NOR is not Associative

Theorem
Let $\downarrow$ signify the NOR operation.

Then there exist propositions $p,q,r$ such that:
 * $p \downarrow \left({q \downarrow r}\right) \not \vdash \left({p \downarrow q}\right) \downarrow r$

That is, NOR is not associative.

Proof

 * align="right" | 6 ||
 * align="right" | 1
 * $\neg \left({q \downarrow r}\right)$
 * Definition of Logical NOR
 * 5
 * 5


 * align="right" | 9 ||
 * align="right" | 1
 * $p \downarrow \left({q \downarrow r}\right)$
 * Definition of Logical NOR
 * 8
 * 8


 * align="right" | 12 ||
 * align="right" | 1
 * $\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)$
 * Definition of Logical NOR
 * 11
 * 11

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

Hence the result.

Proof by Truth Table
Apply the Method of Truth Tables:


 * $\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.