Biconditional Introduction/Sequent Form/Proof 2
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Theorem
- $p \implies q, q \implies p \vdash p \iff q$
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 $p \iff q$.
$\blacksquare$