Biconditional Elimination/Sequent Form

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 1

Form 1

By the tableau method of natural deduction:

$p \iff q \vdash p \implies q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	Premise	(None)
2		1	$p \implies q$	Biconditional Elimination: $\iff \EE_1$	1

$\blacksquare$

Form 2

By the tableau method of natural deduction:

$p \iff q \vdash q \implies p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \iff q$	Premise	(None)
2		1	$q \implies p$	Biconditional Elimination: $\iff \EE_2$	1

$\blacksquare$

Proof by Truth Table

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 $q \implies p$.

$\blacksquare$

Sources

• 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.5$: An aside: proof by contradiction: Exercises $1.6: \ 7$