Non-Equivalence as Equivalence with Negation/Formulation 1/Proof by Truth Table

Theorem

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

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.

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