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.


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

$\blacksquare$