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

## Theorem

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

## Proof

By the tableau method of natural deduction:

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

$\blacksquare$