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

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## Theorem

- $\neg \left({p \implies q}\right) \vdash p \land \neg q$

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $\neg \left({p \implies q}\right)$ | Premise | (None) | ||

2 | 2 | $\neg \left({p \land \neg q}\right)$ | Assumption | (None) | ||

3 | 2 | $p \implies q$ | Sequent Introduction | 2 | Implication Equivalent to Negation of Conjunction with Negative | |

4 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \mathcal E$ | 3, 1 | ||

5 | 1 | $p \land \neg q$ | Reductio ad Absurdum | 2 – 4 | Assumption 2 has been discharged |

$\blacksquare$

## Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Reductio ad Absurdum.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this theorem from an intuitionistic perspective.