Case disjunction according to the value of \( \varphi \). (Contributed by BJ,
6-Apr-2019.) (Proof shortened by Wolf Lammen, 28-Feb-2022.)
\( \vdash \varphi \wedge \psi \vee \neg \varphi \wedge \chi \leftrightarrow (\varphi \rightarrow \psi) \wedge (\neg \varphi \rightarrow \chi) \)
\( \vdash (\varphi \rightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi) \wedge (\neg \varphi \rightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi) \leftrightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi \)
\( \vdash (\varphi \rightarrow \psi) \wedge (\neg \varphi \rightarrow \chi) \leftrightarrow (\varphi \rightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi) \wedge (\neg \varphi \rightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi) \)
\( \vdash \varphi \wedge \psi \vee \neg \varphi \wedge \chi \leftrightarrow \psi \wedge \varphi \vee \chi \wedge \neg \varphi \)