Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \paren {p \iff q}$	Premise	(None)
2		1	$\paren {p \lor q} \land \neg \paren {p \land q}$	Sequent Introduction	1	Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
3		1	$\paren {p \lor q} \land \paren {\neg p \lor \neg q}$	Sequent Introduction	2	De Morgan's Laws: Disjunction of Negations

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \lor q} \land \paren {\neg p \lor \neg q}$	Premise	(None)
2		1	$\paren {p \lor q} \land \neg \paren {p \land q}$	Sequent Introduction	1	De Morgan's Laws: Disjunction of Negations
3		1	$\neg \paren {p \iff q}$	Sequent Introduction	2	Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction

$\blacksquare$