Rule of Transposition/Variant 2/Formulation 1

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Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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


Proof 1

Proof of Forward Implication

By the tableau method of natural deduction:

$\neg p \implies q \vdash \neg q \implies p$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg p \implies q$ Premise (None)
2 2 $\neg q$ Assumption (None)
3 1, 2 $\neg \neg p$ Modus Tollendo Tollens (MTT) 1, 2
4 1, 2 $p$ Double Negation Elimination: $\neg \neg \mathcal E$ 3
5 1 $\neg q \implies p$ Rule of Implication: $\implies \mathcal I$ 2 – 4 Assumption 2 has been discharged

$\blacksquare$

Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle, by way of Double Negation Elimination.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.


Proof of Reverse Implication

By the tableau method of natural deduction:

$\neg q \implies p \vdash \neg p \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg q \implies p$ Premise (None)
2 2 $\neg p$ Assumption (None)
3 1, 2 $\neg \neg q$ Modus Tollendo Tollens (MTT) 1, 2
4 1, 2 $q$ Double Negation Elimination: $\neg \neg \mathcal E$ 3
5 1 $\neg p \implies q$ Rule of Implication: $\implies \mathcal I$ 2 – 4 Assumption 2 has been discharged

$\blacksquare$

Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle, by way of Double Negation Elimination.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.



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.


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

$\blacksquare$