Rule of Transposition/Variant 1/Formulation 2

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Theorem

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


Forward Implication

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

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:

$\vdash \left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$
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: $\neg \neg \mathcal I$ 2
4 1, 2 $\neg p$ Modus Tollendo Tollens (MTT) 1, 3
5 1 $q \implies \neg p$ Rule of Implication: $\implies \mathcal I$ 2 – 4 Assumption 2 has been discharged
6 $\left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 5 Assumption 1 has been discharged

$\blacksquare$


Proof of Reverse Implication

By the tableau method of natural deduction:

$\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$
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: $\neg \neg \mathcal I$ 2
4 1, 2 $\neg q$ Modus Tollendo Tollens (MTT) 1, 3
5 1 $p \implies \neg q$ Rule of Implication: $\implies \mathcal I$ 2 – 4 Assumption 2 has been discharged
6 $\left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 5 Assumption 1 has been discharged

$\blacksquare$


By the tableau method of natural deduction:

$\vdash \left({p \implies \neg q}\right) \iff \left({q \implies \neg p}\right)$
Line Pool Formula Rule Depends upon Notes
1 $\left({p \implies \neg q}\right) \implies \left({q \implies \neg p}\right)$ Theorem Introduction (None) Rule of Transposition: Forward Implication
2 $\left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$ Theorem Introduction (None) Rule of Transposition: Reverse Implication
3 $\left({p \implies \neg q}\right) \iff \left({q \implies \neg p}\right)$ Biconditional Introduction: $\iff \mathcal I$ 1, 2

$\blacksquare$


Sources