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

Jump to navigation
Jump to search

## Theorem

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

## Proof

By the tableau method of natural deduction:

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

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

2 | 1 | $p \land \neg q$ | Sequent Introduction | 1 | Conjunction with Negative Equivalent to Negation of Implication: Formulation 1 | |

3 | $\left({\neg \left({p \implies q}\right)}\right) \implies \left({p \land \neg q}\right)$ | Rule of Implication: $\implies \mathcal I$ | 1 – 2 | Assumption 1 has been discharged |

$\blacksquare$

## Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Conjunction with Negative Equivalent to Negation of Implication/Formulation 1/Reverse Implication.

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.