (' ) '). By (A2) nd Lemm 5.1, `n 8 vi (' ) ') ) (' ) 8 vi '). So by Lemm 5.3, `n ' ) 8 vi '. As v i =2 free(:'), this lso gives `n :' ) 8 vi :'. By Lemm 5.3, we obtin `n :8 vi :' ) ', i.e. `n 9 vi ' ) '. Lemm 5.5. If '; 2 Fm n nd i < n then `n 8 vi (' ) ) ) (8 vi ' ) 8 vi ) : Proof. By (A3) nd Lemm 5.1, So by Lemm 5.3, `n 8 vi (' ) ) ) (' ) ) nd `n 8 vi ' ) ': `n 8 vi (' ) ) ) (8 vi ' ) ): By Lemm 5.2, `n 8 vi (8 vi (' ) ) ) (8 vi ' ) )): Since v i is not free in 8 vi (' ) ), we cn use (A2) nd Lemm 5.3 to get `n 8 vi (' ) ) ) 8 vi (8 vi ' ) ): By (A2) nd Lemm 5.1, `n 8 vi (8 vi ' ) ) ) (8 vi ' ) 8 vi ). By Lemm 5.3, we obtin proving the lemm. `n 8 vi (' ) ) ) (8 vi ' ) 8 vi ); Lemm 5.6. Assume tht '; ' 0 ; 2 Fm n nd tht 0 is the result of replcing n occurrence of ' in by ' 0. If ' n ' 0, then n 0. Hence, if `n then `n 0. Proof. By induction on. If is ', the result is cler. Inductively ssume the result for, so tht 0 n, i.e. `n, 0. For 2 Fm n, consider ). By (A1) nd Lemm 5.1, `n (, 0 ) ) (( ) ), ( 0 ) )). By Lemm 5.3, `n ( ) ), ( 0 ) ), so tht ) 0 n ). The cses ) nd : re similr. Finlly, consider 8 vi. We hve `n ) 0. Lemm 5.2 gives `n 8 vi ( ) 0 ), nd Lemms 5.5 nd 5.3 give `n 8 vi ) 8 0 vi. Similrly, `n 8 0 vi ) 8 vi, giving 8 vi n 8 0 vi vi Lemm 5.3, s required. Lemm 5.7. If '; 2 Fm n nd i < n then `n 8 vi (' ) ) ) (9 vi ' ) 9 vi ) :

11 PROVABILITY WITH FINITELY MANY VARIABLES 11 Proof. Observe tht Lemms 5.3, 5.5, nd 5.6 give `n 8 vi (' ) ) ) 8 vi (: ) :'); `n 8 vi (: ) :') ) (8 vi : ) 8 vi :'); `n (8 vi : ) 8 vi :') ) (:8 vi :' ) :8 vi : ): The lemm now follows by Lemm 5.3. We will nd the following `monotonicity' lemm hndy. Lemm 5.8. Assume ' i ; i ; 2 Fm n nd `n ' i ) i for ech i 2 I. Then `n V W i2i ' i ) W i2i i. Furthermore, for i2i ' i ) V i2i i nd `n ech i 2 I; j < n, `n 9 vj ' i ) 9 vj i, nd if `n ) ' i then `n ) i. Proof. All prts but the third follow immeditely from Lemm 5.3. For the third prt, if `n ' i ) i then Lemm 5.2 gives `n 8 vj (' i ) i ), nd Lemms 5.7 nd 5.3 give `n 9 vj ' i ) 9 vj i. Lemm 5.9. Let i; j < n nd '; 2 Fm n. Then 9 vi (' ) n 9 vi ' 9 vi ; 9 vj (' ) n ' 9 vj if v j =2 free('); 8 vi (' ) n 8 vi ' 8 vi ; 8 vj (' ) n ' 8 vj if v j =2 free('): Proof. By Lemms 5.8 nd 5.3, `n (9 vi ' 9 vi ) ) 9 vi (' ). For the converse, (A1) nd Lemm 5.1 give `n :' ) (: ) :(:' ) )): Lemm 5.2 yields `n 8 vi (:' ) (: ) :(:' ) ))), nd two pplictions of Lemms 5.5 nd 5.3 give us `n 8 vi :' ) (8 vi : ) 8 vi :(:' ) )): As (p ) (q ) r)) ) (:r ) (::p ) :q)) is tutology, Lemm 5.3 now yields `n :8 vi :(:' ) ) ) (::8 vi :' ) :8 vi : ), which, bbrevited, is `n 9 vi (' ) ) (9 vi ' 9 vi ): For the second prt, (A3) nd Lemm 5.1 give `n 8 vj : ) :. An ppliction of Lemm 5.3 yields `n (' ) 8 vj : ) ) (' ) : ), nd by Lemm 5.2, `n 8 vj ((' ) 8 vj : ) ) (' ) : )): As v j is not free in ' ) 8 vj :, (A2) nd Lemms 5.1 nd 5.3 give us `n (' ) 8 vj : ) ) 8 vj (' ) : ): (A2) nd Lemm 5.1 immeditely give the converse of this, so Lemms 5.3 nd 5.6 now yield (' ) 8 vj : ) n 8 vj (' ) : ): :(' ) ::8 vj : ) n :8 vj ::(' ) : );

