Rule of Material Implication/Formulation 2/Proof 2

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Theorem

$\vdash \left({p \implies q}\right) \iff \left({\neg p \lor q}\right)$


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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