Rule of Exportation/Reverse Implication/Formulation 2/Proof by Truth Table

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Theorem

$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$


Proof

We apply the Method of Truth Tables to the proposition.

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

$\begin{array}{|ccccc|c|ccccc|} \hline (p & \implies & (q & \implies & r)) & \implies & ((p & \land & q) & \implies & r) \\ \hline \F & \T & \F & \T & \F & \T & \F & \F & \F & \T & \F \\ \F & \T & \F & \T & \T & \T & \F & \F & \F & \T & \T \\ \F & \T & \T & \F & \F & \T & \F & \F & \T & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \T \\ \T & \T & \F & \T & \F & \T & \T & \F & \F & \T & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \F & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$


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