NAND is not Associative/Proof by Truth Table

Theorem

$p \uparrow \paren {q \uparrow r} \not \vdash \paren {p \uparrow q} \uparrow r$

Proof

We apply the Method of Truth Tables:

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

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

As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.

$\blacksquare$