Theorem 103: Biconditional elimination (inference rule)
(a) Given that the equivalence \phi \Leftrightarrow \psi holds, we may infer \phi entails \psi : \cfrac{\vdash (\phi \Leftrightarrow \psi)}{\phi \vdash \psi}. (b) Given that the equivalence \phi \Leftrightarrow \psi holds, we may infer \psi entails \phi : \cfrac{\vdash (\phi \Leftrightarrow \psi)}{\psi \vdash \phi}.
Proof: (a): The equivalence \phi \Leftrightarrow \psi is shorthand for the conjunction \big( (\phi \Rightarrow \psi) \wedge (\psi \Rightarrow \phi) \big) of implications. By conjunction elimination, we infer \phi \Rightarrow \psi . Using [99], we may thus infer \phi entails \psi . \square
(b): This follows from commutativity of conjunction together with (a). \square