# Rule of Transposition/Variant 1/Formulation 2/Proof

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## Theorem

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

## Proof

### Proof of Forward Implication

By the tableau method of natural deduction:

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

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

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

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

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

5 | 1 | $q \implies \neg p$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |

6 | $\paren {p \implies \neg q} \implies \paren {q \implies \neg p}$ | Rule of Implication: $\implies \II$ | 1 – 5 | 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 | $q \implies \neg p$ | Assumption | (None) | ||

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

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

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

5 | 1 | $p \implies \neg q$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |

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

$\blacksquare$

By the tableau method of natural deduction:

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

1 | $\paren {p \implies \neg q} \implies \paren {q \implies \neg p}$ | Theorem Introduction | (None) | Rule of Transposition: Forward Implication | ||

2 | $\paren {q \implies \neg p} \implies \paren {p \implies \neg q}$ | Theorem Introduction | (None) | Rule of Transposition: Reverse Implication | ||

3 | $\paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | Biconditional Introduction: $\iff \II$ | 1, 2 |

$\blacksquare$