Filtrative and multiplicatively monotone Euclidean implies uniquely Euclidean

Statement
Suppose $$R$$ is an integral domain with a fact about::Euclidean norm $$N$$ satisfying the following two conditions:


 * $$N$$ is a fact about::filtrative norm, so is a fact about::filtrative Euclidean norm: For any natural number $$n$$, the set $$\{ 0 \} \cup \{ a \in R \mid N(a) < n \}$$ is an additive subgroup of $$R$$.
 * $$N$$ is a fact about::multiplicatively monotone norm, so is a fact about::multiplicatively monotone Euclidean norm: If $$ab \ne 0$$, we have $$N(ab) \ge \max \{ N(a), N(b) \}$$.

Then, $$N$$ is a fact about::uniquely Euclidean norm: for any $$a,b \in R$$ with $$b \ne 0$$, there exists a unique pair $$(q,r)$$ satisfying $$a = bq + r$$ and $$r = 0$$ or $$N(r) < N(b)$$.

Proof
Given: Integral domain $$R$$, Euclidean norm $$N: R \setminus \{ 0 \} \to \mathbb{N}_0$$ that is filtrative and multiplicatively monotone.

To prove: $$N$$ is uniquely Euclidean: for any $$a,b \in R$$ with $$b \ne 0$$, there exists a unique pair $$(q,r)$$ for which $$a = bq + r$$ and $$r = 0$$ or $$N(r) < N(b)$$.

Proof: Suppose there are two solution pairs: $$(q_1,r_1)$$ and $$(q_2,r_2)$$. Then, we have:

$$a = bq_1 + r_1 = bq_2 + r_2$$.


 * 1) We have $$b(q_2 - q_1) = r_1 - r_2$$: This follows by manipulating the equation $$bq_1 + r_1 = bq_2 + r_2$$.
 * 2) Either $$r_1 = r_2$$ or $$N(r_1 - r_2) < N(b)$$: By the definition of Euclidean norm, both $$r_1$$ and $$r_2$$ belong to the set $$\{ 0 \} \cup \{ r \in R \mid N(r) < N(b) \}$$. Since $$N$$ is filtrative, this set is a subgroup, so $$r_1 - r_2$$ also belongs to this set.
 * 3) Either $$q_1 = q_2$$ or $$N(b(q_2 - q_1)) \ge N(b)$$: If $$q_1 \ne q_2$$, then $$q_2 - q_1 \ne 0$$. Since $$R$$ is an integral domain, the product $$b(q_2 - q_1)$$ is nonzero, so by the fact that $$N$$ is multiplicatively monotone, we get $$N(b(q_2 - q_1)) \ge N(b)$$.
 * 4) $$(q_1,r_1) = (q_2,r_2)$$: Combining steps (1), (2) and (3), we obtain that we cannot have both $$r_1 \ne r_2$$ and $$q_1 \ne q_2$$. Thus, either $$q_1 = q_2$$ or $$r_1 = r_2$$. But $$q_1 = q_2$$ yields, by step (1), that $$r_1 = r_2$$. Similarly $$r_1 = r_2$$, along with step (1) and the fact that $$R$$ is an integral domain, yields $$q_1 = q_2$$. Hence, $$(q_1,r_1) = (q_2,r_2)$$.