On the complexity of finding falsifying assignments for Herbrand disjunctions

Abstract

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