12 12 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX tht is, ' 9 vj n 9 vj (' ), s required. The lst two sttements follow from the rst two by Lemms 5.3 nd 5.6. Now we mke use of the equlity xioms. Note tht we never use (A5). Lemm If i; j < n, ' 2 Fm n, nd v i ; v j S ij '. =2 free('), then ' n Proof. If i = j then the result is obvious, so ssume i 6= j. By (A6) nd Lemm 5.1, `n (v i =v j ) ) (' ) S ij '): By Lemm 5.2, `n 8 vi ((v i =v j ) ) (' ) S ij ')) : By Lemms 5.7 nd 5.3, By (A4) nd Lemms 5.1 nd 5.3, `n 9 vi (v i =v j ) ) 9 vi (' ) S ij '): `n 9 vi (' ) S ij '): Clerly, v i =2 free(' ) S ij '), so by Lemms 5.4 nd 5.3, `n ' ) S ij ': Similrly, `n :' ) :S ij '; since :S ij ' = S ij (:') nd v i ; v j =2 free(:'). So, by Lemm 5.3, `n S ij ' ) ', whence ' n S ij '. x6. A specil lnguge. Now we put restrictions on n nd L. For the reminder of the pper, x n 4 nd > 2(n? 3) (here, is s in x2), nd let L be the rst-order lnguge with the following set of binry reltion symbols: Rel = fid; E; P 1 ; : : :; P n?3 ; Q 0 ; : : :; Q?1 g: The smllest choices for n nd re n = 4 nd = 1 + 2(1) = 5, nd Rel = fid; E; P 1 ; Q 0 ; : : :; Q 4 g. It cn be checked tht? 1 n? 3 for ll n 4. Prtition Rel 3 = Rel Rel Rel into two prts, s follows. A triple hr; S; Ti 2 Rel 3 is sid to be forbidden if one of the following conditions holds: (F1) R = Id nd S 6= T; or S = Id nd R 6= T; or T = Id nd R 6= S; (F2) fr; S; Tg fe; Q 0 g; (F3) fr; S; Tg fq i : i < g; (F4) fr; S; Tg fp i ; Q i g for some i with 1 i n? 3; (F5) fr; S; Tg = fe; Q i ; Q j g for some i; j < with ji? jj > 1: Notice tht ny permuttion of forbidden triple is lso forbidden. Any triple tht is not forbidden is sid to be mndtory. Let Mnd be the set of mndtory triples, nd let Forb be the set of forbidden triples. Then

13 PROVABILITY WITH FINITELY MANY VARIABLES 13 Rel 3 = Mnd [ Forb nd ; = Mnd \ Forb. Now we dene some specil 3- sentences tht mke ssertions bout the binry reltions denoted by the reltion symbols of L. They sy, mong other things, tht the reltions re symmetric nd form prtition of the universl reltion, nd tht Id is congruence reltion (n identity for reltive multipliction from either side). Any two objects re relted by some reltion: 1 = 8 v0 8 v1 R2Rel v 0 Rv 1 : Forbidden triples do not occur: 2 = hr;s;t i2forb 8 v0 8 v1 8 v2 : v 0 Rv 1 v 0 Sv 2 v 2 T v 1 : Mndtory triples must occur wherever possible: Now let 3 = hr;s;t i2mnd = : 8 v0 8 v1 v 0 Rv 1 ) 9 v2 (v 0 Sv 2 v 2 T v 1 ) Of course, depends implicitly on n. 2 cn be expressed in form similr to 3. By Lemms 5.1, 5.3, 5.6, nd 5.9, 8 v0 8 v1 v 0 Rv 1 ) :9 v2 (v 0 Sv 2 v 2 T v 1 ) : 2 3 hr;s;t i2forb Using only Lemms 5.1, 5.3, 5.6, nd 5.9, we cn derive useful but longer lterntive formultions of these sentences (13) S;T 2Rel S;T 2Rel S;T 2Rel v0 8 v1 hr;s;t i2forb 8 v0 8 v1 hr;s;t i2forb 8 v0 8 v1 S;T 2Rel S;T 2Rel hr;s;t i2mnd v 0 Rv 1 hr;s;t i2mnd : v 0 Rv 1 v 0 Rv 1 ) :9v2 (v 0 Sv 2 v 2 T v 1 ) 8 v0 8 v1 :9v2 (v 0 Sv 2 v 2 T v 1 ), 8 v0 8 v1 9v2 (v 0 Sv 2 v 2 T v 1 ), v 0 Rv 1 ) 9v2 (v 0 Sv 2 v 2 T v 1 ) hr;s;t i2forb hr;s;t i2mnd v 0 Rv 1 v 0 Rv 1 :

