P不等于NP的证明

AI总结：

本文通过构造一个可定义的序列，证明在PA中P≠NP可被证明，且逻辑强度增长迅速。

摘要：

我们展示在PA中可以证明存在一个可定义的序列{ϕₙ:n∈ω} of Π²⁰-sentences，使得PRA+{ϕₙ:n∈ω}是Π²⁰-sound且Π¹⁰-complete，且ϕₙ的长度被一个多项式函数上界所限制，且PRA+ϕₙ+1总是证明PRA+ϕₙ的1-一致性。进一步，逻辑强度的增长在某种意义上是“尽可能快”的，体现在真正由序列中的Π²⁰-sentences所断言的总递归函数的总量在所有总递归函数的集合中是cofinal growth-rate-wise。我们然后发展了一个论证，利用由对角线引理应用所构造的句子序列，这些句子在广义上是Hugh Woodin的“塔罗”构造的广义扩展，如其在《数学中的真理》一书中第18章所概述。该论证证明了在PA中可以证明P≠NP。我们指示如何将该论证全部降至SEFA。

英文摘要：

We show that it is provable in PA that there is an arithmetically definable sequence $\{ϕ_{n}:n \in ω\}$ of $Π^{0}_{2}$-sentences, such that - PRA+$\{ϕ_{n}:n \in ω\}$ is $Π^{0}_{2}$-sound and $Π^{0}_{1}$-complete - the length of $ϕ_{n}$ is bounded above by a polynomial function of $n$ with positive leading coefficient - PRA+$ϕ_{n+1}$ always proves 1-consistency of PRA+$ϕ_{n}$. 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 $Π^{0}_{2}$-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.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。