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.
翻译:我们整合了两种广泛相信的关于调制语言的推测 -- -- 没有最佳证明系统, 多数需要任何系统中的超球体大小证明 -- -- 以所有可加压的$\textbf{coNP} $- 完整的语言满足的价格或无。 条件是: 对于任何接受该语言的图灵机器(TM) $M$, 需要超球体时间的美元=( 相当于第一个测算) 。 大多数都需要任何系统中的超球体大小证明 -- -- 在任何系统中, 超球体大小证明 -- -- 超球体积值证明( 缩数) -- -- 在输入组的查点中( 缩数) 以美元表示超球体大小。 在这样的情况下,这种语言在平均情况下( 美元\ textbf{AvgP} 完全满足了对硬体重量的分布。 证明 tautbormologies的难度很可能是相关的。 令你难以理解的是,因为算句子的编码“ $xormorovs is so non.