Biconditional Elimination/Sequent Form/Proof by Truth Table

Theorem

\(\text {(1)}: \quad\)

\(\ds p \iff q\)

\(\vdash\)

\(\ds p \implies q\)

\(\text {(2)}: \quad\)

\(\ds p \iff q\)

\(\vdash\)

\(\ds q \implies p\)

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 $q \implies p$.

$\blacksquare$