NAND is not Associative/Proof 1

Theorem
Let $\uparrow$ signify the NAND operation.

Then there exist propositions $p,q,r$ such that:


 * $p \uparrow \left({q \uparrow r}\right) \not \vdash \left({p \uparrow q}\right) \uparrow r$

That is, NAND is not associative.

Proof
Taking $p = \top$, $r = \bot$, we find $\vdash p \land \neg r$, and conclude our initial assumption was false.