Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1

Theorem

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

This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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

Proof by Truth Table

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.

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

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Exercise $5 \ \text{(g)}$