14 14 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX These conclusions follow only from the ssumption tht Mnd nd Forb prtition Rel 3 ; their ctul structure hs not mttered, but it does in the next lemm. Let 4 = 8 v0 9 v1 v 0 Idv 1 : Lemm 6.1. `3 1 3 ) 4. Proof. Let R 2 Rel. Since hr; Id; Ri 2 Mnd, we pply Lemm 5.3 to get `3 3 ) 8 v0 8 v1 [v 0 Rv 1 ) 9 v2 (v 0 Idv 2 v 2 Rv 1 )]. By xiom (A3) nd Lemm 5.3, `3 3 ) [v 0 Rv 1 ) 9 v2 (v 0 Idv 2 v 2 Rv 1 )]. As `3 v 0 Idv 2 v 2 Rv 1 ) v 0 Idv 2, Lemms 5.8 nd 5.3 give `3 3 ) [v 0 Rv 1 ) 9 v2 v 0 Idv 2 ]. By Lemm 5.3, we obtin `3 3 ) V R2Rel [v 0Rv 1 ) 9 v2 v 0 Idv 2 ]. Now by (A3) nd Lemm 5.3 gin, `3 1 ) W R2Rel v 0Rv 1. Hence by Lemm 5.3, `3 ( 1 3 ) ) 9 v2 v 0 Idv 2. Lemm 5.10 shows tht 9 v2 v 0 Idv v1 v 0 Idv 1. So by Lemm 5.6, `3 ( 1 3 ) ) 9 v1 v 0 Idv 1. By Lemm 5.2, `3 8 v0 (( 1 3 ) ) 9 v1 v 0 Idv 1 ), nd by (A2) nd Lemm 5.3, `3 ( 1 3 ) ) 8 v0 9 v1 v 0 Idv 1, s required. Remrk 6.2. It is not hrd to show tht the 3-sentence is unstisble. Suppose, to the contrry, tht hd model M, nd pick n element of M. By Lemm 6.1, M j= Id b for some b. Let = fi < : i is eveng. As M j= 3, the fct tht hid; Q i ; Q i i is mndtory triple for ll i 2 forces M to contin points c i (i 2 ) such tht M j= Q i c i c i Q i b. Let i < j in. By 1, M j= c i R c j for some R 2 Rel. Choose such n R. So M j= Q j c j Q i c i c i R c j : From 2 nd the denition of Forb, we see tht R 6= Id since Q i 6= Q j, R 6= E since ji? jj > 1, nd R 6= Q k for every k < since hq i ; Q j ; Q k i is forbidden. Hence, M j= c i P l c j for some l with 1 l n? 3. But, by the choice of, jj > (n? 3), so by Proposition 2.1, there is no (n? 3)-coloring of. There must therefore be i < j < k in such tht M j= c i P l c k c i P l c j c j P l c k for some l. However, this contrdicts 2, since hp l ; P l ; P l i is forbidden. Hence hs no model. This rgument cn be esily formlized to give proof of : using 2 + jj vribles. Remrk 6.3. We cn trnslte : into n equivlent eqution in the clculus of reltions in the wy described in the introduction. (Of course, the preceding remrk shows tht 0 = 0 is lso equivlent to :.) 1 sttes tht P R2Rel R = 1, or equivlently tht Q R2Rel R = 0. 2 sttes tht R (S ;T ) = 0 whenever hr; S; Ti 2 Forb. 3 sttes tht R S ;T = 0 whenever hr; S; Ti 2 Mnd. For ll R; S; T 2 Rel, let ( R S ;T (R; S; T ) = R S ;T if hr; S; Ti 2 Forb; if hr; S; Ti 2 Mnd:

15 PROVABILITY WITH FINITELY MANY VARIABLES 15 Q So sttes tht ( R2Rel R) + (P R;S;T 2Rel (R; S; T )) = 0. The desired eqution equivlent to : is therefore Y X (14) 1; R + (R; S; T ) ;1 = 1: R2Rel R;S;T 2Rel From (F1){(F5) it cn be shown tht for every R 2 Rel there is term t R (x) in the clculus of reltions such tht if is the elementry 3-vrible trnsltion of t R (E) = R (cf. the introduction) then `3 ). For exmple, t Q0 = x;x x. Terms like this re used in x9 to eliminte ll reltion symbols other thn E from. Remrk 6.4. If we were only interested in 3-sentences tht require n vribles to prove, without regrd for the number of binry reltion symbols used in such 3-sentences, then we could simplify the denition of Forb nd Mnd, lthough these chnges hve lmost no eect on the length of : or the corresponding eqution (14). First, delete E from the lnguge. Conditions (F2) nd (F5) in the denition of forbidden triples fll wy. If 0 is the resulting 3-sentence, then : 0 still requires n vribles to prove. Next, drop Q i from (F4) nd let 00 be the resulting 3-sentence. Agin : 00 requires n vribles to prove. At this point the denition of forbidden nd mndtory triples mtches reltion lgebrs tht rise from Monk's cylindric lgebrs. These lgebrs cn lso be obtined by the process of splitting one tom [1, Ex 6], strting with the lgebrs clled E k (f2; 3g) in [15, p 510]. These lgebrs re symmetric integrl reltion lgebrs whose forbidden triples re exctly the ones of the form h; ; i, for every nonidentity tom. Sentences similr to : 00 nd the corresponding equtions (14) re given by Gordeev [5], where it is shown inter li tht for every n 4 there is 3-sentence tht cn be proved in proof theory with equlity nd with n vribles, but not with n? 2. If we wish to construct the tht rises from the lgebrs E k (f2; 3g), we my chieve the sme eect by deleting the reltion symbols Q i nd condition (F3), leving only Id, the symbols P j, (F1), nd the simplied (F4). It is not known, however, except for smll n, whether : hs no model in this cse [3], [23, Prob. 5]. We will show in the next section, by formlising the proof of Proposition 2.1, tht the 3-sentence : dened ccording to (F1){(F5) is n- provble (Theorem 7.5). Then, in x8, we show tht is vlid in certin (n? 1)-dimensionl, non-stndrd models; it follows (Theorem 8.9) tht : is not (n? 1)-provble. In x9 we lter : by replcing ll tomic formul other thn those contining E with formul contining only E. The resulting 3-sentences (for n = 4; 5; : : :) still require n vribles to prove, nd the corresponding equtions (14) cn be used to continue the sequence begun with (1) nd (12).

