Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication

Theorem

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

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

$\blacksquare$