Modus Tollendo Ponens/Sequent Form/Case 1/Proof by Truth Table

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Theorem

\(\ds p \lor q\) \(\) \(\ds \)
\(\ds \neg p\) \(\) \(\ds \)
\(\ds \vdash \ \ \) \(\ds q\) \(\) \(\ds \)


Proof

We apply the Method of Truth Tables to the proposition.

$\begin{array}{|ccc|cc||c|} \hline p & \lor & q & \neg & p & q\\ \hline \F & \F & \F & \T & \F & \F \\ \F & \T & \T & \T & \F & \T \\ \T & \T & \F & \F & \T & \F \\ \T & \T & \T & \F & \T & \T \\ \hline \end{array}$

As can be seen, when $p \lor q$ is true, and so is $\neg p$, then $q$ is also true.

$\blacksquare$


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