16 16 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX t9 vm+1 S 0 = Q i S 1 = P (1) : : : S m = P (m) (some even i < ) t t t v 0 v 1 : : : : : : : : : Figure 1. The conjunct of corresponding to S 2. x7. : is n-provble. To help the induction in the proof tht `n :, we introduce structure clled spectrum. A 0-spectrum is subset of fq i : i <, i eveng. If 1 m n? 3, we sy tht is n m-spectrum with color sequence if is one-to-one function : f1; : : :; mg! f1; : : :; n? 3g, nd v m fq i : i < ; i eveng fp (1) g fp (m) g = Q i ; P (1) ; : : :; P (m) : i < ; i even : Distinct sequences S; S 0 in spectrum hve distinct rst elements (S 0 6= S0 ), but they gree on ll other entries, if ny (S i = S 0 i for 1 i m). For ech m-spectrum we lso dene formul = 9 vm+1 v i S i v m+1 ; S2 sying tht for every sequence S 2 there is v m+1 such tht v 0 ; : : :; v m nd v m+1 `relize' S, s shown in Figure 1. The following two propositions show tht : is n-provble. Proposition 7.1. Let be ny 0-spectrum. Then `n 3 4 ). Proof. Let Q i 2. By (A1) nd Lemm 5.1, `n 3 ) 8 v0 8 v1 (v 0 Idv 1 ) 9 v2 (v 0 Q i v 2 v 2 Q i v 1 )): By (A3) nd Lemms 5.1 nd 5.3, By Lemms 5.7 nd 5.3, `n 4 ) 9 v1 (v 0 Idv 1 ); `n 3 ) 8 v1 (v 0 Idv 1 ) 9 v2 (v 0 Q i v 2 v 2 Q i v 1 )): `n 3 ) (9 v1 (v 0 Idv 1 ) ) 9 v1 9 v2 (v 0 Q i v 2 v 2 Q i v 1 )); `n 3 4 ) 9 v1 9 v2 (v 0 Q i v 2 v 2 Q i v 1 ): As `n v 0 Q i v 2 v 2 Q i v 1 ) v 0 Q i v 2, by Lemms 5.3 nd 5.8 we get `n 3 4 ) 9 v1 9 v2 v 0 Q i v 2 :

17 PROVABILITY WITH FINITELY MANY VARIABLES 17 sv m+1 A AA??? S 0 = Q i S 1 S m s s s v 0 v 1 : : : S 0 0 = Q i 0 S0 1 S 0 m A AA A A????? s v m+2 v m Figure 2. The mening of (S; S 0 ). By Lemm 5.4, `n 9 v1 9 v2 v 0 Q i v 2 ) 9 v2 v 0 Q i v 2, so by Lemm 5.3, `n 3 4 ) 9 v2 v 0 Q i v 2 : But v 1 ; v 2 re not free in 9 v2 v 0 Q i v 2, so Lemm 5.10 gives `n 9 v2 v 0 Q i v 2 ) 9 v1 v 0 Q i v 1. By Lemm 5.3, `n 3 4 ) 9 v1 v 0 Q i v 1 : This holds for ech Q i 2, so by Lemm 5.3 gin we obtin `n 3 4 ), s required. Proposition 7.2. Suppose tht 0 m n?3 nd is n m-spectrum with jj > (n? 3? m). Then `n 1 2 ) : : Proof. The proof will proceed by downwrd induction on m, strting with m = n? 3 nd ending t m = 0. Therefore, we will need to construct n (m + 1)-spectrum from n m-spectrum. The two preliminry lemms 7.3 nd 7.4 llow us to do this. For m; s in the proposition, nd S; S 0 2, let (S; S 0 ) be the following Fm n -formul, illustrted in Figure 2: Lemm 7.3. n (S; S 0 ) = S2 9 vm+1? vi S i v m+1 v i S 0 i v m+2 : S 0 2nfSg 9 vm+2 (S; S 0 ).

18 18 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX Proof. Lemm 5.1, n S2 p i, i2i 9vm+1 By Lemm 5.10, we hve 9 vm+1 S 0 2nfSg i2i So by Lemm 5.6, we obtin n S2 9vm+1 p i j2infig v i S i v m+1 v i S 0 i v m+1 n v i S i v m+1 p j is tutology, so by (A1) nd S 0 2nfSg S 0 2nfSg S 0 2nfSg 9 vm+1 9 vm+2 9 vm+2 v i S 0 i v m+1 : v i S 0 i v m+2: v i S 0 i v m+2 : Distribute 9 vm+1 over the conjuncts in which v m+1 does not pper free (Lemms 5.9 nd 5.6): v i S i v m+1 9 vm+2 n S2 9 vm+1 S 0 2nfSg v i S 0 i v m+2 : Distribute conjunctions over conjunctions, using (A1) nd Lemms 5.1 nd 5.3: n S2 9 vm+1 S 0 2nfSg v i S i v m+1 9vm+2 v i S 0 i v m+2 : Now distribute 9 vm+2 over the conjuncts in which v m+2 does not pper free, using Lemms 5.9 nd 5.6: 9 vm+1 9 vm+2 v i S i v m+1 n S2 S 0 2nfSg v i S 0 i v m+2 : Use the ssocitivity nd commuttivity of conjunction ((A1), Lemm 5.1, Lemm 5.6). Conclude tht n 9 vm+1 9 vm+2 v i S i v m+1 v i S 0 i v m+2 ; s required. (15) S2 S 0 2nfSg {z } (S;S 0 ) Let hve color sequence : f1; : : :; mg! f1; : : :; n? 3g, nd let T = fp j : 1 j n? 3; j =2 rng()g; where rng() is the rnge of. Note tht T = ; if m = n? 3.

