# Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Forward Implication

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## Theorem

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

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \land q$ | Premise | (None) | ||

2 | 2 | $p \implies \neg q$ | Assumption | (None) | Assume the opposite of what is to be proved ... | |

3 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||

4 | 1 | $q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||

5 | 1, 2 | $\neg q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||

6 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 4, 5 | ... and demonstrate a contradiction | |

7 | 1 | $\neg \paren {p \implies \neg q}$ | Proof by Contradiction: $\neg \II$ | 2 – 6 | Assumption 2 has been discharged |

$\blacksquare$