Rule of Transposition/Formulation 2/Proof by Truth Table

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Theorem

$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$


Proof

We apply the Method of Truth Tables to the proposition.

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

$\begin{array}{|ccc|c|ccccc|} \hline p & \implies & q) & \iff & (\neg & q & \implies & \neg & p) \\ \hline \F & \T & \F & \T & \T & \F & \T & \T & \F \\ \F & \T & \T & \T & \F & \T & \T & \T & \F \\ \T & \F & \F & \T & \T & \F & \F & \F & \T \\ \T & \T & \T & \T & \F & \T & \T & \F & \T \\ \hline \end{array}$

$\blacksquare$


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