Threshold logic proof systems

Threshold logic proof systems Samuel Buss Peter Clote May 19, 1995 In this note, we show the intersimulation of three threshold logics within a polynomial size and constant depth factor. The logics are PTK, PTK and FC, the latter introduced by J. Krajíče in [2]. Definition 1 Propositional threshold logic is given as follows. Formula depth and size are defined inductively by: i. a propositional variable x i, i N, is a formula of depth 0 and size 1. 1 ii. if F is a formula then F is a formula of depth 1 + dp(f ) and size 1 + size(f ). iii. if F 1,...,F n are formulas and 1 n then T n (F 1,...,F n )isaformula of depth 1 + max{depth(f i ):1 i n} and size (n + ) +1+ 1 i n size(f i). Propositional threshold logic can be viewed as an extension of propositional logic in the connectives,,, the latter two connectives being defined by F i T1 n (F 1,...,F n ) 1 i n 1 i n F i T n n (F 1,...,F n ) A cedent is any sequence F 1,...,F n of formulas separated by commas. Cedents are sometimes designated by Γ, Δ,...(capital Gree letters). A sequent is given by Γ Δ, where Γ, Δ are arbitrary cedents. The size [resp. depth] of a cedent F 1,...,F n is 1 i n size(f i) [resp. max 1 i n (depth(f i ))]. The size [resp. depth] of a sequent Γ Δ is size(γ) + size(δ) [resp. max(depth(γ),depth(δ))]. The intended interpretation of the sequent Γ Δis Γ Δ. An initial sequent is of the form F F where F is any formula of propositional threshold logic. The rules of inference of PTK, the sequent calculus of 1 One could as well allow propositional constants 1 (true) and0(false)ofdepth0and size 1. 1

propositional threshold logic, are as follows. 2 By convention, T n m(a 1,...,A n )is only defined if 1 m n. structural rules wea left: Γ, Δ Γ Γ Γ, Δ Γ,A,Δ Γ wea right: Γ Γ,A,Δ contract left: Γ,A,A,Δ Γ Γ Γ,A,A,Δ Γ,A,Δ Γ contract right: Γ Γ,A,Δ permute left: cut rule Γ,A,B,Δ Γ Γ Γ,A,B,Δ Γ,B,A,Δ Γ permute right: Γ Γ,B,A,Δ Γ,A Δ Γ A, Δ Γ, Γ Δ, Δ logical rules -left: A, Γ Δ Γ A, Δ -right: Γ A, Δ A, Γ Δ A 1,...,A n, Γ Δ -left: for n 1 T n n (A 1,...,A n ), Γ Δ Γ A 1, Δ Γ A n, Δ -right: for n 1 Γ T n n (A 1,...,A n ), Δ A 1, Γ Δ A n, Γ Δ -left: for n 1 T n 1 (A 1,...A n ), Γ Δ Γ A 1,...,A n, Δ -right: for n 1 Γ T n 1 (A 1,...,A n ), Δ 2 Gentzen s original sequent calculus for first order logic was called LK (Logischer Kalül). The propositional sequent calculus with connectives,, has sometimes been called PK (propositional Kalül), so our propositional threshold Kalül is denoted PTK. 2

T n -left: T n 1 (A 2,...,A n ), Γ Δ A 1,T n 1 1 (A 2,...,A n ), Γ Δ T n(a 1,...,A n ), Γ Δ for 2 <n T n -right: Γ A 1,T n 1 (A 2,...,A n ), Δ Γ T n 1 1 (A 2,...,A n ), Δ Γ T n(a 1,...,A n ), Δ for 2 <n The structural rules, cut rule, rules, rules and rules are the same as for PTK. However, in place of the T n rules of PTK, PTK has the following rules. T n -left1: T n (A 1,...,A n ), Γ Δ T n +l (A 1,...,A n ), Γ Δ for 1 <+ l n T n -left2: T n (A 1,...,A n ), Γ Δ T n+m +m (A 1,...,A n,b 1,...,B m ), Γ Δ for 1 n<n+ m T n -left3: A 1,..., A n,t m (B 1,...,B m ), Γ Δ A 1,..., A n,t m+n (A 1,...,A n,b 1,...,B m ), Γ Δ for 1 m<m+ n T n -right1: Γ T n (A 1,...,A n ), Δ Γ T n+m (A 1,...,A n,b 1,...,B m ), Δ for 1 n<n+ m T n -right: Γ T n (A 1,...,A n ), Δ Γ T m l (B 1,...,B m ), Δ Γ T n+m +l (A 1,...,A n,b 1,...,B m ), Δ for 1 m<m+ n In [2], J. Krajíče introduced an extension of the Frege system F, called FC for Frege with counting. In addition to the usual connectives of F, counting connectives C n, (x 1,...,x n ) are admitted, whose interpretation is that exactly of the x i equal 1. Definition 2 FC is the propositional proof system having connectives,,,, together with infinitely many new connectives C n, (φ 1,...,φ n ), for 1 n and n. The axioms of FC are those of F (see [1]) together with the new axioms: 1. A C 1,1 (A) 2. C n,0 (A 1,...,A n ) ( A 1... A n ) 3

