NAND is Commutative/Proof by Truth Table

Theorem

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

Proof

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.

$\blacksquare$