Rule of Material Implication/Formulation 2/Proof by Truth Table

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Theorem

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


Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.

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

$\blacksquare$


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