Rule of Transposition/Variant 1/Formulation 1

Theorem

$p \implies \neg q \dashv \vdash q \implies \neg p$

This can be expressed as two separate theorems:

Forward Implication

$p \implies \neg q \vdash q \implies \neg p$

Reverse Implication

$q \implies \neg p \vdash \neg p \implies q$

Proof 1

Proof of Forward Implication

By the tableau method of natural deduction:

$p \implies \neg q \vdash q \implies \neg p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \implies \neg q$	Premise	(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

$\blacksquare$

Proof of Reverse Implication

By the tableau method of natural deduction:

$q \implies \neg p \vdash p \implies \neg q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$q \implies \neg p$	Premise	(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

$\blacksquare$

Proof 2

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|cccc||cccc|} \hline p & \implies & \neg & q & q & \implies & \neg & p \\ \hline F & T & T & F & F & T & T & F \\ F & T & F & T & T & T & T & F \\ T & F & T & F & F & F & F & T \\ T & T & F & T & T & T & F & T \\ \hline \end{array}$

$\blacksquare$