19 PROVABILITY WITH FINITELY MANY VARIABLES 19 Lemm 7.4. For ll distinct S; S 0 2, `n 1 2 (S; S 0 ) ) R2T? vi S 0 i v m+2 v m+1 Rv m+2 : Proof. Let S; S 0 2 with S 6= S 0. It follows by Lemm 5.10 tht 1 n 8 vm+1 8 vm+2 WR2Rel v m+1rv m+2. So by xiom (A3) nd Lemm 5.3,? `n 1 (S; S 0 ) ) v m+1 Rv m+2 vi S i v m+1 v i S 0 i v m+2 : By Lemm 5.3, (16) R2Rel `n 1 (S; S 0 ) ) R2Rel? vi S 0 i v m+2 v i S i v m+1 v m+1 Rv m+2 : Now using Lemm 5.10 to chnge the vribles of 2, nd then (A3) nd Lemm 5.3 to remove the outer quntiers nd irrelevnt conjuncts, we obtin (17) `n 2 ) :(v i P v m+2 v i Qv m+1 v m+1 Rv m+2 ) for ny i < m + 1; hp; Q; Ri 2 Forb: From these lst two sttements we see tht every conjunct in (16) of the form v i S 0 i v m+2v i S i v m+1v m+1 Rv m+2 is contrdicted by 2 ccording to (17) whenever hs 0 i ; S i; Ri 2 Forb. Since S 0 ; S0 0 2 fq i : i < ; i eveng nd S 0 6= S0, the denition of Forb implies tht hs 0; S0 ; Ri 2 Forb whenever R 2 fid; E; Q i : i < g. Also, for 1 i m; S i = S 0 i = P (i) so hs i ; S 0 i ; Ri 2 Forb if R = P (i). So (15), (17), nd Lemm 5.3 give (18) if R 2 Rel n T then `n 2 ) Combining (16) nd (18) by Lemm 5.3 gives `n 1 2 (S; S 0 ) ) R2T nd Lemm 5.8 (or 5.3) now yields `n 1 2 (S; S 0 ) ) :? v i S 0 i v m+2 v i S i v m+1 v m+1 Rv m+2 :? vi S 0 i v m+2 v i S i v m+1 v m+1 Rv m+2 ; R2T? vi S 0 i v m+2 v m+1 Rv m+2 ; s required. If T = ;, this is `n 1 2 (S; S 0 ) )?. The proof of Proposition 7.2 now proceeds by induction on n? 3? m. The bse cse is n? 3? m = 0. In the bse cse we hve n 4, m = n? 3 1, is surjective, (n? 3? m) = 1, nd jj 2. It follows tht there re distinct S; S 0 2 giving non-empty conjunctions

20 20 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX in Lemm 7.3, nd the disjunction over T in Lemm 7.4 is empty, so Lemms 5.8 nd 5.3 give us (19) (20) `n ) 9 vm+1 9 vm+2 (S; S 0 ); `n 1 2 (S; S 0 ) )?: Hence, in the usul wy, `n 1 2 ) : (S; S 0 ) by (20) nd Lemm 5.3; `n 8 vm+2 ( 1 2 ) : (S; S 0 )) by Lemm 5.2; `n 1 2 ) 8 vm+2 : (S; S 0 ) by (A2) nd Lemm 5.3; `n 1 2 ) :9 vm+2 (S; S 0 ) by Lemm 5.3; `n 1 2 ) :9 vm+1 9 vm+2 (S; S 0 ) similrly; `n 1 2 ) : by (19) nd Lemm 5.3: This completes the proof of the bse cse. Assume tht n? 3? m > 0. The inductive hypothesis is tht `n 1 2 ) : whenever is n (m + 1)-spectrum of size lrger thn (n? 3? (m + 1)) = (n? 4? m). We will show from this ssumption tht `n 1 2 ) : whenever is n m-spectrum of size greter thn (n? 3? m). Fix S 2. By Lemms 7.3 nd 5.3, `n 1 2 ) vm+1 Using Lemms 5.3, 5.6, nd 5.9, we obtin `n 1 2 ) 9 vm+1 By Lemms 7.4 nd 5.8, `n 1 2 ) 9 vm+1 S 0 2nfSg S 0 2nfSg 9 vm+2 R2T S 0 2nfSg 9 vm+2 (S; S 0 ): 9 vm+2 ( 1 2 (S; S 0 )): (v i S 0 i v m+2 v m+1 Rv m+2 ) : Distribute the existentil quntiers 9 vm+2 over disjunctions (Lemms 5.9 nd 5.6): `n 1 2 ) 9 vm+1 S 0 2nfSg 9 vm+2 R2T v i S 0 i v m+2 v m+1 Rv m+2 :

