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

Theorem

$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$

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 boolean interpretations.

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

$\blacksquare$

Sources

• 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 7$: Determining the Truth Values of Complex Propositions