Modus Ponendo Ponens/Variant 3/Proof by Truth Table

Theorem

 * $\vdash \left({\left({p \implies q}\right) \land p}\right) \implies q$

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.

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