Dedekind-Hasse norm implies principal ideal ring

Statement
A commutative unital ring that admits a fact about::Dedekind-Hasse norm must be a fact about::principal ideal ring.

In particular, an fact about::integral domain that admits a Dedekind-Hasse norm must be a fact about::principal ideal domain.

Dedekind-Hasse norm
A Dedekind-Hasse norm on a commutative unital ring $$R$$ is a function $$N$$ from the nonzero elements of $$R$$ to the nonnegative integers with the property that for any elements $$a,b \in R$$ with $$b \ne 0$$, it is true that either $$a \in (b)$$ or there exists an element $$r$$ in the ideal $$(a,b)$$ such that $$N(r) < N(b)$$.

Principal ideal ring
A commutative unital ring $$R$$ is termed a principal ideal ring if every ideal of $$R$$ is principal.

Proof
Given: A commutative unital ring $$R$$ with a Dedekind-Hasse norm $$N$$. An ideal $$I$$ of $$R$$.

To prove: There exists $$b \in I$$ such that $$I = (b)$$.

Proof: If $$I = 0$$, we can take $$b = 0$$, and we are done.

So, suppose $$I \ne 0$$. Consider the function $$N$$ on the nonzero elements of $$I$$. Since $$N$$ maps to a well-ordered set, there is an element $$b \in I \setminus \{ 0 \}$$ such that $$N(b) \le N(r)$$ for all $$r \in I \setminus \{ 0 \}$$.

Pick any $$a \in I$$. If $$a = 0$$, then $$a \in (b)$$ or there exists $$r \in (a,b)$$ with $$N(r) < N(b)$$.

Note that in the latter case, we have $$r \in (a,b) \subseteq I$$, so $$r \in I$$, and $$N(r) < N(b)$$, contradicting the assumption that $$b$$ has minimum norm among the nonzero elements of $$I$$. Hence, we have the case $$a \in (b)$$. Thus, $$I \subseteq (b)$$. Conversely, we clearly have $$(b) \subseteq I$$, so $$I = (b)$$.