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

Theorem

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

Proof
We apply the Method of Truth Tables.

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

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