Peirce's Law/Strong Form/Formulation 1

Theorem

 * $((p \implies q) \implies p) \dashv \vdash p$

Proof
We apply the Method of Truth Tables.

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

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