Relating the Bounded Arithmetic and Polynomial ... - Semantic Scholar
Relating the Bounded Arithmetic and Polynomial Time Hierarchies Samuel R. Buss
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Department of Mathematics University of California, San Diego La Jolla, CA 92093-0112 [email protected] May 1993, last revision November 1994
Abstract
The bounded arithmetic theory S2 is nitely axiomatized if and only if the polynomial hierarchy provably collapses. If T 2 equals S2+1 then T2 is equal to S2 and proves that the polynomial time hierarchy collapses to 6 +3 , and, in fact, to the Boolean hierarchy over 6 +2 and to 6 +1 =poly . i
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1 Introduction Theories of bounded arithmetic are theories of arithmetic obtained by putting restrictions on induction axioms; namely, allowing induction only for certain classes, 6 , of bounded formulas, and using polynomial, or length, induction (PIND or LIND) in place of successor induction (IND). The most important subtheories of bounded arithmetic are the theories S2 , axiomatized with 6 PIND (or equivalently, 6 -LIND, if i 1), and the theories T2 , axiomatized by 6 -IND. The following inclusions are known for these theories: b i
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S20 ( T20 S21 T21 S22 T22 1 1 1
3 Supported in part by NSF grant DMS-9205181
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and their union is the theory S2 = T2 [2]. However, with the exception of S20 6= T20 (see [13]), it is not known whether the rest of the theories of bounded arithmetic are distinct. It is a well-known fact that S2 and T2 are nitely axiomatized for i > 0, and thus it is immediate that this hierarchy of theories collapses if and only if S2 is nitely axiomatized. This latter condition is equivalent to I 1 0 + 1 being nitely axiomatized (see Parikh [11] and Wilkie-Paris [14] for this alternate, and original, approach to bounded arithmetic). There are close connections between theories of bounded arithmetic and the polynomial hierarchy. First, the class of predicates de nable by 6 (or 5 ) formulas is precisely the class of predicates in the i -th level 6 (or 5 , respectively) of the polynomial hierarchy. For instance, S 21 and T21 are axiomatized with their induction axioms restricted to NP-predicates (since NP = 61 is the class of predicates de nable by 61 -formulas). Second, it is known that the 6 -de nable functions of S1 are precisely the -functions, which are the functions which are polynomial time computable with an oracle for 6 01 . For instance, the 61 -de nable functions of S21 are precisely the polynomial time computable functions. Since it is open whether the polynomial time hierarchy collapses, it is natural to ask whether there is any connection between the possible collapses of the hierarchy of bounded arithmetic theories and the polynomial hierarchy. This question has already been partially answered by the work of KrajcekPudlak-Takeuti [10] who showed that if T2 = S2+1 for any i 1, then the polynomial hierarchy collapses with 6 +2 = 5 +2 (in fact, they show that in this case, 6 +1 1 =poly ). The main results of this paper strengthen the results of Krajcek-PudlakTakeuti by proving that if T2 = S2+1 holds, then the following conditions must hold: (1) T2 = S2 , so that the hierarchy of bounded arithmetic theories collapses, and (2) T2 can provethat the polynomial time hierarchycollapses to B(6 +2) and to 6 +1=poly , where B(6 +2) is the class of Boolean combinations of 6 +2 -predicates. Our proofs are easier, in a combinatorial sense, than the proofs of [10] and this makes it possible to formalize them in T2 . We believe that the results of this paper are nearly the strongest that are obtainable relating the possibility that T2 = S2+1 to the possible collapse of i
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the polynomial time hierarchy | at least with current techniques. To support this belief, consider the three conditions: ( ) The polynomial hierarchy collapses ( ) S2 proves that the polynomial hierarchy collapses ( ) S2 is nitely axiomatized Our results show that ( ) and ( ) are equivalent; however, we do not expect to show that ( ) is equivalent to ( ) using current techniques. The reason for this is that ( ) is a 602 -condition whereas, since ( ) is a 601 -condition, the results of the current paper show that ( ) is a 601 -condition; and, based on the history of attempts to solve the P versus NP problem, it seems to be dicult even to establish that the collapse of the polynomial time hierarchy is equivalent to a natural 601 -condition like ( ). It is known that S2+1 is conservative over T2 with respect to 86 +1 sentences [3]. On the other hand, the axioms of S2+1 can be expressed as 85 +2 -sentences (in this formulation, an induction axiom of S2+1 will become a 85 +2 -formula with a sharply bounded existential quanti er in its outermost block of bounded universal quanti ers). Thus saying S2+1 is 5 +2 -conservative over T2 is equivalent to saying that S2+1 = T2 . An open problem is to try to relate the condition S2 = T2 to the possible collapse of the polynomial hierarchy. Krajcek [9] shows that if S2 = T2 , then the set (6 ) of predicates logspace, Turing reducible to 6 is equal to the set (6 ) of predicatespolynomial time, Turingreducibleto 6 . However,it is open whether this last condition implies the polynomial hierarchy collapses. See [5,4,6] for more on this connection. The prerequisites for reading this paper are a basic knowledge of bounded arithmetic theories as contained in [2]. The reader would also bene t from knowledge of [10] and [3]. In the next section we will review the necessary background material needed from [10]. After preparing the rst draft of this paper, we learned that D. Zambella has independently discovered the main results of this paper [15]. i
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2 The KPT Witnessing Theorem for T2i There are two important witnessing theorems for T2 . The rst follows from the `Main Theorem' for S2+1 and the fact that S2+1 is 6 +1 -conservative over T2 : this witnessing theorem states that the 6 +1 -de nable functions of T2 are precisely the functions which can be computed in polynomial time with a 6 -oracle (i.e., the +1 -functions). The second witnessing theorem puts a necessary condition on the 6 +2 - and 6 +3 -de nable functions of T2 ; we call this the `KPT witnessing theorem'. It is this latter witnessing theorem that we need for our proofs: Theorem 1 Let i 1 . Suppose T2 ` (8x)(9y )(8z )B (x; y; z ) , where B is a 95 -formula, with only x; y; z as free variables. There exists k > 0 and functions f1; ... ; f such that each f is m -ary and is 6 +1 -de nable by T2 i
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and such that
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` B (a; f1(a); b1) _ B (a; f2(a; b1); b2) _ B (a; f3(a; b1; b2); b3 ) _ 1 1 1 1 1 1 _ B (a; f (a; b1; b2; ... ; b 01); b ):
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For i = 0 , the same result holds for P V1 in place of T20 . As usual, P V1 denotes the conservative extension of P V to rst-order logic, or equivalently, P V1 is S20 or T20 enlarged to have function symbols and their de ning equations for all polynomial time functions.
