Two ways to express exclusive disjunction ( [df-xor] ). Theorem *5.22 of
[WhiteheadRussell] p. 124. (Contributed by NM, 3-Jan-2005.) (Proof
shortened by Wolf Lammen, 22-Jan-2013.)
\( \vdash \neg (\varphi \leftrightarrow \psi) \leftrightarrow \varphi \wedge \neg \psi \vee \psi \wedge \neg \varphi \)
\( \vdash (\varphi \rightarrow \psi) \wedge (\psi \rightarrow \varphi) \leftrightarrow \neg (\varphi \wedge \neg \psi) \wedge \neg (\psi \wedge \neg \varphi) \)
\( \vdash \varphi \leftrightarrow \psi \leftrightarrow (\varphi \rightarrow \psi) \wedge (\psi \rightarrow \varphi) \)
\( \vdash \neg (\varphi \wedge \neg \psi \vee \psi \wedge \neg \varphi) \leftrightarrow \neg (\varphi \wedge \neg \psi) \wedge \neg (\psi \wedge \neg \varphi) \)