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

From ProofWiki
Jump to navigation Jump to search

Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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


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.

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

$\blacksquare$


Sources