Principle of Dilemma/Formulation 1/Proof
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Theorem
- $\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$
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.
$\begin{array}{|cccccccc||c|} \hline
(p & \implies & q) & \land & (\neg & p & \implies & q) & q \\
\hline
F & T & F & F & T & F & F & F & F \\
F & T & T & T & T & F & T & T & T \\
T & F & F & F & F & T & T & F & F \\
T & T & T & T & F & T & T & T & T \\
\hline
\end{array}$
$\blacksquare$