# De Morgan's Laws (Logic)/Disjunction/Formulation 2

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

$\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
Line Pool Formula Rule Depends upon Notes
1 1 $p \lor q$ Assumption (None)
2 1 $\neg \paren {\neg p \land \neg q}$ Sequent Introduction 1 De Morgan's Laws (Logic): Disjunction: Formulation 1
3 $\paren {p \lor q} \implies \paren {\neg \paren {\neg p \land \neg q} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $\neg \paren {\neg p \land \neg q}$ Assumption (None)
5 4 $p \lor q$ Sequent Introduction 4 De Morgan's Laws (Logic): Disjunction: Formulation 1
6 $\paren {\neg \paren {\neg p \land \neg q} } \implies \paren {p \lor q}$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$

## Proof by Truth Table

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.

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

$\blacksquare$