# On existence of complete predicate calculus in metamathematics without exponentiation

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## Keywords

Polynomial Time Predicate Calculus Proof System Binary String Predicate Logic
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## References

- 1.Cook, S. and Reckhow, R., On the Lengths of Proofs in the Propositional Calculus. Preliminary Version. Conference Record of 6th ACM STOC (1974), 135–148.Google Scholar
- 2.Sazonov, V. Yu., A Logical Approach to the Problem "P = NP?", Proc. of MFCS'80, Lecture Notes in Computer Sci. 88 (1980), 562–575. (Corrections. 1. In this paper Lemma 5.4(4) is given without formal proof, and, as the author has understood, he cannot give such a proof at this moment. Therefore, the strong form of Main Result represented by theorems 1.1, 1.3, 3.7 and 5.1 and depending on this Lemma is actually not proved. However, its weakened form given by Theorem 3.8 is proved correctly. Theorems 1.1, 1.3 and 5.1 should also be correspondingly weakened. (Substitute T
_{Q}in place of T.) Note that the damage proof of 5.1 contains useful Lemma 5.2 and demonstrates what can be done in theories T_{O}and T. 2. In the proof of theorem 2.3 it is not exactly said that ξ_{n}^{x}may be defined as a subsequence of \(\tilde \xi _n^x\). It should be taken \(\xi _\Lambda ^x = \tilde \xi _\Lambda ^x\)and \(\xi _{n + 1}^x = \underline {if} \tilde \xi _{n + 1}^x \notin \{ \xi _i^x /i \leqslant n\}\) then \(\tilde \xi _{n + 1}^x\) else the first B_{j}∉{ξ_{i}^{x}| i⩽n} (it is obvious that j⩽n+1). Then \(\tilde \xi _n^x = \xi _{q(n,x)}^x\)for some P-function q(n,x)⩽n. Addition. With the contrast to the hypothesis on page 565 it can actually be proved that the linear and the lexicographical inductions for quantifier free formulas are equivalent.)Google Scholar - 3.Sazonov, V. Yu., Polynomial Computability and Recursivity in Finite Domains. EIK 16 (1980) 7, 319–323.Google Scholar
- 4.Kleene, S. C., Mathematical Logic, J.Wiley & Sons, INC, New York-London-Sydney, 1967.Google Scholar

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© Springer-Verlag Berlin Heidelberg 1981