Disjunction of Conditional and Converse/Proof by Truth Table

Theorem

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

Proof

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations, proving a tautology.

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

$\blacksquare$