Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof

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


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