Peirce's Law/Strong Form/Formulation 1

Theorem

 * $\paren {\paren {p \implies q} \implies p} \dashv \vdash p$

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.

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