Biconditional as Disjunction of Conjunctions/Formulation 1/Proof

Theorem

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

Proof
We apply the Method of Truth Tables.

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

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