Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction/Proof by Truth Table

Theorem

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

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.

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