Note that since the functions f are 6 +1 -de nable by T2 , they must be +1 -functions. Theorem 1 is due to [10]; some later, related results can be found in [8,12,1]. We do not include a proof here. We next use Theorem 1 to establish a consequence of the condition T2 = S2+1 . We assume that i 0 and work with the theory T2 ; when i = 0 our results are intended to hold for P V1 in place of T20 . De nition A quanti ed Boolean formula is a formula constructed from Boolean connectives (say, ^ , _ and : ) and quanti ers ranging over Boolean values. A quanti er (8p) or (9p) indicates quanti cation allowing p to range over the values True and False. b
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Given a truth assignment to the free variables of a quanti ed Boolean formula, it is obvious how the truth valueof the formula should be de ned. A quanti ed Boolean formula is satis able if there is some truth assignment to its free variables which gives it value True. A 5 -formula is a quanti ed Boolean formula which is in prenex form with i blocks of like quanti ers starting with a universal block. It is well-known that the set of satis able 5 -formulas is 6 +1 -complete. De nition Let i 0. TRU and SAT are bounded arithmetic formulas which express: TRU ('; w) , ' codes a 5 -formula and w codes a satisfying assignment of ' SAT (') , (9w ')TRU ('; w): In the de nition of TRU and SAT we presume that quanti ed Boolean formulas and truth assignments are coded in some natural and ecient way by integers; we use Greek letters '; ... as variables that range over integers which are intended to code quanti ed Boolean formulas. Since the code of a truth assignment can w.l.o.g. always be less than the code of a formula, SAT (') expresses the condition that ' is satis able. Standard bootstrapping techniques allow TRU to be a 1 +1 -formula with respect to the theory T2 ; in fact, for i 1, TRU is a 5 -formula. Hence SAT is a 6 +1 -formula. Also, T2 can prove basic properties of the TRU and SAT predicates. Most importantly, T2 can provethat SAT is many-one complete for 6 +1 -formulas; i.e., for any 6 +1 -formula A(~b), there is a polynomial time function f so that A(~b) is T2 -provably equivalent to SAT (f (~b)). As an application of Theorem 1, consider the formula 8h'0; ... ; ' i(9` n)(9hw0 ; ... ; w i) (1) h i (8j `)TRU (' ; w ) ^ (` < n ! :(9w +1)TRU (' +1; w +1)) : The meaning of formula (1) requires some explanation. First, a notation like (8h'0; ... ; ' i)B ('~ ; `) means the same as \there is an integer '3 which codes a sequence of 5 -formulas '0; ... ; ' so that B ('~ ; `) holds". The B i
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quanti er (9` n) is a sharply bounded quanti er since ` can be bounded by the length of the code for h'~ i , and the quanti ers (9hw~ i) and (9w +1) are bounded quanti ers since each w may be bounded by ' . By using prenex operations and using the fact that ` can be computed from hw0; ... ; w i , formula (1) is equivalent to the formula (8h'0; ... ; ' i)(9hw0; ... ; w i)(8w +1) (2) i h (8j `)TRU (' ; w ) ^ (` < n ! :TRU (' +1; w +1)) : which is a 8981 +1 -formula. The intuitive meaning of formula (1) or (2) is, of course, that every sequence '0; ... ; ' of 5 -formulas has an initial sequence of maximal length ` of satis able formulas. Furthermore, the formula (1) is a theorem of S2+1 . This is because S2+1 can use length induction on the 6 +1 -formula S (h'~ i; `) expressing the condition that the rst ` formulas of the sequence are satis able. (An equivalent way to see this is to note that S2+1 can prove the 6 +1 -length-maximization principle.) Now suppose T2 is equal to S2+1 ; in particular, T2 proves the formula (2). By Theorem 1, this means that there is an integer k 0 and there are 6 +1 -de ned functions f0; ... ; f so that, letting A(h'~ i; hw~ i; w +1) be the subformula of (2) enclosed in square brackets, we have that T2 ` (8h'~ i)[A(h'~ i; f0(h'~ i); b0) _ A(h'~ i; f1 (h'~ i; b0 ); b1) _ 1 1 1 (3) 1 1 1 _ A(h'~ i; f (h'~ i; b0; b1 ; ... ; b 01); b )] We henceforth shall use (3) restricted to the case where n = k , so that the sequence '~ is '0; ... ; ' . Without loss of generality, each f satis es the following property (provably in T2 ): whenever T RU (' ; b ) holds for r = 0; ... ; j 0 1, then the value f (h'~ i; b0; ... ; b 01) is the Godel number of a sequence hv0 ; ... ; v 01i of length ` j so that T RU (' ; v ) holds for all r = 0; ... ; ` 0 1. Recall that represents the Godel function so that (i; w) is equal to the i -th integer in the sequence coded by w . De ne g ('0 ; ... ; ' ; w0; ... ; w 01 ) = (j + 1; f (h'0 ; ... ; ' i; w0 ; ... ; w 01)): `
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Suppose that '0; ... ; ' are codes for satis able 5 -Boolean formulas and let w0 ; ... ; w be satisfying assignments. De ne b0 ; b1; ... inductively as follows: if f (h'~ i; b0; ... ; b 01) is a sequenceof length ` + 1 k , then let b equal w +1 . It is obvious that whenever f (h'~ i; b0; ... ; b 01) has length ` + 1 k then b givesa \counterexample"so that A(h'~ i; f (h'~ i; b0; ... ; b 01); b ) is false. Now, by (2), there is some j k for which f (h'~ i; b0; ... ; b 01) has length k + 1. Let j0 be the least value such that f (h'~ i; b0; ... ; b 01) has length j0 + 1. It must be that TRU (' ; g ('~ ; w0; ... ; w 01)) holds. This argument formalizes in T2 and thus we have proven: Lemma 2 Suppose T2 = S2+1 . Then there is k 0 and there are 6 +1 de nable functions g0 ; ... ; g of T2 so that B i
