Principle of Dilemma/Formulation 1/Proof

Theorem

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

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.

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