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

Theorem

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

Proof

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

$\blacksquare$