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

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## Theorem

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

## Proof

By the tableau method of natural deduction:

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 |

$\blacksquare$