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3 Collapsing Bounded Arithmetic In this and the next section, we examine consequences of the condition T2 = S2+1 . In this section we show that this implies that S2 collapses to T2 . Our point of departure is Lemma 2 above; we henceforth x k and g0 ; ... ; g . This lemma states that at least one of the functions g can nd a satisfying assignment for ' using only the vector '~ and arbitrary satisfying assignments w0; ... ; w 01 . However, it need not always be the same g that succeeds in this way; di erent vectors of formulas '~ and even di erent witnesses w~ may cause di erent g 's to succeed. We de ne SucceedBy (`; '~ ; w~ ) to be the following formula which states that one of the i
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rst ` + 1 g 's succeeds in this way; namely, it is de ned as SucceedBy (`; '~ ; w~ )
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Our rst goal is to show that 6 +1 = 5 +1=poly where the \ poly " means that polynomial amount of advice is needed. As a preliminary to de ning what constitutes advice, we de ne \preadvice" by letting P reAdvice (a; h' +1; ... ; ' i) be the formula i
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(8h'0; ... ; ' i)(8hw0; ... ; w i) VTRU (h'~ i; hw~ i; a) ! SucceedBy(`; '~ ; w~ ) ; where VTRU (h'0; ... ; ' i; hw0; ... ; w i; a) abbreviates (8j `)(TRU (' ; w ) ^ w ' ^ ' < 2j j ): Several points to note are: rstly, in de ning P reAdvice we are continuing our practice of letting variables ' represent integers that must code 5 formulas; secondly, the value of ` is determined by the second argument to P reAdvice ( k is xed and ` varies, namely, ` equals k + 1 minus the length of the sequence coded by the second argument of P reAdvice ); thirdly, the quanti ers are bounded quanti ers since the ' 's and w 's are bounded by 2j j . The reason for bounding everything by 2j j is that we need only de ne \advice" that works for ' 's with j'j a for a an arbitrary integer. Also note that P reAdvice is a 5 +1 -formula. We can now de ne \advice" for formulas of length jaj by Advice (a; h' +1; ... ; ' i) , P reAdvice (a; h' +1; ... ; ' i) ^ :(9' )P reAdvice (a; h' ; ... ; ' i): Note that ' is bounded by 2j j ; thus Advice is a 5 +2 formula. The next lemma shows that T2 can prove that there always does exist advice: Lemma 3 Suppose T2 = S2+1 . Then T2 ` (8a)(9h'~ i)Advice (a; h'~ i): `
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Proof First, note that Lemma 2 implies that T2 proves that P reAdvice (a; hi) holds. Since k is a constant, it follows (without using induction) that there is a least ` such that (9h' +1; ... ; ' i)P reAdvice (a; h'~ i) holds. For this ` , any `preadvice' is actually advice. 2 i
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Next we give the key lemma that shows how `advice' can be used to make 6 +1 -IND hold and the polynomial time hierarchy collapse, provably in T 2 . Lemma 4 Suppose T2 = S2+1 . Then T2 proves Advice (a; h' +1; ... ; ' i) ^ ' < 2j j ! h :SAT (' ) $ (9h'0 ; ... ; ' 01i)(9hw0 ; ... ; w 01i) n VTRU (h' ~ i; hw0 ; ... ; w 01i; a) ^:SucceedBy(` 0 1; '~ ; w~ ) oi ^:TRU (' ; g ('0 ; ... ; ' ; w0 ; ... ; w 01)) : b i
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Let RHS(' ; h' +1 ; ... ; ' i; a)) denote the formula on the righthand side of the $ connective in the formula above; we often suppress the variables ' +1; ... ; ' and a that occur freely in RHS and write just RHS(' ). We shall argue informally in T2 to prove the lemma. Suppose ' ; ... ; ' 2j j are formulas and that Advice (a; h' +1 ; ... ; ' i) holds. The latter condition obviously implies that :P reAdvice (a; h' ; ... ; ' i). By the de nition of P reAdvice , there must exist 5 -formulas '0; ... ; ' 01 satis ed by witnesses w0; ... ; w 01 such that SucceedBy(` 0 1; '~ ; w~ ) is forced to be false. First suppose that ' is not satis able. Then clearly TRU (' ; g ('~ ; w~ )) must be false. Thus RHS(' ) follows from :SAT (' ). Second, suppose that ' is satis able. By P reAdvice (a; h' +1 ; ... ; ' i), it must be that SucceedBy (`; '~ ; w~ ) holds. On the other hand, SucceedBy (` 0 1; '~ ; w~ ) is false. Thus TRU (' ; g ('~ ; w~ )) is forced to be true and we have shown that SAT (' ) implies :RHS(' ). 2 In the subformula RHS , the leading existential quanti ers are actually
Proof
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bounded existential quanti ers since the formulas ' and their witnesses w are bounded by 2j j . This means that RHS(' ) is a 6 +1 -formula. Lemma 5 Suppose T2 = S2+1 . Then T2 ` 6 +1 -IND and T2 = T2+1 . Proof The proof is based on the fact that SAT (1 1 1) is complete for 6 +1 -formulas and is also equivalent on bounded ranges to the 5 +1 -formula :RHS(1 1 1) (under the assumption that T2 = S2+1 , as always). Indeed, for any 6 +1 -formula B (c; d~), there is a polynomial time and 61 -computable function f (c; d~) so that B (c; d~) is T2 -provably equivalent to SAT (f (c; d~)). The induction axiom for the formula B (c; d~) can be expressed as B (0; d~) ^ (8x)(B (x; d~) ! B (x + 1; d~)) ! B (c; d~): Let us prove this by reasoning informally in T2 which is presumed to equal S2+1 . Considering particular values for c and d~, there is a value a so that f (x; d~) < 2j j for all x c . Let ' +1; ... ; ' be formulas such that Advice (a; h' +1; ... ; ' i) holds. Then, with these parameters, Lemma 4, we have that the 6 +1 -formula B (x; d~) is equivalent to the 5 +1 -formula :RHS(f (x; d~)) for all x c . Now, it is known that S2+1 proves 1 +1 -IND and the usual proof (see Theorem 2.22 of [2]) shows that T2 = S2+1 proves induction for B , since B is \1 +1 with parameters" on the range 0 x c . j
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Iterating the method of this proof, we obtain: Theorem 6 If T2 = S2+1 , then T2 = S2 . Thus, if T2 = S2+1 , then S2 i
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nitely axiomatized. Also, if P V1 = S21 (P V ) , then P V1 = S2 (P V ) .
