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 $1.6: \ 7$
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