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

Theorem

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

Proof
We apply the Method of Truth Tables.

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

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