# De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

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

$\blacksquare$