NAND is Commutative/Proof 1

Theorem

$p \uparrow q \dashv \vdash q \uparrow p$

Proof

By the tableau method of natural deduction:

$p \uparrow q \vdash q \uparrow p$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \uparrow q$	Premise	(None)
2		1	$\neg \paren {p \land q}$	Sequent Introduction	1	Definition of Logical NAND
3		1	$\neg \paren {q \land p}$	Sequent Introduction	2	Conjunction is Commutative
4		1	$q \uparrow p$	Sequent Introduction	3	Definition of Logical NAND

$\Box$

By the tableau method of natural deduction:

$q \uparrow p \vdash p \uparrow q$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$q \uparrow p$	Premise	(None)
2		1	$\neg \paren {q \land p}$	Sequent Introduction	1	Definition of Logical NAND
3		1	$\neg \paren {p \land q}$	Sequent Introduction	2	Conjunction is Commutative
4		1	$p \uparrow q$	Sequent Introduction	3	Definition of Logical NAND

$\blacksquare$