Theorem 82: Law of Excluded Middle (inference rule)
We may (from no premises at all) infer that the disjunction \phi \vee \neg \phi of \phi and its negation \neg \phi holds: \cfrac{}{\vdash (\phi \vee \neg \phi)}.
Proof: Assume \neg (\phi \vee \neg \phi) . By, (b) in [81], the conjunction \neg \phi \wedge \neg \neg \phi holds. Thus, \neg (\phi \vee \neg \phi) entails this conjunction. But \neg (\neg \phi \wedge \neg \neg \phi) holds by the Law of Noncontradiction; using weakening, \neg (\phi \vee \neg \phi) entails this negation. By negation introduction, we can therefore infer that \neg \neg (\phi \vee \neg \phi) holds. We conclude using double negation elimination that \phi \vee \neg \phi holds. \square