21 PROVABILITY WITH FINITELY MANY VARIABLES 21 Then, in the criticl step, distribute conjunctions over disjunctions using Lemm 5.6 nd (A1) with the tutology?? p ij, i2i j2j g:i!j i2i p i;g(i) ; exponentilly expnding the length of the formul: `n 1 2 ) 9 vm+1 g:nfsg!t S 0 2nfSg 9 vm+2 v i S 0 i v m+2 v m+1 g(s 0 )v m+2 : Finlly, distribute 9 vm+1 over the resulting disjunctions (Lemms 5.9 nd 5.6): `n 1 2 ) g:nfsg!t 9 vm+1 S 0 2nfSg 9 vm+2 v i S 0 i v m+2 v m+1 g(s 0 )v m+2 : Now ech function g : n fsg! T induces prtition of n fsg whose pieces re the preimges of the symbols in T. The number of pieces is t most jt j = n? 3? m. For ech g, let P(g) be one of the lrgest pieces. Restricting the conjunctions over n fsg to P(g) we obtin, by Lemm 5.8, (21) `n 1 2 ) g:nfsg!t 9 vm+1 S 0 2P(g) 9 vm+2 v i S 0 i v m+2 v m+1 g(s 0 )v m+2 : For ech g : nfsg! T, let (g) = fhs0 0 ; S0 1 ; : : :; S0 m ; g(s0 )i : S 0 2 P(g)g: Since g is constnt on P(g), (g) is n (m + 1)-spectrum. So with this nottion, we cn rewrite (21): `n 1 2 ) tht is, (22) g:nfsg!t 9 vm+1 `n 1 2 ) v i S + i v m+2 9 vm+2 S + 2(g) i<m+2 {z } (g) g:nfsg!t 9 vm+1 (g) : Now prtitioned set cnnot hve more elements thn the number of pieces times the size of lrgest piece, so for ny g : n fsg! T we hve (23) (n? 3? m)jp(g)j jj? 1:

22 22 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX Our ssumption on the size of is tht (24) jj > (n? 3? m): Since j(g)j = jp(g)j, the denition of nd inequlities (23) nd (24) give 1 + (n? 3? m)j(g)j jj > (n? 3? m) = 1 + (n? 3? m) (n? 4? m); so j(g)j > (n?3?(m+1)): Thus, ech (g) is suciently lrge (m+1)- spectrum tht, by the inductive hypothesis, `n 1 2 ) : (g). As in the bse cse, we deduce from this tht `n 1 2 ) :9 vm+1 (g) : This holds for ll g, so by (22) nd Lemm 5.3, `n 1 2 ) :. This completes the induction nd the proof of the proposition. Theorem 7.5. `n :. Proof. Let be the 0-spectrum fq i : i < ; i eveng. By Proposition 7.1, `n 3 4 ). By Proposition 7.2 since jj > (n? 3), `n 1 2 ) :. Also, by Lemm 6.1, `n 1 3 ) 4. So by Lemm 5.3, we hve `n :( ), or `n :, s required. Thus the 3-sentence : is n-provble without using (A5). In the next section we show tht : is not (n? 1)-provble. x8. Alterntive semntics. To show tht : is not (n?1)-provble, we will use non-clssicl semntics for which `n?1 is sound nd is vlid. Suppose A is n (n? 1)-by-(n? 1) mtrix of reltion symbols in Rel. For i; j < n? 1, the symbol in row i nd column j is denoted by A ij or A i;j. Thus, 2 3 A 00 A 01 A 02 A 0;n?2 A 10 A 11 A 12 A 1;n?2 6 A 20 A 21 A 22 A 2;n?2 7 A = : A n?2;0 A n?2;1 A n?2;2 A n?2;n?2 Think of A s list of binry reltions tht hold mong n objects, tht is, A is just specil kind of quntier-free n-type hving the sme `mening' s the conjunction of the tomic formul v i A ij v j : `A, i;j<n?1 v i A ij v j ': Let M be the set of (n? 1)-by-(n? 1) mtrices A of reltion symbols in Rel tht stisfy the following conditions: (M 1 ) A ii = Id for ll i < n? 1, 7

23 PROVABILITY WITH FINITELY MANY VARIABLES 23 (M 2 ) ha ij ; A ik ; A kj i 2 Mnd for ll i; j; k < n? 1. The mtrices in M re clled tomic mtrices. Lemm 8.1. If i; j; k < n? 1, A 2 M, nd A ij = Id, then A ik = A kj. Hence, ny tomic mtrix A 2 M is symmetric: it stises (M 3 ) A ij = A ji for ll i; j < n? 1. Proof. By (M 2 ), ha ij ; A ik ; A kj i = hid; A ik ; A kj i 2 Mnd. From the denition of Mnd we see tht A ik = A kj. By (M 1 ), A ii = Id for ny A 2 M nd i; j < n? 1, so by the foregoing, A ij = A ji, giving (M 3 ). For ny i; j < n? 1, let [i; j] nd [i=j] be the functions mpping n? 1 to n? 1 dened by the following conditions for ll k < n? 1: [i; j]i = j [i; j]j = i [i; j]k = k if k 6= i; j [i=j]i = j [i=j]k = k if k 6= i [i; j] is trnsposition nd [i=j] is replcement. If A is n tomic mtrix nd g is ny function mpping n? 1 into n? 1, then Ag is the mtrix of reltion symbols dened by (Ag) ij = A g(i);g(j) for ll i; j < n? 1. It is esy to check tht Ag is n tomic mtrix. In prticulr, A[i; j] nd A[i=j] re tomic mtrices. Let k; l < n? 1. Two tomic mtrices A; B re sid to gree up to k, written A =k B, if A ij = B ij whenever i; j < n? 1 nd i; j 6= k, nd they gree up to k; l, written A =kl B, if A ij = B ij whenever i; j < n?1 nd i; j =2 fk; lg. We will need to know tht M hs the following properties. Lemm 8.2. (C 1 ) if A; C 2 M, i; j < n? 1, i 6= j, nd A =ij C, then there is some B 2 M such tht A =i B nd B =j C, (C 2 ) if A 2 M nd i; j < n? 1 then A[i=j] 2 M, (C 3 ) if A 2 M nd i; j < n? 1 then A[i; j] 2 M. Proof. (C 2 ) nd (C 3 ) were discussed bove. The hrd prt is (C 1 ). Assume A; C 2 M nd A =ij C. We wish to nd some B 2 M such tht A =i B nd B =j C. The requirements A =i B, B =j C determine B kl whenever fk; lg 6= fi; jg, so we only need to choose B ij = B ji so tht hb ji ; A jk ; C ki i 2 Mnd for every k < n? 1, k 6= i; j. It is esily seen tht this is sucient to ensure tht B 2 M. Let K = fk : i; j 6= k < n? 1g. If A jk = Id for some k 2 K then we let B ij = B ji = C ik. Then by Lemm 8.1, hb ji ; A jl ; C li i = hc ki ; C kl ; C li i 2 Mnd for ll l 2 K. Similrly, if C ik = Id for some k 2 K then we my let B ij = B ji = A kj. We my therefore ssume tht A jk 6= Id 6= C ik for ll k 2 K. Suppose tht for every k 2 K there is some l 2 f1; : : :; n?3g such tht fa jk ; C ki g fp l ; Q l g. Then he; A jk ; C ki i is mndtory whenever k 2 K, so we my let B ij = B ji = E. If not, then becuse jkj = n? 3

