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
Proceed by the Tableau method.

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

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

Proof by Truth Table
We apply the Method of Truth Tables to the propositions in turn.

As can be seen by inspection, in the first two cases, the truth values under the main connectives match for all models.

$\begin{array}{|ccc||cc|} \hline p & \uparrow & p & \neg & p \\ \hline F & T & F & T & F \\ T & F & T & F & T \\ \hline \end{array}$

Proof of commutativity:

$\begin{array}{|ccc||ccc|} \hline p & \uparrow & q & q & \uparrow & p \\ \hline F & T & F & F & T & F \\ F & T & T & T & T & F \\ T & T & F & F & T & T \\ T & F & T & T & F & T \\ \hline \end{array}$

Proof of non-associativity:

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

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