Non-Equivalence as Disjunction of Negated Conditionals/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

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	$\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$	Sequent Introduction	2	De Morgan's Laws: Disjunction of Negations

$\Box$

By the tableau method of natural deduction:

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

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

$\blacksquare$