Biconditional as Disjunction of Conjunctions/Formulation 2/Proof by Truth Table

Theorem

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

Proof
We apply the Method of Truth Tables.

As can be seen by inspection, in all cases the truth values under the main connective is true for all models.

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