De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Proof by Truth Table
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Theorem
- $\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \land & \neg & q & \iff & \neg & (p & \lor & q) \\ \hline \T & \F & \T & \T & \F & \T & \T & \F & \F & \F \\ \T & \F & \F & \F & \T & \T & \F & \F & \T & \T \\ \F & \T & \F & \T & \F & \T & \F & \T & \T & \F \\ \F & \T & \F & \F & \T & \T & \F & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}$: Exercise $14 \ \text{(b)}$
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms