De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Proof by Truth Table

From ProofWiki
Jump to navigation Jump to search

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