De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1

Theorem

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

This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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

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.

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

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text {(g)}$
• 2000: Michael R.A. Huth and Mark D. Ryan: Logic in Computer Science: Modelling and reasoning about systems ... (previous) ... (next): $\S 1.2.4$: Provable equivalence