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 by Tableau

Proceed by the Tableau method:

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

Line		Pool	Formula	Rule	Depends upon
1		1	$p \uparrow q$	$\mathrm P$	(None)
2		1	$\neg \left({p \land q}\right)$	By definition	1
3		1	$\neg \left({q \land p}\right)$	$\mathrm {Comm}$	2
4		1	$q \uparrow p$	By definition	3

$q \uparrow p \vdash p \uparrow q$ is proved similarly.

$\blacksquare$

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 models.

$\blacksquare$