Biconditional Introduction/Sequent Form/Proof 2

Theorem

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

Proof
We apply the Method of Truth Tables.

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

As can be seen, only when both $p \implies q$ and $q \implies p$ are true, then so is $p \iff q$.