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

Theorem

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

This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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

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.

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

$\blacksquare$

Sources

• 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 2.4$: Relations between Truth-Functions
• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text{(e)}$
• 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$