Properties of NAND

Theorem
Let $$\uparrow$$ signify the NAND operation.

The following results hold.

NAND with itself is the Not operation:


 * $$p \uparrow p \dashv \vdash \neg p$$

NAND is commutative:


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

NAND is not associative:


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

Proof by Natural deduction
Commutativity is proved by the Tableau method:

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

Proof by Truth Table
Let $$v: \left\{{p}\right\} \to \left\{{T, F}\right\}$$ be an interpretation for a boolean variable $$p$$.

As can be seen, $$v \left({p \uparrow \left({q \uparrow r}\right)}\right) \ne v \left({\left({p \uparrow q}\right) \uparrow r}\right)$$ except in four combinations of values, that is, when $$v \left({p}\right) = v \left({r}\right)$$.

So $$p \uparrow \left({q \uparrow r}\right) \not \vdash \left({p \uparrow q}\right) \uparrow r$$ nor the other way about.