NAND is Commutative

Theorem

Let $\uparrow$ signify the NAND operation.

Then, for any two propositions $p$ and $q$:

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

That is, NAND is commutative.

Proof 1

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

$\blacksquare$

Proof by Truth Table

We apply the Method of Truth Tables:

$\begin{array}{|ccc||ccc|} \hline

p & \uparrow & q & q & \uparrow & p \\ \hline \F & \T & \F & \F & \T & \F \\ \F & \T & \T & \T & \T & \F \\ \T & \T & \F & \F & \T & \T \\ \T & \F & \T & \T & \F & \T \\ \hline \end{array}$

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\blacksquare$

Also see

• NOR is Commutative

Sources

• 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$