Biconditional Elimination/Sequent Form/Proof 2

Theorem

 * $(1): \quad p \iff q \vdash p \implies q$
 * $(2): \quad p \iff q \vdash q \implies p$

Proof
We apply the Method of Truth Tables.

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

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