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.
$\blacksquare$
Sources
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables