# Conjunction Equivalent to Negation of Implication of Negative/Formulation 2

## Theorem

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

## Proof 1

By the tableau method of natural deduction:

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

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

$\blacksquare$