Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Proof 1

Theorem

$\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$

Proof

Forward Implication: Proof

By the tableau method of natural deduction:

$\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} $

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \paren {p \iff q}$	Premise	(None)
2		1	$\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$	Sequent Introduction	1	Rule of Material Equivalence
3		1	$\neg \paren {\paren {\neg p \lor q} \land \paren {\neg q \lor p} }$	Sequent Introduction	2	Rule of Material Implication (twice)
4		1	$\neg \paren {\neg p \lor q} \lor \neg \paren {\neg q \lor p}$	Sequent Introduction	3	De Morgan's Laws: Disjunction of Negations
5		1	$\paren {\neg \neg p \land \neg q} \lor \paren {\neg \neg q \land \neg p}$	Sequent Introduction	4	De Morgan's Laws: Conjunction of Negations
6		1	$\paren {p \land \neg q} \lor \paren {q \land \neg p}$	Double Negation Elimination: $\neg \neg \EE$	5
7		1	$\paren {p \land \neg q} \lor \paren {\neg p \land q}$	Sequent Introduction	6	Conjunction is Commutative
8		1	$\paren {\neg p \land q} \lor \paren {p \land \neg q}$	Sequent Introduction	7	Disjunction is Commutative

$\blacksquare$

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.

Reverse Implication: Proof

By the tableau method of natural deduction:

$\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} $

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {\neg p \land q} \lor \paren {p \land \neg q}$	Premise	(None)
2		1	$\paren {p \land \neg q} \lor \paren {\neg p \land q}$	Sequent Introduction	1	Disjunction is Commutative
3		1	$\paren {p \land \neg q} \lor \paren {q \land \neg p}$	Sequent Introduction	2	Conjunction is Commutative
4		1	$\paren {\neg \neg p \land \neg q} \lor \paren {\neg \neg q \land \neg p}$	Double Negation Introduction: $\neg \neg \II$	3
5		1	$\neg \paren {\neg p \lor q} \lor \neg \paren {\neg q \lor p}$	Sequent Introduction	4	De Morgan's Laws: Conjunction of Negations
6		1	$\neg \paren {\paren {\neg p \lor q} \land \paren {\neg q \lor p} }$	Sequent Introduction	5	De Morgan's Laws: Disjunction of Negations
7		1	$\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$	Sequent Introduction	6	Rule of Material Implication (twice)
8		1	$\neg \paren {p \iff q}$	Sequent Introduction	7	Rule of Material Equivalence

$\blacksquare$