# Rule of Transposition/Formulation 2/Proof 1

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## Theorem

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

## Proof

### Proof of Forward Implication

By the tableau method of natural deduction:

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

1 | 1 | $p \implies q$ | Assumption | (None) | ||

2 | 2 | $\neg q$ | Assumption | (None) | ||

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

4 | 1 | $\neg q \implies \neg p$ | Rule of Implication: $\implies \II$ | 2 – 3 | Assumption 2 has been discharged | |

5 | $\paren {p \implies q} \implies \paren {\neg q \implies \neg p}$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged |

$\blacksquare$

### Proof of Reverse Implication

By the tableau method of natural deduction:

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

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

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

3 | 2 | $\neg \neg p$ | Double Negation Introduction: $\neg \neg \II$ | 2 | ||

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

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

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

7 | $\left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)$ | Rule of Implication: $\implies \II$ | 1 – 6 | Assumption 1 has been discharged |

$\blacksquare$

#### Law of the Excluded Middle

This proof 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 proof from an intuitionistic perspective.