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

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


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