24 24 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX nd there re n? 3 pirs of the form fp l ; Q l g, 1 l n?3, there must be some l 2 f1; : : :; n? 3g such tht for every k 2 K, fa jk ; C ki g 6 fp l ; Q l g. Then since A jk ; C ki 6= Id, hp l ; A jk ; C ki i is mndtory for every k 2 K, so let B ij = B ji = P l. The stisfction reltion j= ssocited with M is the unique binry reltion between mtrices nd (n? 1)-formul tht stises the following conditions for ll A 2 M, '; 2 Fm n?1, i; j < n? 1, nd R 2 Rel: A j= v i =v j i A ij = Id; A j= v i Rv j i A ij = R; A j= :' i A 6j= '; A j= ' ) i either A 6j= ' or A j= or both; A j= 8 vi ' i B j= ' whenever B 2 M nd B =i A: A formul tht is M-stised by every tomic mtrix is sid to be M- vlid. Lemm 8.3. If ' 2 Fm n?1, v i =2 free('), nd A =i B, then A j= ' i B j= '. Proof. The proof is by induction on '. Suppose v i =2 free(') nd A =i B. If ' is tomic, then ' is v j =v k or v j Rv k with R 2 Rel, nd i 6= j; k. Then A jk = B jk since A =i B, hence A j= ' i B j= '. Assume the lemm holds for, nd let ' = :. Then v i =2 free(') = free( ), so inductively, A j= ' i A 6j= i B 6j= i B j= '. It is eqully esy to show tht if the lemm holds for nd, then it holds for ). Assume the lemm holds for, nd let ' = 8 vj. We only show tht B j= ' whenever A j= ' nd A =i B, since the converse follows by interchnging A nd B. Assume A j= ' nd A =i B. To show B j= ', it suces to ssume B =j C nd show C j=. If j = i, then C j= becuse A =i C, so ssume j 6= i. From A =i B nd B =j C we conclude tht A =ij C. Since M hs property (C 1 ), there must be some D 2 M such tht A =j D nd D =i C. Then D j= since A j= 8 vj nd A =j D. Now v i =2 free( ) since j 6= i, so it follows from the inductive hypothesis nd D =i C tht C j=, s desired. Lemm 8.4. For ll i; j < n? 1, A 2 M, nd ' 2 Fm n?1, we hve A j= S ij ' i A[i; j] j= '. Proof. The proof is gin by induction. If ' = xry, where x; y 2 Vr n?1 nd R 2 Rel, then there re severl cses to be distinguished ccording to whether x or y is either v i, v j, or neither. We will tret only one cse, s n exmple. Suppose x = v i, y = v k, nd k 6= i; j. Then, since S ij ' = S ij v i Rv k = v j Rv k, nd A[i; j] ik = A jk, we hve A j= S ij ' i A j= v j Rv k i A jk = R i A[i; j] ik = R i A[i; j] j= v i Rv k i A[i; j] j= '. (Notice tht (C 3 ) of Lemm 8.2, giving A[i; j] 2 M, is needed here.)