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Analogous to the method of proof of Lemma 5, we must show that any bounded formula is equivalent to a 6 +1 -formula with parameters, where the parameters vary with the range of the induction variable. From this, using Lemma 5, it will follow that T2 proves induction for any bounded formula. We do the case of B (c; d~) 2 6 +2 in some detail. We may suppose that B (x; d~) is of the form (9y t(x; d~))C (x; y; d~) for some 5 +1 -formula C . Proof
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We argue informally in T2 . By the method of Lemma 5, there is an a , given by a polynomial time function a = a(c; d~) of c and d~, and there is a polynomial time function f , so that for all x c , and y t(x; d~) and for advice h'~ i satisfying Advice , the 5 +1 -formula C (x; y; d~) is equivalent to the 6 +1 -formula RHS(f (x; y; d~); h'~ i; a(c; d~)). Thus, for 0 x c , B (x; d~) is equivalent to a 6 +1 -formula, and full induction holds for B up to c by Lemma 5. Hence T2 = T2+2 . A slight modi cation of the construction of the last paragraph shows that if A(~x) is a 6 +2 -formula (respectively, a 5 +2 -formula, then there is a polynomial growth rate function a(c) and a 6 +1 -formula (respectively, 5 +1 -formula) A3(~x; '3; a(c)) such that for all ~x such that maxf~xg c and all h'~ i such that Advice (a(c); h'~ i), A(~x) is equivalent to A 3(~x; h'~ i; a(c)), provably in T2 . This further implies that if A(~x) is a 6 +3 -formula, then A3 may be taken to be a 6 +2 -formula, because, if A(~x) is (9y t(~x))B (~x; y), then there is a 6 +2 -formula B 3 so that A(~x) will be equivalent to (9y t(~x))B 3(~x; h'~ i; a) for a given by a polynomial growth rate function of c max ~x and for h'~ i such that Advice (a; h'~ i). This fact is sucient to imply that T2 = T2+3 . By iterating the abovemethod of proof, one can show that T2 is equal to all of S2 . We shall leave the details of this to the reader, and remark instead that an alternative proof is given by Theorem 7 below where it is shown that T2 can prove that every bounded formula is equivalent to a Boolean combination of 6 +2 -formulas without any additional parameters or advice. Then since T2 = T2+2 = S2+2 and S2+2 proves induction for Boolean combinations of 6 +2 -formulas [3], it follows that T2 = S2 . 2 i
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4 Collapsing the Polynomial Hierarchy All the work of this section is predicated on the condition that T 2 = S2+1 . We have shown above that if T2 = S2+1 , then T2 proves that the 6 +2 predicates are contained in 6 +1 =poly . From this, the methods of KarpLipton [7] imply that the entire polynomial time hierarchy is contained in 6 +1=poly and in 5 +1=poly ; furthermore, the proof of this containment can be formalizedin T2 . The methods of Karp-Lipton also implyimmediatelythat i
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the polynomial hierarchy collapses to 6 +3 = 5 +3 . However, we shall prove an somewhat stronger result; namely, if T2 = S2+1 , then every polynomial hierarchy predicate (i.e., bounded formula) is T 2 -provably equivalent to a Boolean combination of 6 +2 -formulas. To prove this, it will suce to prove that every 6 +3 -formula is equivalent to a Boolean combination of 6 +2 -formulas. Let A(b) be an arbitrary 6 +3 formula. From the previous section, we know that T2 proves that A(b) is equivalent to (9h'~ i)[Advice (a(b); h'~ i) ^ A3(b; h'~ i)] (4) and to (8h'~ i)[Advice (a(b); h'~ i) ! A3(b; h'~ i)]; (5) where A3 is a 6 +2 -formula and a = a(b) is function of suciently large polynomial growth rate. Unfortunately, Advice is a 5 +2 -formula and the quanti er complexity of these equivalent formulations of A(b) is higher than we desire; namely, formula (4) is a 6 +3 -formula and formula (5) is a 5 +3 formula. This implies that every bounded formula is 1 +3 with respect to T2 , but we wish to prove a yet stronger result. To reduce the complexity of these formulas we would like to use P reAdvice in place of Advice . However, this can not be done directly since if h'~ i satis es P reAdvice , then it is not necessarily true that A3(b; h'~ i) is equivalent to A(b). Instead, we look for a longest vector h'~ i which satis es P reAdvice ; namely, consider the formula A 0 (b) de ned as: h i (9h'1; ... ; ' i) P reAdvice (a(b); h'1; ... ; ' i) ^ A3(b; h'1; ... ; ' i) p
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_ WW :(9h' 01; ... ; ' i)P reAdvice (a(b); h' 01; ... ; ' i) =2 o ^(9h' ; ... ; ' i)[P reAdvice (a(b); h' ; ... ; ' i) ^ A3 (b; h' ; ... ; ' i)] We claim that A0(b) is equivalent to A(b). The proof of this now quite easy. First, there must exist a least ` 1 such that there exists h' ; ... ; ' i which satis es P reAdvice . Second, if P reAdvice (h' ; ... ; ' i) holds and if there k
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is no h'0 01 ; ... ; '0 i which satis es P reAdvice , then clearly h' ; ... ; ' i satis es Advice . And for this advice, A3(b; h'~ i) is equivalent to A(b). Since P reAdvice is a 5 +1 -formula and A3 is a 6 +2 -formula, A0 is a Boolean combination of 6 +2 -formulas. This establishes: Theorem 7 If T2 = S2+1 , then every bounded formula is T2 -provably equivalent to a Boolean combination of 6 +2 formulas. In other words, if T2 = S2+1 , i
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then the polynomial hierarchy T2 -provably collapses to (a nite level of) the Boolean hierarchy over 6 +2 . Also, in this case, T2 proves that the polynomial time hierarchy collapses to 6 +1 =poly . If P V1 = S21 (P V ) , then every bounded formula is P V1 -provably equivalent to a Boolean combination of 62 -formulas, so the polynomial time hierarchy provably collapses to the Boolean hierarchy over 62 . Also, in this case, P V1 proves that the polynomial time hierarchy collapses to NP=poly . i
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It should be noted again that [10] have shown that if T2+1 = S2+2 then the polynomial hierarchy collapses to 6 +2 = 5 +2 and to 1 +1=poly : we do not know how to prove that this stronger collapse would be T2 -provable. i
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References [1] , The witness function method and fragments of Peano arithmetic. To appear in the Proceedings of the Ninth International Congress on Logic, Methodology and Philosophy of Science. [2] , Bounded Arithmetic, Bibliopolis, 1986. Revision of 1985 Princeton University Ph.D. thesis. [3] , Axiomatizations and conservation results for fragments of bounded arithmetic, in Logic and Computation, proceedings of a Workshop held Carnegie-Mellon University, 1987, vol. 106 of Contemporary Mathematics, American Mathematical Society, 1990, pp. 57{84. [4] , On truth-table reducibility to SAT, Information and Computation, 91 (1991), pp. 86{102. S. R. Buss
S. R. Buss and L. Hay
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[5]
, An application of Boolean complexity to separation problems in bounded arithmetic, Proc. London Math. Society, 69 (1994), pp. 1{21. [6] , The boolean hierarchy and the polynomial hierarchy: a closer connection, in Proceedings Fifth Annual Structure in Complexity Conference, IEEE Computer Society Press, 1990, pp. 169{ 178. [7] , Turing machines that take advice, L'Enseignement Mathematique, 28 (1982), pp. 191{209. [8] , No counter-example interpretation and interactive computation, in Logic From Computer Science: Proceedings of a Workshop held November 13-17, 1989, Mathematical Sciences Research Institute Publication #21, Springer-Verlag, 1992, pp. 287{293. [9] , Fragments of bounded arithmetic and bounded query classes, Transactions of the A.M.S., (1993), pp. 587{598. [10] , Bounded arithmetic and the polynomial hierarchy, Annals of Pure and Applied Logic, 52 (1991), pp. 143{153. [11] , Existence and feasibility in arithmetic, Journal of Symbolic Logic, 36 (1971), pp. 494{508. [12] , Some relations between subsystems of arithmetic and the complexity of computations, in Logic From Computer Science: Proceedings of a Workshop held November 13-17, 1989, Mathematical Sciences Research Institute Publication #21, Springer-Verlag, 1992, pp. 499{519. [13] , Sharply bounded arithmetic and the function a 0 1, in Logic and Computation, proceedings of a Workshop held Carnegie-Mellon University, 1987, vol. 106 of Contemporary Mathematics, American Mathematical Society, 1990, pp. 281{288. ek S. R. Buss and J. Kraj c
