Modus Ponendo Tollens/Variant/Formulation 1/Proof
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Theorem
- $\neg \paren {p \land q} \dashv \vdash p \implies \neg q$
Proof
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline \neg & (p & \land & q) & p & \implies & \neg & q \\ \hline T & F & F & F & F & T & T & F \\ T & F & F & T & F & T & F & T \\ T & T & F & F & T & T & T & F \\ F & T & T & T & T & F & F & T \\ \hline \end{array}$
$\blacksquare$