Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \paren {p \land \neg q}$	Premise	(None)
2		1	$\neg p \lor \neg \neg q$	Sequent Introduction	1	De Morgan's Laws: Disjunction of Negations
3		1	$p \implies \neg \neg q$	Sequent Introduction	2	Rule of Material Implication
4		4	$p$	Assumption	(None)
5		1, 4	$\neg \neg q$	Modus Ponendo Ponens: $\implies \mathcal E$	3, 4
6		1, 4	$q$	Double Negation Elimination: $\neg \neg \EE$	5
7		1	$p \implies q$	Rule of Implication: $\implies \II$	4 – 6	Assumption 4 has been discharged

$\blacksquare$

Law of the Excluded Middle

This theorem 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 theorem from an intuitionistic perspective.

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorems $35 \ \text{(b)}$