One Succeeds Zero in Well-Ordered Integral Domain

Theorem
Let $\left({D, +, \times, \le}\right)$ be a well-ordered integral domain.

Let $0$ and $1$ be the zero and unity respectively of $D$.

Then $0$ is the immediate predecessor of $1$:
 * $0 < 1$
 * $\neg \exists a \in D: 0 < a < 1$

Proof
Suppose there exists an element $a \in D$ such that $0 < a < 1$.

Let us create the set $S$ of all such elements:
 * $S = \left\{{x \in D: 0 < x < 1}\right\}$

We know $S$ is not empty because we've already supposed $a$ is in it.

And all the elements in $S$ are positive by definition.

Because $D$ is well-ordered it follows that $S$ has a minimal value, which we will call $m$.

Thus we have $0 < m < 1$.

Then from Square of Element Less than Unity in Ordered Integral Domain we have:
 * $0 < m \times m < m$

Thus $m \times m \in S$ is an element of $S$ which is smaller than $m$.

But $m$ was supposed to be the minimal element of $S$.

From this contradiction we deduce that there can be no such element $a$ such that $0 < a < 1$.