Non-Equivalence as Disjunction of Negated Conditionals/Proof by Truth Table

Theorem

$\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|cccc||ccccccccc|} \hline \neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\ \hline \F & \F & \T & \F & \F & \F & \T & \F & \F & \F & \F & \T & \F \\ \T & \F & \F & \T & \F & \F & \T & \T & \T & \T & \T & \F & \F \\ \T & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \T & \T \\ \F & \T & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$