# Rule of Transposition/Formulation 1/Reverse Implication/Proof

## Theorem

- $\neg q \implies \neg p \vdash p \implies q$

## Proof

By the tableau method of natural deduction:

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

1 | 1 | $\neg q \implies \neg p$ | Premise | (None) | ||

2 | 2 | $p$ | Assumption | (None) | ||

3 | 2 | $\neg \neg p$ | Double Negation Introduction: $\neg \neg \mathcal I$ | 2 | ||

4 | 1, 2 | $\neg \neg q$ | Modus Tollendo Tollens (MTT) | 1, 3 | ||

5 | 1, 2 | $q$ | Double Negation Elimination: $\neg \neg \mathcal E$ | 4 | ||

6 | 1 | $p \implies q$ | Rule of Implication: $\implies \mathcal I$ | 2 – 5 | Assumption 2 has been discharged |

$\blacksquare$

## Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Double Negation Elimination.

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.

## Sources

- 1996: H. Jerome Keisler and Joel Robbin:
*Mathematical Logic and Computability*: $\S 1.13$ - 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*: $\S 1.2.1$: Exercise $1.5: \ 1 \ \text{(a)}$