Conjunction Equivalent to Negation of Implication of Negative/Formulation 1/Reverse Implication

Theorem

$\neg \left({p \implies \neg q}\right) \vdash p \land q$

Proofs

By the tableau method of natural deduction:

$\neg \left({p \implies \neg q}\right) \vdash p \land q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \left({p \implies \neg q}\right)$	Premise	(None)
2		2	$\neg \left({p \land q}\right)$	Assumption	(None)	Assume the negation of what is to be proved ...
3		2	$p \implies \neg q$	Sequent Introduction	2	Modus Ponendo Tollens
4		1, 2	$\bot$	Principle of Non-Contradiction: $\neg \EE$	3, 1	... and demonstrate a contradiction
5		1	$p \land q$	Reductio ad Absurdum	2 – 4	Assumption 2 has been discharged

$\blacksquare$

Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Reductio ad Absurdum.

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 theorem from an intuitionistic perspective.