Modulo Multiplication is Associative/Proof 2

Proof
Let $j$ be the largest integer such that:
 * $j m \le x y$

Let $p$ be the largest integer such that:
 * $p m \le y z$

By definition of multiplication modulo $m$:
 * $x \cdot_m y = x y - j m$
 * $y \cdot_m z = y z - p m$

Let $k$ be the largest integer such that:
 * $k m \le \paren {x y - j m} z$

Let $q$ be the largest integer such that:
 * $q m \le x \paren {y z - p m}$

Then:
 * $\paren {j z + k} m \le \paren {x y} z$
 * $\paren {q + x p} m \le x \paren {y z}$

Thus:

But suppose that there exists an integer $s$ such that:
 * $s m \le \paren {x y} z$

and:
 * $j z + k < s$

Then:
 * $\paren {j z + k + 1} m \le \paren {x y} z$

from which:
 * $\paren {k + 1} m \le \paren {x y - j m} z$

But this contradicts the definition of $k$.

Thus $j z + k$ is the largest of those integers $i$ such that $i m \le \paren {x y} z$.

Similarly, $q + x p$ is the largest of those integers $i$ such that $i m \le x \paren {y z}$.

From Integer Multiplication is Associative:
 * $\paren {x y} z = x \paren {y z}$

Thus $j z + k = q + x p$ and so: