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" | 4 ||
 * align="right" | 1
 * $q \lor r$
 * $\lor \mathcal I_2$
 * 3
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \neg \left({q \lor r}\right)$
 * $\neg \neg \mathcal I$
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $\neg \left({q \downarrow r}\right)$
 * By definition
 * 5
 * align="right" | 6 ||
 * align="right" | 1
 * $\neg \left({q \downarrow r}\right)$
 * By definition
 * 5


 * align="right" | 8 ||
 * align="right" | 1
 * $\neg \left({p \lor \left({q \downarrow r}\right) }\right)$
 * $\mathrm {DM}$
 * 7
 * align="right" | 9 ||
 * align="right" | 1
 * $p \downarrow \left({q \downarrow r}\right)$
 * By definition
 * 8
 * align="right" | 10 ||
 * align="right" | 1
 * $\left({p \downarrow q}\right) \lor r$
 * $\lor \mathcal I_2$
 * 3
 * align="right" | 11 ||
 * align="right" | 1
 * $\neg \neg \left({\left({p \downarrow q}\right) \lor r}\right)$
 * $\neg \neg \mathcal I$
 * 10
 * align="right" | 12 ||
 * align="right" | 1
 * $\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)$
 * By definition
 * 11
 * align="right" | 13 ||
 * align="right" | 1
 * $\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)$
 * $\neg \neg \mathcal I$
 * 9
 * align="right" | 1
 * $\neg \left({\left({p \downarrow q}\right) \downarrow r}\right)$
 * By definition
 * 11
 * align="right" | 13 ||
 * align="right" | 1
 * $\neg \neg \left({p \downarrow \left({q \downarrow r}\right)}\right)$
 * $\neg \neg \mathcal I$
 * 9
 * $\neg \neg \mathcal I$
 * 9


 * align="right" | 15 ||
 * align="right" | 1
 * $\neg \left({\neg \left({p \downarrow \left({q \downarrow r}\right)}\right) \lor \left({\left({p \downarrow q}\right) \downarrow r}\right)}\right)$
 * $\mathrm {DM}$
 * 14
 * align="right" | 16 ||
 * align="right" | 1
 * $\neg \left({ p \downarrow \left({q \downarrow r}\right) \implies \left({p \downarrow q}\right) \downarrow r }\right)$
 * $\mathrm {TI}$
 * 15
 * Rule of Material Implication
 * $\mathrm {TI}$
 * 15
 * Rule of Material Implication

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.