3. C n+1,+1 (A 1,...,A n+1 ) [(C n, (A 1,...,A n ) A n+1 ) (C n,+1 (A 1,...,A n ) A n+1 )] if <n 4. C n+1,n+1 (A 1,...,A n+1 ) [(C n,n (A 1,...,A n ) A n+1 )]. We intend to show the relation between FC and our threshold proof systems PTK and PTK ; namely that constant depth polynomial size FC proofs correspond to polynomial size constant depth PTK and PTK proofs, and vice versa. We begin by simulating FC within PTK. Definition 3 Translate the FC formula A by the PTK formula A as follows: FC formula PTK formula x x n i=1 A i Tn n (A 1,...,A n ) n i=1 A i T1 n (A 1,...,A n ) A B T1 2 ( A, B) A B T2 2 (A B,B A) C n, (A 1,...,A n ), 0 <<n T2 2 (T n(a 1,...,A n ), T+1 n (A 1,...,A n )) C n,n (A 1,...,A n ) Tn n (A 1,...,A n ) C n,0 (A 1,...,A n ) T1 n (A 1,...,A n ) For each axiom scheme A of FC,wesetchthePTK proof of A (usually the last few steps from the formula à proved to the equivalent A are easy and left to the reader). In our notation, C n, ( A) abbreviates C n, (A 1,...,A n ), and T n( A) abbreviates T n(a 1,...,A n ). We often abbreviate A n+1 by A, so that for instance in the first subclaim appearing in the proof of Axiom 3 below, abbreviates +1 ( A) T n ( A) A, T n +1( A) A +1 (A 1,...,A n+1 ) T n (A 1,...,A n ) A n+1,t n +1(A 1,...,A n ) A n+1 Axiom 1 x C 1,1 (x) 4

x x x T 1 1 (x) x, T 1 1 (x) x T 1 1 (x) This completes the proof of axiom 1. x x T 1 1 (x) x T 1 1 (x),x T 1 1 (x) x ( x T 1 1 (x)) ( T 1 1 (x) x) Axiom 2 C n,0 (A 1,...,A n ) A 1... A n (Recall that, associate to the left.) Claim C n,0 (A, B, C) ( A B) C Pf A T 1 1 (A) T 3 1 (A, B, C) A A T 3 1 (A, B, C) T 3 1 (A, B, C) ( A B) B B B T 1 1 (B) T 3 1 (A, B, C) B B T 3 1 (A, B, C) T 3 1 (A, B, C) ( A B) C C C C T 1 1 (C) T 3 1 (A, B, C) C C T 3 1 (A, B, C) Claim ( A B) C C 3,0 (A, B, C) Pf A, A A, A, B A, ( A B) A, ( A B), C A, ( A B C) This completes the proof of axiom 2. B B B, B B, A, B B,( A B) B,( A B), C B,( A B C) T 3 1 (A, B, C), ( A B C) A B C T 3 1 (A, B, C) C C C, C C, A, B, C C, ( A B), C C, ( A B C) Axiom 3 C n+1,+1 ( A) (C n, ( A) A n+1 ) (C n,+1 ( A, A n+1 ) A n+1 Claim PTK proves C n+1,+1 ( A) (C n, ( A) A n+1 ) (C n,+1 ( A, A n+1 ) A n+1 The claim follows from two subclaims. 5

Subclaim +1 ( A) T n ( A) A, T n +1 ( A) A Pf T n +1 ( A) T n +1 ( A) A, T n +1 ( A) T n +1 ( A) A, +1 ( A) T n +1 ( A) +1 ( A) T n +1 ( A), A +1 ( A) A, T n +1 ( A) A, A +1 ( A) A, T n +1 ( A) A T n ( A) T n ( A) +1 ( A) T n ( A) A, A +1 ( A) A, A +1 ( A) T+1 n ( A) A, T n( A) Combining the last lines of the previous two proofs using -right, we have which establishes the subclaim. +1 ( A) T n ( A) A, T n +1( A) A Subclaim +2 ( A) T n +2 ( A) A, T n +1 A Pf First we prove the following. T n +2 ( A) T n +2 ( A) T n +2 ( A) +2 ( A) T n +2 ( A) +2 ( A), T n +1 +2 ( A) T n +2 ( A), T n +1 ( A) T n +2 ( A) T n +2 ( A) T n +2 ( A) +2 ( A) +2 ( A) T n +2 ( A) +2 ( A) T n +2 ( A), T n +1 ( A),A +2 ( A) T n +2 ( A),A Second we prove the following. T n +1 ( A) T n +1 ( A) A, T n +1 ( A) T n +1 ( A) A, T n +1 ( A) +2 ( A) A, +2 ( A) T n +1 ( A) A, T n +1 ( A) T 1 1 (A) A, +2 ( A) T n +1 ( A) A +2 ( A) A, T n +1 ( A) A A, +2 A 6

