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On the complexity of finding falsifying assignments for Herbrand disjunctions

Published 12 Nov 2014 in cs.LO and math.LO | (1411.3304v1)

Abstract: Suppose that $\Phi$ is a consistent sentence. Then there is no Herbrand proof of $\neg \Phi$, which means that any Herbrand disjunction made from the prenex form of $\neg \Phi$ is falsifiable. We show that the problem of finding such a falsifying assignment is hard in the following sense. For every total polynomial search problem $R$, there exists a consistent $\Phi$ such that finding solutions to $R$ can be reduced to finding a falsifying assignment to an Herbrand disjunction made from $\neg \Phi$. It has been conjectured that there are no complete total polynomial search problems. If this conjecture is true, then for every consistent sentence $\Phi$, there exists a consistence sentence $\Psi$, such that the search problem associated with $\Psi$ cannot be reduced to the search problem associated with $\Phi$.

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