Cartesian Product is not Associative/Proof 2

Formal Proof
Assign to every set $X$ the following number $n \left({X}\right) \in \N$:


 * $n \left({X}\right) = \begin{cases}

0 & : X = \varnothing \\ \displaystyle 1 + \max_{Y \mathop \in X} \ n \left({Y}\right) & : \text{ otherwise} \end{cases}$

From the Axiom of Foundation:
 * $\forall X \in \N: n \left({X}\right) < \infty$

Now let $a \in A$ be such that:
 * $\displaystyle n \left({a}\right) = \min_{b \mathop \in A} \ n \left({b}\right)$

Suppose that:
 * $\exists a' \in A, b \in B: a = \left({a', b}\right)$

That is, that $a$ equals the ordered pair of $a'$ and $b'$.

Then it follows that:

That is:
 * $n \left({a'}\right) < n \left({a}\right)$

contradicting the assumed minimality of the latter.

Therefore:
 * $a \notin A \times B$

and hence:
 * $A \nsubseteq A \times B$

It follows from Equality of Cartesian Products that:
 * $A \times \left({B \times C}\right) \ne \left({A \times B}\right) \times C$