Modus Ponendo Ponens/Variant 2/Proof 2

Theorem

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

Proof
We apply the Method of Truth Tables.

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

As can be seen by inspection, the main connective is true throughout.