Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication

Theorem

$\neg p \iff \neg q \vdash p \iff q$

Proof

By the tableau method of natural deduction:

$\neg p \iff \neg q \vdash p \iff q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg p \iff \neg q$	Premise	(None)
2		1	$\neg p \implies \neg q$	Biconditional Elimination: $\iff \EE_1$	1
3		1	$\neg \neg q \implies \neg \neg p$	Sequent Introduction	2	Rule of Transposition
4		1	$q \implies p$	Double Negation Elimination: $\neg \neg \EE$	3	(twice)
5		1	$\neg q \implies \neg p$	Biconditional Elimination: $\iff \EE_2$	1
6		1	$\neg \neg p \implies \neg \neg q$	Sequent Introduction	5	Rule of Transposition
7		1	$p \implies q$	Double Negation Elimination: $\neg \neg \EE$	3	(twice)
8		1	$p \iff q$	Biconditional Introduction: $\iff \II$	7, 4

$\blacksquare$

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 $1$: $4$ The Biconditional: Exercise $1 \ \text{(d)}$