Rule of Exportation/Formulation 1/Proof by Truth Table

Theorem

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

Proof

We apply the Method of Truth Tables to the proposition.

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

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

$\blacksquare$