# Proof by Cases/Formulation 2/Forward Implication/Proof by Truth Table

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## Theorem

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

## 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.

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

$\blacksquare$