Modus Ponendo Tollens/Variant/Formulation 1/Proof

Theorem

 * $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$

Proof
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

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