Rule of Material Equivalence/Formulation 1/Proof 2

Theorem

 * $p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$

Proof
We apply the Method of Truth Tables.

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

$\begin{array}{|ccc|ccccccc|} \hline p & \iff & q & (p & \implies & q) & \land & (q & \implies & p) \\ \hline \F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\ \F & \F & \T & \F & \T & \T & \F & \T & \F & \F \\ \T & \F & \F & \T & \F & \F & \F & \F & \T & \T \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$