R. Chang and J. Kadin
R. M. Karp and R. J. Lipton
ek J. Kraj c
ek, P. Pudl a k, and G. Takeuti J. Kraj c
R. Parikh
k P. Pudl a
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G. Takeuti
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[14]
, On the scheme of induction for bounded arithmetic formulas, Annals of Pure and Applied Logic, 35 (1987), pp. 261{302. [15] , Notes on polynomially bounded arithmetic. Submitted for publication. A. J. Wilkie and J. B. Paris
D. Zambella
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Department of Mathematics University of California, San Diego La Jolla, CA 92093-0112 [email protected] May 1993, last revision November 1994
Abstract
The bounded arithmetic theory S2 is nitely axiomatized if and only if the polynomial hierarchy provably collapses. If T 2 equals S2+1 then T2 is equal to S2 and proves that the polynomial time hierarchy collapses to 6 +3 , and, in fact, to the Boolean hierarchy over 6 +2 and to 6 +1 =poly . i
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1 Introduction Theories of bounded arithmetic are theories of arithmetic obtained by putting restrictions on induction axioms; namely, allowing induction only for certain classes, 6 , of bounded formulas, and using polynomial, or length, induction (PIND or LIND) in place of successor induction (IND). The most important subtheories of bounded arithmetic are the theories S2 , axiomatized with 6 PIND (or equivalently, 6 -LIND, if i 1), and the theories T2 , axiomatized by 6 -IND. The following inclusions are known for these theories: b i
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S20 ( T20 S21 T21 S22 T22 1 1 1
3 Supported in part by NSF grant DMS-9205181
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and their union is the theory S2 = T2 [2]. However, with the exception of S20 6= T20 (see [13]), it is not known whether the rest of the theories of bounded arithmetic are distinct. It is a well-known fact that S2 and T2 are nitely axiomatized for i > 0, and thus it is immediate that this hierarchy of theories collapses if and only if S2 is nitely axiomatized. This latter condition is equivalent to I 1 0 + 1 being nitely axiomatized (see Parikh [11] and Wilkie-Paris [14] for this alternate, and original, approach to bounded arithmetic). There are close connections between theories of bounded arithmetic and the polynomial hierarchy. First, the class of predicates de nable by 6 (or 5 ) formulas is precisely the class of predicates in the i -th level 6 (or 5 , respectively) of the polynomial hierarchy. For instance, S 21 and T21 are axiomatized with their induction axioms restricted to NP-predicates (since NP = 61 is the class of predicates de nable by 61 -formulas). Second, it is known that the 6 -de nable functions of S1 are precisely the -functions, which are the functions which are polynomial time computable with an oracle for 6 01 . For instance, the 61 -de nable functions of S21 are precisely the polynomial time computable functions. Since it is open whether the polynomial time hierarchy collapses, it is natural to ask whether there is any connection between the possible collapses of the hierarchy of bounded arithmetic theories and the polynomial hierarchy. This question has already been partially answered by the work of KrajcekPudlak-Takeuti [10] who showed that if T2 = S2+1 for any i 1, then the polynomial hierarchy collapses with 6 +2 = 5 +2 (in fact, they show that in this case, 6 +1 1 =poly ). The main results of this paper strengthen the results of Krajcek-PudlakTakeuti by proving that if T2 = S2+1 holds, then the following conditions must hold: (1) T2 = S2 , so that the hierarchy of bounded arithmetic theories collapses, and (2) T2 can provethat the polynomial time hierarchycollapses to B(6 +2) and to 6 +1=poly , where B(6 +2) is the class of Boolean combinations of 6 +2 -predicates. Our proofs are easier, in a combinatorial sense, than the proofs of [10] and this makes it possible to formalize them in T2 . We believe that the results of this paper are nearly the strongest that are obtainable relating the possibility that T2 = S2+1 to the possible collapse of i
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the polynomial time hierarchy | at least with current techniques. To support this belief, consider the three conditions: ( ) The polynomial hierarchy collapses ( ) S2 proves that the polynomial hierarchy collapses ( ) S2 is nitely axiomatized Our results show that ( ) and ( ) are equivalent; however, we do not expect to show that ( ) is equivalent to ( ) using current techniques. The reason for this is that ( ) is a 602 -condition whereas, since ( ) is a 601 -condition, the results of the current paper show that ( ) is a 601 -condition; and, based on the history of attempts to solve the P versus NP problem, it seems to be dicult even to establish that the collapse of the polynomial time hierarchy is equivalent to a natural 601 -condition like ( ). It is known that S2+1 is conservative over T2 with respect to 86 +1 sentences [3]. On the other hand, the axioms of S2+1 can be expressed as 85 +2 -sentences (in this formulation, an induction axiom of S2+1 will become a 85 +2 -formula with a sharply bounded existential quanti er in its outermost block of bounded universal quanti ers). Thus saying S2+1 is 5 +2 -conservative over T2 is equivalent to saying that S2+1 = T2 . An open problem is to try to relate the condition S2 = T2 to the possible collapse of the polynomial hierarchy. Krajcek [9] shows that if S2 = T2 , then the set (6 ) of predicates logspace, Turing reducible to 6 is equal to the set (6 ) of predicatespolynomial time, Turingreducibleto 6 . However,it is open whether this last condition implies the polynomial hierarchy collapses. See [5,4,6] for more on this connection. The prerequisites for reading this paper are a basic knowledge of bounded arithmetic theories as contained in [2]. The reader would also bene t from knowledge of [10] and [3]. In the next section we will review the necessary background material needed from [10]. After preparing the rst draft of this paper, we learned that D. Zambella has independently discovered the main results of this paper [15]. i
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2 The KPT Witnessing Theorem for T2i There are two important witnessing theorems for T2 . The rst follows from the `Main Theorem' for S2+1 and the fact that S2+1 is 6 +1 -conservative over T2 : this witnessing theorem states that the 6 +1 -de nable functions of T2 are precisely the functions which can be computed in polynomial time with a 6 -oracle (i.e., the +1 -functions). The second witnessing theorem puts a necessary condition on the 6 +2 - and 6 +3 -de nable functions of T2 ; we call this the `KPT witnessing theorem'. It is this latter witnessing theorem that we need for our proofs: Theorem 1 Let i 1 . Suppose T2 ` (8x)(9y )(8z )B (x; y; z ) , where B is a 95 -formula, with only x; y; z as free variables. There exists k > 0 and functions f1; ... ; f such that each f is m -ary and is 6 +1 -de nable by T2 i
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` B (a; f1(a); b1) _ B (a; f2(a; b1); b2) _ B (a; f3(a; b1; b2); b3 ) _ 1 1 1 1 1 1 _ B (a; f (a; b1; b2; ... ; b 01); b ):
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For i = 0 , the same result holds for P V1 in place of T20 . As usual, P V1 denotes the conservative extension of P V to rst-order logic, or equivalently, P V1 is S20 or T20 enlarged to have function symbols and their de ning equations for all polynomial time functions.
