Rule of Transposition/Variant 1/Formulation 1

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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: $\neg \neg \mathcal I$ 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

$\blacksquare$


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: $\neg \neg \mathcal I$ 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

$\blacksquare$


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.


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

$\blacksquare$