Rule of Material Equivalence/Formulation 1/Proof 2

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Theorem

$p \iff q \dashv \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$


Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

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

$\blacksquare$


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