Note that since the functions f are 6 +1 -de nable by T2 , they must be +1 -functions. Theorem 1 is due to [10]; some later, related results can be found in [8,12,1]. We do not include a proof here. We next use Theorem 1 to establish a consequence of the condition T2 = S2+1 . We assume that i 0 and work with the theory T2 ; when i = 0 our results are intended to hold for P V1 in place of T20 . De nition A quanti ed Boolean formula is a formula constructed from Boolean connectives (say, ^ , _ and : ) and quanti ers ranging over Boolean values. A quanti er (8p) or (9p) indicates quanti cation allowing p to range over the values True and False. b
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Given a truth assignment to the free variables of a quanti ed Boolean formula, it is obvious how the truth valueof the formula should be de ned. A quanti ed Boolean formula is satis able if there is some truth assignment to its free variables which gives it value True. A 5 -formula is a quanti ed Boolean formula which is in prenex form with i blocks of like quanti ers starting with a universal block. It is well-known that the set of satis able 5 -formulas is 6 +1 -complete. De nition Let i 0. TRU and SAT are bounded arithmetic formulas which express: TRU ('; w) , ' codes a 5 -formula and w codes a satisfying assignment of ' SAT (') , (9w ')TRU ('; w): In the de nition of TRU and SAT we presume that quanti ed Boolean formulas and truth assignments are coded in some natural and ecient way by integers; we use Greek letters '; ... as variables that range over integers which are intended to code quanti ed Boolean formulas. Since the code of a truth assignment can w.l.o.g. always be less than the code of a formula, SAT (') expresses the condition that ' is satis able. Standard bootstrapping techniques allow TRU to be a 1 +1 -formula with respect to the theory T2 ; in fact, for i 1, TRU is a 5 -formula. Hence SAT is a 6 +1 -formula. Also, T2 can prove basic properties of the TRU and SAT predicates. Most importantly, T2 can provethat SAT is many-one complete for 6 +1 -formulas; i.e., for any 6 +1 -formula A(~b), there is a polynomial time function f so that A(~b) is T2 -provably equivalent to SAT (f (~b)). As an application of Theorem 1, consider the formula 8h'0; ... ; ' i(9` n)(9hw0 ; ... ; w i) (1) h i (8j `)TRU (' ; w ) ^ (` < n ! :(9w +1)TRU (' +1; w +1)) : The meaning of formula (1) requires some explanation. First, a notation like (8h'0; ... ; ' i)B ('~ ; `) means the same as \there is an integer '3 which codes a sequence of 5 -formulas '0; ... ; ' so that B ('~ ; `) holds". The B i
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quanti er (9` n) is a sharply bounded quanti er since ` can be bounded by the length of the code for h'~ i , and the quanti ers (9hw~ i) and (9w +1) are bounded quanti ers since each w may be bounded by ' . By using prenex operations and using the fact that ` can be computed from hw0; ... ; w i , formula (1) is equivalent to the formula (8h'0; ... ; ' i)(9hw0; ... ; w i)(8w +1) (2) i h (8j `)TRU (' ; w ) ^ (` < n ! :TRU (' +1; w +1)) : which is a 8981 +1 -formula. The intuitive meaning of formula (1) or (2) is, of course, that every sequence '0; ... ; ' of 5 -formulas has an initial sequence of maximal length ` of satis able formulas. Furthermore, the formula (1) is a theorem of S2+1 . This is because S2+1 can use length induction on the 6 +1 -formula S (h'~ i; `) expressing the condition that the rst ` formulas of the sequence are satis able. (An equivalent way to see this is to note that S2+1 can prove the 6 +1 -length-maximization principle.) Now suppose T2 is equal to S2+1 ; in particular, T2 proves the formula (2). By Theorem 1, this means that there is an integer k 0 and there are 6 +1 -de ned functions f0; ... ; f so that, letting A(h'~ i; hw~ i; w +1) be the subformula of (2) enclosed in square brackets, we have that T2 ` (8h'~ i)[A(h'~ i; f0(h'~ i); b0) _ A(h'~ i; f1 (h'~ i; b0 ); b1) _ 1 1 1 (3) 1 1 1 _ A(h'~ i; f (h'~ i; b0; b1 ; ... ; b 01); b )] We henceforth shall use (3) restricted to the case where n = k , so that the sequence '~ is '0; ... ; ' . Without loss of generality, each f satis es the following property (provably in T2 ): whenever T RU (' ; b ) holds for r = 0; ... ; j 0 1, then the value f (h'~ i; b0; ... ; b 01) is the Godel number of a sequence hv0 ; ... ; v 01i of length ` j so that T RU (' ; v ) holds for all r = 0; ... ; ` 0 1. Recall that represents the Godel function so that (i; w) is equal to the i -th integer in the sequence coded by w . De ne g ('0 ; ... ; ' ; w0; ... ; w 01 ) = (j + 1; f (h'0 ; ... ; ' i; w0 ; ... ; w 01)): `
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Suppose that '0; ... ; ' are codes for satis able 5 -Boolean formulas and let w0 ; ... ; w be satisfying assignments. De ne b0 ; b1; ... inductively as follows: if f (h'~ i; b0; ... ; b 01) is a sequenceof length ` + 1 k , then let b equal w +1 . It is obvious that whenever f (h'~ i; b0; ... ; b 01) has length ` + 1 k then b givesa \counterexample"so that A(h'~ i; f (h'~ i; b0; ... ; b 01); b ) is false. Now, by (2), there is some j k for which f (h'~ i; b0; ... ; b 01) has length k + 1. Let j0 be the least value such that f (h'~ i; b0; ... ; b 01) has length j0 + 1. It must be that TRU (' ; g ('~ ; w0; ... ; w 01)) holds. This argument formalizes in T2 and thus we have proven: Lemma 2 Suppose T2 = S2+1 . Then there is k 0 and there are 6 +1 de nable functions g0 ; ... ; g of T2 so that B i
