Multiplicatively monotone Euclidean norm admits unique Euclidean division for exact divisor

Statement
Suppose $$R$$ is an commutative unital ring with a fact about::Euclidean norm $$N$$ that is multiplicatively monotone: $$N(ab) \ge \max \{ N(a), N(b) \}$$ whenever $$ab \ne 0$$.

Suppose $$a = bc$$ for $$b \ne 0$$. Then, we cannot write:

$$a = bq + r$$

with $$r \ne 0$$ and $$N(r) < N(b)$$.

Related facts

 * Filtrative and multiplicatively monotone implies uniquely Euclidean

Proof
Given: An integral domain $$R$$ with multiplicatively monotone Euclidean norm $$N$$. $$b \ne 0$$ and $$a = bc$$.

To prove: We cannot write $$a = bq + r$$ with $$N(r) < N(b)$$.

Proof: Suppose $$a = bq + r$$ with $$N(r) < N(b)$$. Then, since $$a = bc$$, we get:

$$bc = bq + r \implies b(c - q) = r$$.

Take $$N$$ both sides, to get:

$$N(b(c-q)) = N(r)$$.

Since $$N$$ is multiplicatively monotone, we get:

$$N(b(c-q)) \ge \max \{ N(b), N(c - q) \}$$.

On the other hand, we have, by assumption:

$$N(r) < N(b)$$.

This is a contradiction, completing the proof.