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Count(q) versus the Pigeon-Hole Principle
, 1996
"... For each p 2 there exists a model M of I \Delta 0 (ff) which satisfies the Count(p) principle. Furthermore, if p contains all prime factors of q there exist n; r 2 M and a bijective map f 2 dom(M ) mapping f1; 2; :::; ng onto f1; 2; :::; n+ q r g. A corollary is a complete classificati ..."
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Cited by 4 (1 self)
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For each p 2 there exists a model M of I \Delta 0 (ff) which satisfies the Count(p) principle. Furthermore, if p contains all prime factors of q there exist n; r 2 M and a bijective map f 2 dom(M ) mapping f1; 2; :::; ng onto f1; 2; :::; n+ q r g. A corollary is a complete classification of the Count(q) versus Count(p) problem. Another corollary shows that the pigeon-hole principle for injective maps does not follow from any of the Count(q) principles. This solves an open question [Ajtai 94]. 1 Introduction The most fundamental questions in the theory of the complexity of calculations are concerned with complexity classes in which `counting' is only possible in a quite restricted sense. Thus it is not surprising that many elementary counting principles are unprovable in systems of Bounded Arithmetic. These are axiom systems where the induction axiom schema is restricted to predicates of low syntactic complexity. For a good basic reference see [Krajicek 95]. The status of...
On the complexity of proof deskolemization
- J. Symbolic Logic
"... Abstract. We consider the following problem: Given a proof of the Skolemization of a formula F, what is the length of the shortest proof of F? For the restriction of this question to cut-free proofs we prove corresponding exponential upper and lower bounds. §1. Introduction. The Skolemization of for ..."
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Abstract. We consider the following problem: Given a proof of the Skolemization of a formula F, what is the length of the shortest proof of F? For the restriction of this question to cut-free proofs we prove corresponding exponential upper and lower bounds. §1. Introduction. The Skolemization of formulas is a standard technique in logic. It consists of replacing existential quantifiers by new function symbols whose arguments reflect the dependencies of the quantifier. The Skolemization of a formula is satisfiability-equivalent to the original formula. This transformation has a number of applications, it is for example crucial for automated theorem
Are There Hard Examples for Frege Proof Systems?
, 1995
"... It is generally conjectured that there is an exponential separation between Frege and extended Frege systems. This paper reviews and introduces some candidates for families of combinatoriM tautologies for which Frege proofs might need to be superpolynomially longer than extended Frege proofs. Surpri ..."
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Cited by 1 (0 self)
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It is generally conjectured that there is an exponential separation between Frege and extended Frege systems. This paper reviews and introduces some candidates for families of combinatoriM tautologies for which Frege proofs might need to be superpolynomially longer than extended Frege proofs. Surprisingly, we conclude that no particularly good or convincing examples are known. The examples of combinatorial tautologies that we consider seem to give at most a quasipolynomial speed-up of extended Frege proofs over Frege proofs, with the sole possible exception of tautologies based on a theorem of Frankl.
Σ_n-Bounding and Δ_n-Induction
- Proc. Amer. Math. Soc
"... Working in the base theory of PA - exp, we show that for all n #, the bounding principle for # n -formulas (B# n ) is equivalent to the induction principle for # n -formulas (I# n ). This partially answers a question of J. Paris; see Clote and Kraj cek (1993). ..."
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Working in the base theory of PA - exp, we show that for all n #, the bounding principle for # n -formulas (B# n ) is equivalent to the induction principle for # n -formulas (I# n ). This partially answers a question of J. Paris; see Clote and Kraj cek (1993).
INTRODUCTION TO THE COMBINATORICS AND COMPLEXITY OF CUT ELIMINATION
"... Abstract. Modus Ponens says that if you know A and you know that A implies B, then you know B. This is a basic rule that we take for granted ..."
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Abstract. Modus Ponens says that if you know A and you know that A implies B, then you know B. This is a basic rule that we take for granted