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3 Collapsing Bounded Arithmetic In this and the next section, we examine consequences of the condition T2 = S2+1 . In this section we show that this implies that S2 collapses to T2 . Our point of departure is Lemma 2 above; we henceforth x k and g0 ; ... ; g . This lemma states that at least one of the functions g can nd a satisfying assignment for ' using only the vector '~ and arbitrary satisfying assignments w0; ... ; w 01 . However, it need not always be the same g that succeeds in this way; di erent vectors of formulas '~ and even di erent witnesses w~ may cause di erent g 's to succeed. We de ne SucceedBy (`; '~ ; w~ ) to be the following formula which states that one of the i
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Our rst goal is to show that 6 +1 = 5 +1=poly where the \ poly " means that polynomial amount of advice is needed. As a preliminary to de ning what constitutes advice, we de ne \preadvice" by letting P reAdvice (a; h' +1; ... ; ' i) be the formula i
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(8h'0; ... ; ' i)(8hw0; ... ; w i) VTRU (h'~ i; hw~ i; a) ! SucceedBy(`; '~ ; w~ ) ; where VTRU (h'0; ... ; ' i; hw0; ... ; w i; a) abbreviates (8j `)(TRU (' ; w ) ^ w ' ^ ' < 2j j ): Several points to note are: rstly, in de ning P reAdvice we are continuing our practice of letting variables ' represent integers that must code 5 formulas; secondly, the value of ` is determined by the second argument to P reAdvice ( k is xed and ` varies, namely, ` equals k + 1 minus the length of the sequence coded by the second argument of P reAdvice ); thirdly, the quanti ers are bounded quanti ers since the ' 's and w 's are bounded by 2j j . The reason for bounding everything by 2j j is that we need only de ne \advice" that works for ' 's with j'j a for a an arbitrary integer. Also note that P reAdvice is a 5 +1 -formula. We can now de ne \advice" for formulas of length jaj by Advice (a; h' +1; ... ; ' i) , P reAdvice (a; h' +1; ... ; ' i) ^ :(9' )P reAdvice (a; h' ; ... ; ' i): Note that ' is bounded by 2j j ; thus Advice is a 5 +2 formula. The next lemma shows that T2 can prove that there always does exist advice: Lemma 3 Suppose T2 = S2+1 . Then T2 ` (8a)(9h'~ i)Advice (a; h'~ i): `
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Proof First, note that Lemma 2 implies that T2 proves that P reAdvice (a; hi) holds. Since k is a constant, it follows (without using induction) that there is a least ` such that (9h' +1; ... ; ' i)P reAdvice (a; h'~ i) holds. For this ` , any `preadvice' is actually advice. 2 i
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Next we give the key lemma that shows how `advice' can be used to make 6 +1 -IND hold and the polynomial time hierarchy collapse, provably in T 2 . Lemma 4 Suppose T2 = S2+1 . Then T2 proves Advice (a; h' +1; ... ; ' i) ^ ' < 2j j ! h :SAT (' ) $ (9h'0 ; ... ; ' 01i)(9hw0 ; ... ; w 01i) n VTRU (h' ~ i; hw0 ; ... ; w 01i; a) ^:SucceedBy(` 0 1; '~ ; w~ ) oi ^:TRU (' ; g ('0 ; ... ; ' ; w0 ; ... ; w 01)) : b i
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Let RHS(' ; h' +1 ; ... ; ' i; a)) denote the formula on the righthand side of the $ connective in the formula above; we often suppress the variables ' +1; ... ; ' and a that occur freely in RHS and write just RHS(' ). We shall argue informally in T2 to prove the lemma. Suppose ' ; ... ; ' 2j j are formulas and that Advice (a; h' +1 ; ... ; ' i) holds. The latter condition obviously implies that :P reAdvice (a; h' ; ... ; ' i). By the de nition of P reAdvice , there must exist 5 -formulas '0; ... ; ' 01 satis ed by witnesses w0; ... ; w 01 such that SucceedBy(` 0 1; '~ ; w~ ) is forced to be false. First suppose that ' is not satis able. Then clearly TRU (' ; g ('~ ; w~ )) must be false. Thus RHS(' ) follows from :SAT (' ). Second, suppose that ' is satis able. By P reAdvice (a; h' +1 ; ... ; ' i), it must be that SucceedBy (`; '~ ; w~ ) holds. On the other hand, SucceedBy (` 0 1; '~ ; w~ ) is false. Thus TRU (' ; g ('~ ; w~ )) is forced to be true and we have shown that SAT (' ) implies :RHS(' ). 2 In the subformula RHS , the leading existential quanti ers are actually
Proof
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bounded existential quanti ers since the formulas ' and their witnesses w are bounded by 2j j . This means that RHS(' ) is a 6 +1 -formula. Lemma 5 Suppose T2 = S2+1 . Then T2 ` 6 +1 -IND and T2 = T2+1 . Proof The proof is based on the fact that SAT (1 1 1) is complete for 6 +1 -formulas and is also equivalent on bounded ranges to the 5 +1 -formula :RHS(1 1 1) (under the assumption that T2 = S2+1 , as always). Indeed, for any 6 +1 -formula B (c; d~), there is a polynomial time and 61 -computable function f (c; d~) so that B (c; d~) is T2 -provably equivalent to SAT (f (c; d~)). The induction axiom for the formula B (c; d~) can be expressed as B (0; d~) ^ (8x)(B (x; d~) ! B (x + 1; d~)) ! B (c; d~): Let us prove this by reasoning informally in T2 which is presumed to equal S2+1 . Considering particular values for c and d~, there is a value a so that f (x; d~) < 2j j for all x c . Let ' +1; ... ; ' be formulas such that Advice (a; h' +1; ... ; ' i) holds. Then, with these parameters, Lemma 4, we have that the 6 +1 -formula B (x; d~) is equivalent to the 5 +1 -formula :RHS(f (x; d~)) for all x c . Now, it is known that S2+1 proves 1 +1 -IND and the usual proof (see Theorem 2.22 of [2]) shows that T2 = S2+1 proves induction for B , since B is \1 +1 with parameters" on the range 0 x c . j
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Iterating the method of this proof, we obtain: Theorem 6 If T2 = S2+1 , then T2 = S2 . Thus, if T2 = S2+1 , then S2 i
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nitely axiomatized. Also, if P V1 = S21 (P V ) , then P V1 = S2 (P V ) .
