On the complexity of finding falsifying assignments for Herbrand   disjunctions
  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 $\Phi$. 
Pavel Pudlak
cs.LO math.LO