Modus Ponendo Ponens/Variant 1/Proof 2

Theorem

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

Proof
We apply the Method of Truth Tables.

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

As can be seen by inspection, when $p$ is true, the value of the main connective is also true.