Non-Equivalence as Disjunction of Negated Implications/Proof 2

Theorem

 * $\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \lor \neg \left({q \implies p}\right)$

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.

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