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.

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