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

Theorem

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

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.

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

$\blacksquare$

Sources

• 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.4.1$: The meaning of logical connectives: Exercise $1.8: \ 2$