Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Proof

Theorem

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

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.

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