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Analogous to the method of proof of Lemma 5, we must show that any bounded formula is equivalent to a 6 +1 -formula with parameters, where the parameters vary with the range of the induction variable. From this, using Lemma 5, it will follow that T2 proves induction for any bounded formula. We do the case of B (c; d~) 2 6 +2 in some detail. We may suppose that B (x; d~) is of the form (9y t(x; d~))C (x; y; d~) for some 5 +1 -formula C . Proof
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We argue informally in T2 . By the method of Lemma 5, there is an a , given by a polynomial time function a = a(c; d~) of c and d~, and there is a polynomial time function f , so that for all x c , and y t(x; d~) and for advice h'~ i satisfying Advice , the 5 +1 -formula C (x; y; d~) is equivalent to the 6 +1 -formula RHS(f (x; y; d~); h'~ i; a(c; d~)). Thus, for 0 x c , B (x; d~) is equivalent to a 6 +1 -formula, and full induction holds for B up to c by Lemma 5. Hence T2 = T2+2 . A slight modi cation of the construction of the last paragraph shows that if A(~x) is a 6 +2 -formula (respectively, a 5 +2 -formula, then there is a polynomial growth rate function a(c) and a 6 +1 -formula (respectively, 5 +1 -formula) A3(~x; '3; a(c)) such that for all ~x such that maxf~xg c and all h'~ i such that Advice (a(c); h'~ i), A(~x) is equivalent to A 3(~x; h'~ i; a(c)), provably in T2 . This further implies that if A(~x) is a 6 +3 -formula, then A3 may be taken to be a 6 +2 -formula, because, if A(~x) is (9y t(~x))B (~x; y), then there is a 6 +2 -formula B 3 so that A(~x) will be equivalent to (9y t(~x))B 3(~x; h'~ i; a) for a given by a polynomial growth rate function of c max ~x and for h'~ i such that Advice (a; h'~ i). This fact is sucient to imply that T2 = T2+3 . By iterating the abovemethod of proof, one can show that T2 is equal to all of S2 . We shall leave the details of this to the reader, and remark instead that an alternative proof is given by Theorem 7 below where it is shown that T2 can prove that every bounded formula is equivalent to a Boolean combination of 6 +2 -formulas without any additional parameters or advice. Then since T2 = T2+2 = S2+2 and S2+2 proves induction for Boolean combinations of 6 +2 -formulas [3], it follows that T2 = S2 . 2 i
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4 Collapsing the Polynomial Hierarchy All the work of this section is predicated on the condition that T 2 = S2+1 . We have shown above that if T2 = S2+1 , then T2 proves that the 6 +2 predicates are contained in 6 +1 =poly . From this, the methods of KarpLipton [7] imply that the entire polynomial time hierarchy is contained in 6 +1=poly and in 5 +1=poly ; furthermore, the proof of this containment can be formalizedin T2 . The methods of Karp-Lipton also implyimmediatelythat i
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the polynomial hierarchy collapses to 6 +3 = 5 +3 . However, we shall prove an somewhat stronger result; namely, if T2 = S2+1 , then every polynomial hierarchy predicate (i.e., bounded formula) is T 2 -provably equivalent to a Boolean combination of 6 +2 -formulas. To prove this, it will suce to prove that every 6 +3 -formula is equivalent to a Boolean combination of 6 +2 -formulas. Let A(b) be an arbitrary 6 +3 formula. From the previous section, we know that T2 proves that A(b) is equivalent to (9h'~ i)[Advice (a(b); h'~ i) ^ A3(b; h'~ i)] (4) and to (8h'~ i)[Advice (a(b); h'~ i) ! A3(b; h'~ i)]; (5) where A3 is a 6 +2 -formula and a = a(b) is function of suciently large polynomial growth rate. Unfortunately, Advice is a 5 +2 -formula and the quanti er complexity of these equivalent formulations of A(b) is higher than we desire; namely, formula (4) is a 6 +3 -formula and formula (5) is a 5 +3 formula. This implies that every bounded formula is 1 +3 with respect to T2 , but we wish to prove a yet stronger result. To reduce the complexity of these formulas we would like to use P reAdvice in place of Advice . However, this can not be done directly since if h'~ i satis es P reAdvice , then it is not necessarily true that A3(b; h'~ i) is equivalent to A(b). Instead, we look for a longest vector h'~ i which satis es P reAdvice ; namely, consider the formula A 0 (b) de ned as: h i (9h'1; ... ; ' i) P reAdvice (a(b); h'1; ... ; ' i) ^ A3(b; h'1; ... ; ' i) p
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_ WW :(9h' 01; ... ; ' i)P reAdvice (a(b); h' 01; ... ; ' i) =2 o ^(9h' ; ... ; ' i)[P reAdvice (a(b); h' ; ... ; ' i) ^ A3 (b; h' ; ... ; ' i)] We claim that A0(b) is equivalent to A(b). The proof of this now quite easy. First, there must exist a least ` 1 such that there exists h' ; ... ; ' i which satis es P reAdvice . Second, if P reAdvice (h' ; ... ; ' i) holds and if there k
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is no h'0 01 ; ... ; '0 i which satis es P reAdvice , then clearly h' ; ... ; ' i satis es Advice . And for this advice, A3(b; h'~ i) is equivalent to A(b). Since P reAdvice is a 5 +1 -formula and A3 is a 6 +2 -formula, A0 is a Boolean combination of 6 +2 -formulas. This establishes: Theorem 7 If T2 = S2+1 , then every bounded formula is T2 -provably equivalent to a Boolean combination of 6 +2 formulas. In other words, if T2 = S2+1 , i
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then the polynomial hierarchy T2 -provably collapses to (a nite level of) the Boolean hierarchy over 6 +2 . Also, in this case, T2 proves that the polynomial time hierarchy collapses to 6 +1 =poly . If P V1 = S21 (P V ) , then every bounded formula is P V1 -provably equivalent to a Boolean combination of 62 -formulas, so the polynomial time hierarchy provably collapses to the Boolean hierarchy over 62 . Also, in this case, P V1 proves that the polynomial time hierarchy collapses to NP=poly . i
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It should be noted again that [10] have shown that if T2+1 = S2+2 then the polynomial hierarchy collapses to 6 +2 = 5 +2 and to 1 +1=poly : we do not know how to prove that this stronger collapse would be T2 -provable. i
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