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 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.