Rule of Material Equivalence/Formulation 2/Proof by Truth Table

Theorem

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

Proof
We apply the Method of Truth Tables.

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

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