Combining the last lines of the previous two proofs using -right, we have +2 ( A) T n +2( A) A, T n +1( A) A as desired. Now from both subclaims, it can be shown that +1 ( A) +2 ( A) T n ( A) T n +1( A) A, T n +1( A) T n +2( A) A. This establishes the claim that C n+1,+1 ( A) (C n, ( A) A n+1 ) (C n,+1 ( A) A n+1 ) Claim PTK proves the converse of the previous, i.e. This translates to (C n, ( A) A n+1 ) (C n,+1 ( A) A n+1 ) C n+1,+1 ( A) (T n ( A) T n +1( A) A) (T n +1( A) T n +2( A) A) +1 ( A) +2 ( A). The claim follows from two subclaims. Subclaim (T n( A) T+1 n ( A) A) (T+1 n ( A) T+2 n ( A) A) +1 ( A) Pf A T 1 1 (A) A, T n ( A) T 1 1 (A) A, T n ( A) +1 ( A) A, T n ( A), T n +1 ( A) +1 ( A) T n ( A) T n ( A) A, T n ( A) T n ( A) T n +1 ( A) T n +1 ( A) T n +1 ( A) +1 ( A) A, T n +1 ( A) +1 ( A) A, T n +1 ( A), T n +2 ( A) +1 ( A) Now combining the last two proofs using -left, we have (A, T n ( A), T n +1( A)) ( A, T n +1( A), T n +2( A)) +1 ( A) Subclaim (T n ( A) T n +1 ( A) A) (T n +1 ( A) T n +2 ( A) A) +2 ( A) Pf 7

T n +1 ( A) T n +1 ( A) +2 ( A) T n +1 ( A) T n +1 ( A) +2 ( A) A, T n ( A), T n +1 ( A) +2 ( A) T n +2 ( A) T n +2 ( A) A, T n +2 ( A) T n +2 ( A) A, +1 ( A) T n +2 ( A) A, T n +2 ( A) +2 ( A) A, T n +1 ( A), T n +2 ( A) +2 ( A) (A, T n( A), T+1 n ( A)) ( A, T+1 n ( A), T+2 n ( A)) +2 ( A) From the two subclaims, we obtain a proof of (T n ( A) T+1( n A) A) (T+1( n A) T+2( n A) A) +1 ( A) +2 ( A) which establishes (C n, ( A) A) (C n,+1 ( A) A) C n,+1 ( A). This concludes the proof of axiom 3. Axiom 4 C n+1,n+1 ( A) C n,n ( A) A Claim C n+1,n+1 ( A) C n,n ( A) A. Pf Show Tn+1 n+1 ( A) Tn n ( A) A. A 1 A 1 A n A n A 1,...,A n+1 A 1 A 1,...,A n+1 A n n+1 ( A) A 1 This completes the proof of the claim. Claim C n,n ( A) A C n+1,n+1 ( A) Pf Show T n n ( A) A n+1 ( A). n+1 ( A) A n n+1 ( A) T n n ( A) n+1 ( A) T n n ( A) A n+1 A n+1 A n+1 A 1,...,A n+1 A n+1 n+1 ( A) A n+1 A 1 A 1 A 1,...,A n+1 A 1 A n A n A 1,...,A n+1 A n A n+1 A n+1 A 1,...,A n+1 A n+1 T n n ( A),A n+1 A 1 T n n ( A),A n+1 A n T n n ( A),A n+1 A n+1 Tn n ( A),A n+1 Tn+1 n+1 ( A) This completes the proof of the claims and so establishes the provability of the translation of Axiom 4 in PTK. By depth and size of a proof in a propositional proof system such as F, FC, PTK, etc. we mean the maximum depth and size of any formula appearing in the proof (in particular, we do not mean the depth of the proof tree in a sequent calculus proof). 8