25 PROVABILITY WITH FINITELY MANY VARIABLES 25 Suppose the lemm holds for, nd let ' = :. Then, since S ij ' = S ij : = :S ij, we hve A j= S ij ' i A 6j= S ij i A[i; j] 6j= i A[i; j] j= : i A[i; j] j= '. It is eqully esy to tret the cse ' = ( ) ). Finlly, suppose the lemm holds for, nd let ' = 8 vk. We rst ssume A j= S ij 8 vk nd prove A[i; j] j= 8 vk. To get this conclusion, we ssume B =k A[i; j] nd show B j=. Let l = [i; j]k. Then k = [i; j]l, nd it is esy to check tht B[i; j] =l A. Now A j= S ij 8 vk = 8 vl S ij, so B[i; j] j= S ij. But then B = B[i; j][i; j] j= by the inductive hypothesis. Thus A[i; j] j= 8 vk whenever A j= S ij 8 vk. Conversely, ssume A[i; j] j= 8 vk. To show tht A j= S ij 8 vk = 8 vl S ij, we suppose B =l A nd show B j= S ij. From B =l A we get B[i; j] =k A[i; j], so B[i; j] j=, nd hence B j= S ij by the inductive hypothesis. Lemm 8.5. If A 2 M nd ' 2 Ax n?1 then A j= '. Thus, every xiom is M-vlid. Proof. Both the lterntive nd the usul semntics tret the connectives : nd ) in the sme wy, so the proof tht every tutology is semnticlly vlid lso shows tht every tutology is M-vlid. Hence the lemm holds when ' is n instnce of (A1). Suppose ' is n instnce 8 vi ( ) ) ) ( ) 8 vi ) of (A2), where v i =2 free( ). To show tht A j= ', it suces to ssume A j= 8 vi ( ) ), A j=, nd B =i A, nd then show B j=. From the rst nd third of these ssumptions we know tht B j= ), i.e. either B 6j= or else B j=. However, the second nd third ssumptions imply, by Lemm 8.3 nd v i =2 free( ), tht B j=. Hence B j=. Suppose ' is n instnce 8 vi ) of (A3). If A j= 8 vi, then A j= since A =i A. Hence A j= '. If ' is n instnce of (A4), then ' is 9 vi (v i =v j ) for some distinct i; j < n? 1. Now A j= 9 vi (v i =v j ) i there is some B 2 M such tht B =i A nd B j= v i =v j. Let B = A[i=j]. Then B 2 M since M hs property (C 2 ). Furthermore, A =i B, for if k; l 6= i then [i=j]k = k nd [i=j]l = l so A kl = A[i=j] kl. Finlly, B j= v i =v j, since B ij = A[i=j] ij = A jj = Id. We will tret only one type of instnce of (A5), nmely (v i =v j ) ) (v i Rv k ) v j Rv k ). Suppose A j= v i =v j nd A j= v i Rv k. Then A ij = Id nd A ik = R. By Lemm 8.1, A jk = R, hence A j= v j Rv k. The other cses re similr. Finlly, let ' be (v i =v j ) ) ( ) S ij ), n instnce of (A6). Suppose A j= v i =v j. Then A ij = Id, nd it is esy to prove from this tht A = A[i; j]. (For exmple, if k 6= i; j then by Lemm 8.1, A ik = A jk = A[i; j] ik.) If lso A j=, then A[i; j] j=, so A j= S ij by Lemm 8.4. Thus A j= ' for every A 2 M. Lemm 8.6. If '; ' ) 2 Fm n?1 re M-vlid, then so re nd 8 vi ' for ll i < n? 1.

26 26 ROBIN HIRSCH, IAN HODKINSON, AND ROGER D MADDUX Proof. Let A 2 M. Then A j= ' nd A j= ' ) since ' nd ' ) re M-vlid, nd it follows tht A j=. We get A j= 8 vi ' becuse if A =i B then by the M-vlidity of ', B j= '. Lemm 8.7. If `n?1 ' then ' is M-vlid. Proof. The set of M-vlid formul contins Ax n?1 by Lemm 8.5, nd is closed under modus ponens nd (n? 1)-generliztion by Lemm 8.6, so it must therefore include ll (n? 1)-provble formul. Lemm 8.8. is M-vlid. Proof. We use the second lterntive form (13) of : 3 8 v0 8 v1 R2Rel S;T 2Rel v 0 Rv 1 8 v0 8 v1 9v2 (v 0 Sv 2 v 2 T v 1 ), hr;s;t i2mnd v 0 Rv 1 : Certinly, 8 v0 8 v1 WR2Rel v 0Rv 1 is M-vlid. Suppose S; T 2 Rel. Let A 2 M. If A j= 9 v2 (v 0 Sv 2 v 2 T v 1 ) then for some B 2 M with B =2 A, we hve B j= v 0 Sv 2 v 2 T v 1. By (M 2 ), B j= W hr;s;t i2mnd v 0Rv 1, so by Lemm 8.3, A j= W hr;s;t i2mnd v 0Rv 1. Conversely, suppose tht R 2 Rel, hr; S; Ti 2 Mnd, nd A j= v 0 Rv 1. Dene = [3=0][4=0] [n? 2=0] : (n? 1)! f0; 1; 2g, nd dene B by B lm = 8 >< >: Id if (l) = (m); R if f(l); (m)g = f0; 1g; S if f(l); (m)g = f0; 2g; T if f(l); (m)g = f2; 1g; where l; m < n? 1. It is esy to check tht B 2 M. Now evidently, (25) A =3 A[3=0] =4 A[3=0][4=0] =5 =n?2 A[3=0] [n? 2=0] = A =2 B: By (C 2 ), ll mtrices shown re in M. As B 02 = S nd B 21 = T, we hve B j= 9 v2 (v 0 Sv 2 v 2 T v 1 ). But free(9 v2 (v 0 Sv 2 v 2 T v 1 )) = fv 0 ; v 1 g, so by repeted use of Lemm 8.3 on (25), we obtin A j= 9 v2 (v 0 Sv 2 v 2 T v 1 ), s required. Theorem `n?1 :. Proof. By Lemm 8.8, is M-vlid. Hence : is not M-vlid. By Lemm 8.7, 6 `n?1 :. We hve shown tht : is n-provble without (A5), but is not (n? 1)- provble, even using (A5).