Properties of NOR

Theorem
Let $\downarrow$ signify the NOR operation.

The following results hold.

NOR with itself is the Not operation:


 * $p \downarrow p \dashv \vdash \neg p$

NOR is commutative:


 * $p \downarrow q \dashv \vdash q \downarrow p$

NOR is not associative:


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

Proof by Tableau
The three assertions can be proved by the tableau method:

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

Non-associativity
Taking $p = \bot$ and $r = \top$, we have $\vdash \neg p \land r$, discharging the last assumption.

Hence the result.

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.

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

Proof of commutativity:

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

Proof of non-associativity:

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

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