Theorem 4 Suppose that P n : n 1 is a family of FC proofs, where P n is a depth d(n), sizes(n) proof of φ n. Then there exists a constant c for which there exists a family P n : n 1 of PTK proofs, where P n is a depth c + d(n), size c s(n) proof of φ n. Proof. The axioms of FC have previously been treated, and modus ponens (the only rule of inference of FC) is a special case of the cut rule of PTK. Analysis of the previous PTK proofs of the axioms of FC gives appropriate constant c. We now consider the simulation of PTK by FC. Definition 5 Translate the PTK formula A by the FC formula à as follows: PTK formula FC formula x x A à T n(a n 1,...,A n ) i= C n,i(ã1,...,ãn) A PTK sequent Γ Δ, which is equivalent to the formula is translated by the FC formula n m A i i=1 j=1 B j n m à i i=1 j=1 B j. Theorem 6 Suppose that P n : n 1 is a family of PTK proofs, where P n is a depth d(n), sizes(n) proof of φ n. Then there exists a constant c for which there exists a family P n : n 1 of FC proofs, where P n is a depth c + d(n), size s(n) c proof of φn. Proof setch By induction on the number of proof inferences. For each axiom of PTK, the translation of its sequent is easily provable in FC. Similarly, an appropriate translation of each proof rule of PTK is provable in FC. For instance, a binary rule A 1,...,A n1 B 1,...,B n2 C 1,...,C n3 D 1,...,D n4 E 1,...,E n5 F 1,...,F n6 9

is translated into n 1 ( i=1 Ã i n 2 i=1 n 5 ( i=1 n 3 B i ) ( Ẽ i i=1 n 6 i=1 C i To prove in FC the translation of the rule T n -left, begin with the tautology ( C n,i Γ Δ) ( C n,i Γ) Δ i n F i ) i n n 4 Using an axiom of FC, obtain ((A C n 1,i 1 ) ( A C n 1,i )) Γ Δ i n This is equivalent to the following. ((A C n 1,j ) ( A j<n i<n i=1 D i ) C n 1,i )) Γ Δ Using the translation into FC of T n (and for notational simplicity denoting the translation of formulas A by themselves), this yields the following. By distribution of this yields ((A T n 1 n 1 1 ) ( A T )) Γ Δ ((A T n 1 1 By distribution of this yields ((A T n 1 1 Γ) ( A T n 1 Γ)) Δ Γ) Δ) (( A T n 1 Γ) Δ) (T n Γ) Δ It will be shown in the proof of the next theorem that and so From this, since A T n 1 T n 1 T n 1 T n 1 1 Γ A T n 1 1 Γ (A T n 1 ) ( A T n 1 ) 10

it is not hard to see that there is an FC proof of the following. ((A T n 1 1 Γ) Δ) ((T n 1 Γ) Δ) (T n Γ) Δ But this is the translation of rule T n -left into FC.TheFC proofofthetranslation of T n -right is similar. Theorem 7 Suppose that P n : n 1 is a family of PTK proofs, where P n is a depth d(n), sizes(n) proof of φ n. Then there exists a constant c for which there exists a family P n : n 1 of PTK proofs, where P n is a depth c + d(n), size c s(n) proof of φ n. Proof. Note first that T n T n 1 and that T n T n 1 T n T n 1 1 T n T n T n A, T n 1 T n 1 1 T n 1 A, T n 1 1 T n 1 T n 1 Thus the n proof of Tj i T i+1 j yield a proof of and T i j+1 T i j for i<nand j<together (1) T n Case 1: T n-left1 Since T n has size O(n), there is an no(1) size proof of (1). Now T+l n T n T n, Γ Δ T+l n, Γ Δ Case 2: T n-left2 T+1 n T +1 n A, T n A T n +1 A T n,tn +1 11

From this, we obtain and by iteration rule T n-left2. Case 3: T n-left3 +1 T n T n +1, A A T n +1 A, T n, A +1, A A T n +1 by using the T n -left rule of PTK. Iterating this, we have the proof of the T n-left3 rule of PTK. Case 4: T n-right1 Immediate from (1). Case 5: T n-right2 Iterating the idea of proof of case 2, we can show that T n ( A) T m l ( B) T n+m +l ( A, B) From this, case 5 follows. This completes the proof of the theorem. It is not difficult to see that the simulations of FC, PTK and PTK are within a polynomial factor of the size and a constant factor of the depth. 12

References [1] S. A. Coo and R. Rechow. On the relative efficiency of propositional proof systems. Journal of Symbolic Logic, 44:36 50, 1977. [2] J. Krajíče. On Frege and extended Frege systems. In P. Clote and J. Remmel, editors, Feasible Mathematics II, pages 284 319. Birhäuser, 1994. 13