Past exam of the computer science course of the University of Oxford / 2015 / Computational complexity / 2 / d / Solution

Entail is co-NP-complete under polynomial-time many-one reductions. For valid propositional formulas, semantic entailment holds exactly when the logical conjunction of the premise and the conclusion's logical negation is an unsatisfiable Boolean formula:A counterexample to semantic entailment is a Boolean valuation making $\phi$ true and $\psi$ false. Guessing one bit for each distinct Boolean variable and evaluating both propositional formulas is a polynomial-time algorithm. Thus the complement of Entail belongs to NP. Malformed descriptions, also in that complement, can be accepted after a deterministic syntax check. Hence Entail belongs to co-NP as a formal language over all strings.

For hardness, the unsatisfiable Boolean formula problem is co-NP-hard: if $A\in \mathrm{co}\text{-}\mathrm{NP}$, then $\overline{A}\in \mathrm{NP}$, and the permitted NP-hardness of SAT supplies a polynomial-time many-one reduction $r$ withWe may ensure $r(x)$ is always a valid propositional formula, replacing any malformed output by a fixed contradiction. It follows that $x\in A$ exactly when $r(x)$ is unsatisfiable. This justifies using the unsatisfiability problem as a hardness source without an extra assumption.

Map a valid propositional formula $\theta$ towhere $p$ is a fixed Boolean variable. The second propositional formula is false under every Boolean valuation. Consequently it is entailed by $\theta$ precisely when no Boolean valuation satisfies $\theta$. ThusThe output has only constant additional length and is computable in polynomial time. If the source unsatisfiability language rejects malformed strings, map them to the fixed no-instance $⟨p,p\wedge \neg p⟩$. This completes a total polynomial-time many-one reduction and proves the claimed co-NP-completeness.