A proof of P!=NP

TL;DR

This paper presents a proof within PA that P does not equal NP, constructing a sequence of sentences with specific logical properties and growth rates, extending to SEFA.

Contribution

It provides the first proof in PA that P ≠ NP by constructing a special sequence of sentences with maximal growth in logical strength.

Findings

Constructed a sequence of $orall^{0}_{2}$-sentences with polynomial length bounds.

Proved that PA can derive P ≠ NP using these sentences.

Demonstrated the sequence's growth rate matches the cofinal growth of all total recursive functions.

Abstract

We show that it is provable in PA that there is an arithmetically definable sequence $\{\phi_{n}:n \in \omega\}$ of $\Pi^{0}_{2}$-sentences, such that - PRA+$\{\phi_{n}:n \in \omega\}$ is $\Pi^{0}_{2}$-sound and $\Pi^{0}_{1}$-complete - the length of $\phi_{n}$ is bounded above by a polynomial function of $n$ with positive leading coefficient - PRA+$\phi_{n+1}$ always proves 1-consistency of PRA+$\phi_{n}$. One has that the growth in logical strength is in some sense "as fast as possible", manifested in the fact that the total general recursive functions whose totality is asserted by the true $\Pi^{0}_{2}$-sentences in the sequence are cofinal growth-rate-wise in the set of all total general recursive functions. We then develop an argument which makes use of a sequence of sentences constructed by an application of the diagonal lemma, which are generalisations in a broad sense of Hugh Woodin's "Tower of Hanoi" construction as outlined in his essay "Tower of Hanoi" in Chapter 18 of the anthology "Truth in Mathematics". The argument establishes the result that it is provable in PA that $P \neq NP$. We indicate how to pull the argument all the way down into SEFA.