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On the Automatizability of Resolution and Related Propositional Proof Systems

机译:解析和相关命题证明系统的自动化

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Having good algorithms to verify tautologies as efficiently as possible is of prime interest in different fields of computer science. In this paper we present an algorithm for finding Resolution refutations based on finding tree-like Res(k) refutations. The algorithm is based on the one of Beame and Pitassi cite{BP96} for tree-like Resolution, but it is provably more efficient. On the other hand our algorithm is also more efficient than Davis-Putnam and better in the sense of space usage than the one of Ben-Sasson and Wigderson cite{BSW01}. We also analyse the possibility that a system that simulates Resolution is automatizable. We call this notion "weak automatizability". We prove that Resolution is weakly automatizable if and only if Res(2) has feasible interpolation. In order to prove this theorem, we show that Res(2) has polynomial-size proofs of the reflection principle of Resolution (and of any Res(k)), which is a version of consistency. We also show that Resolution proofs of its own reflection principle require slightly subexponential size. This gives a slightly subexponential lower bound for the monotone interpolation of Res(2) and a better separation from Resolution as a byproduct. Finally, the techniques for proving these results give us a way to obtain a large class of examples that have small Resolution refutations but require relatively large width. This answers a question of Alekhnovich and Razborov cite{AR01b} related to whether Resolution is automatizable in quasipolynomial-time.
机译:在计算机科学的不同领域中,拥有最好的算法来尽可能有效地验证重言式是最重要的。在本文中,我们提出了一种基于查找树状Res(k)引用的找到分辨率引用的算法。该算法基于Beame和Pitassi cite {BP96}中的一种,可实现树状分辨率,但可证明效率更高。另一方面,我们的算法也比Davis-Putnam更有效,并且在空间使用方面也比Ben-Sasson和Wigderson cite {BSW01}更好。我们还分析了模拟分辨率的系统可自动化的可能性。我们称这个概念为“弱自动化”。我们证明,只有当Res(2)具有可行的插值时,Resolution才能实现弱自动化。为了证明该定理,我们证明Res(2)具有分辨率(以及任何Res(k))的反射原理的多项式大小证明,这是一致性的一种形式。我们还表明,其自身反射原理的分辨率证明需要略微低于指数的大小。这为Res(2)的单调插值给出了一个略微低于指数的下限,并与作为副产品的分辨率更好地分离了。最后,用于证明这些结果的技术为我们提供了一种方法,该方法可以获取具有少量“分辨率”反驳但需要相对较大宽度的大量示例。这回答了Alekhnovich和Razborov cite {AR01b}的问题,该问题与分辨率是否可以在准多项式时间内自动进行。

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