# Rule of Transposition/Variant 1/Formulation 2

## Theorem

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

### Forward Implication

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

### Reverse Implication

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

## Proof

### Proof of Forward Implication

By the tableau method of natural deduction:

$\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p}$
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:

$\vdash \paren {q \implies \neg p} \implies \paren {p \implies \neg q}$
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:

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

$\blacksquare$