De Morgan's Laws (Logic)/Disjunction/Formulation 2/Proof by Truth Table

Theorem

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

Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.

$\begin{array}{|ccc|c|cccccc|} \hline (p & \lor & q) & \iff & (\neg & (\neg & p & \land & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \T & \F & \T & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \T & \F & \T & \T & \F & \T & \F & \T & \F \\ \T & \T & \T & \T & \T & \F & \T & \F & \F & \T \\ \hline \end{array}$

$\blacksquare$

Sources

• 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 8$