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