De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Proof by Truth Table
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Theorem
- $\neg p \land \neg q \dashv \vdash \neg \paren {p \lor q}$
Proof
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.
$\begin{array}{|ccccc||cccc|} \hline \neg & p & \land & \neg & q & \neg & (p & \lor & q) \\ \hline \T & \F & \T & \T & \F & \T & \F & \F & \F \\ \T & \F & \F & \F & \T & \F & \F & \T & \T \\ \F & \T & \F & \T & \F & \F & \T & \T & \F \\ \F & \T & \F & \F & \T & \F & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1988: Alan G. Hamilton: Logic for Mathematicians (2nd ed.) ... (previous) ... (next): $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Example $1.8 \ \text{(c)}$