Average-Case Hardness of Proving Tautologies and Theorems

Hunter Monroe
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引用次数: 0

Abstract

We consolidate two widely believed conjectures about tautologies -- no optimal proof system exists, and most require superpolynomial size proofs in any system -- into a $p$-isomorphism-invariant condition satisfied by all paddable $\textbf{coNP}$-complete languages or none. The condition is: for any Turing machine (TM) $M$ accepting the language, $\textbf{P}$-uniform input families requiring superpolynomial time by $M$ exist (equivalent to the first conjecture) and appear with positive upper density in an enumeration of input families (implies the second). In that case, no such language is easy on average (in $\textbf{AvgP}$) for a distribution applying non-negligible weight to the hard families. The hardness of proving tautologies and theorems is likely related. Motivated by the fact that arithmetic sentences encoding"string $x$ is Kolmogorov random"are true but unprovable with positive density in a finitely axiomatized theory $\mathcal{T}$ (Calude and J{\"u}rgensen), we conjecture that any propositional proof system requires superpolynomial size proofs for a dense set of $\textbf{P}$-uniform families of tautologies encoding"there is no $\mathcal{T}$ proof of size $\leq t$ showing that string $x$ is Kolmogorov random". This implies the above condition. The conjecture suggests that there is no optimal proof system because undecidable theories help prove tautologies and do so more efficiently as axioms are added, and that constructing hard tautologies seems difficult because it is impossible to construct Kolmogorov random strings. Similar conjectures that computational blind spots are manifestations of noncomputability would resolve other open problems.
证明重言式和定理的平均情形硬度
我们将关于重言式的两个被广泛相信的猜想——不存在最优证明系统,并且大多数在任何系统中都需要超多项式大小的证明——整合到a $p$-同构-所有可填条件满足的不变条件 $\textbf{coNP}$-完整的语言或没有。条件为:对于任意图灵机(TM) $M$ 接受这种语言, $\textbf{P}$-需要超多项式时间的均匀输入族 $M$ 存在(相当于第一个猜想)并以正的上密度出现在输入族的枚举中(暗示第二个猜想)。在这种情况下,一般来说,没有这样的语言是容易的 $\textbf{AvgP}$),这是一个对硬科施加不可忽略的权重的分布。证明重言式和定理的难度可能与此有关。由于算术句子编码“字符串”这一事实 $x$ 在有限公理化理论中,柯尔莫哥洛夫是随机的吗 $\mathcal{T}$ (Calude和J{ü}Rgensen),我们推测任何命题证明系统都需要超多项式大小的证明 $\textbf{P}$-编码“there is no”的同族重言式 $\mathcal{T}$ 尺寸证明 $\leq t$ 显示字符串 $x$ 柯尔莫哥洛夫是随机的吗?”这暗示了上述条件。这一猜想表明,不存在最优证明系统,因为不可判定理论有助于证明重言式,而且随着公理的加入,证明重言式的效率会更高,而且构造硬重言式似乎很困难,因为不可能构造柯尔莫哥罗夫随机字符串。计算盲点是不可计算性表现的类似猜想将解决其他开放问题。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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