Element is Unit iff its Euclidean Valuation equals that of 1

Theorem
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.

Let the valuation function of $D$ be $\nu$.

Let $a \in D$.

Then:
 * $a$ is a unit of $D$ $\map \nu a = \map \nu 1$

Proof
For $a \in D$ we have that:
 * $\map \nu 1 \le \map \nu {1 a} = \map \nu a$

by definition of Euclidean valuation.

Let $a$ be a unit of $D$.

Then:
 * $\exists b \in D: a b = 1$

Then:
 * $\map \nu a \le \map \nu {a b} = \map \nu 1$

and so:
 * $\map \nu a = \map \nu 1$

Let $\map \nu a = \map \nu 1$.

We can write this as:
 * $\map \nu {1 a} = \map \nu 1$

and it follows from Euclidean Valuation of Non-Unit is less than that of Product that $a$ is a unit of $D$.