Theorem 80: Disjunctive syllogism (inference rules)
(a) Given that the disjunction \phi \vee \psi holds and the negation \neg \phi holds, we may infer that \psi holds: \cfrac{\vdash (\phi \vee \psi) \quad \vdash \neg \phi}{\vdash \psi}. (b) Given that the disjunction \phi \vee \psi holds and the negation \neg \phi holds, we may infer that \psi holds: \cfrac{\vdash (\phi \vee \psi) \quad \vdash \neg \psi}{\vdash \phi}.
Proof: (a): By negation elimination, we infer that \phi entails \psi . By the identity rule, \psi entails \psi . Thus, by disjunction elimination, \psi holds.
(b): By commutativity of disjunction, \psi \vee \phi holds. By (a), \phi holds. \square