Reductio ad Absurdum/Variant 2/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg p \implies \left({q \land \neg q}\right)$	Premise	(None)
2		2	$\neg p$	Assumption	(None)
3		1, 2	$q \land \neg q$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 2
4		1, 2	$q$	Rule of Simplification: $\land \EE_1$	3
5		1, 2	$\neg q$	Rule of Simplification: $\land \EE_2$	3
6		1, 2	$\bot$	Principle of Non-Contradiction: $\neg \EE$	4, 5
7		1	$\neg \neg p$	Proof by Contradiction: $\neg \II$	2 – 6	Assumption 2 has been discharged
8		1	$p$	Double Negation Elimination: $\neg \neg \EE$	7

$\blacksquare$

Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

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.