Conjunction Equivalent to Negation of Implication of Negative/Formulation 2/Proof by Truth Table

Theorem

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

Proof
We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective are true for all models.

We apply the Method of Truth Tables.

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

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