Peirce's Law/Formulation 2/Proof by Truth Table

Theorem

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

Proof
We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective are $T$ for all boolean interpretations.

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