# Conjunction with Negative Equivalent to Negation of Implication/Formulation 2/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$
Line Pool Formula Rule Depends upon Notes
1 1 $p \land \neg q$ Assumption (None)
2 1 $\neg \paren {p \implies q}$ Sequent Introduction 1 Conjunction with Negative Equivalent to Negation of Implication: Formulation 1
3 $\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $\neg \paren {p \implies q}$ Assumption (None)
5 4 $p \land \neg q$ Sequent Introduction 4 Conjunction with Negative Equivalent to Negation of Implication: Formulation 1
6 $\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$