Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 2

Theorem

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

Proof
We apply the Method of Truth Tables.

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

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