Minimal Element in Toset is Unique and Smallest

Theorem
Let $\left({S, \preceq}\right)$ be a totally ordered set.

Let $m$ be a minimal element of $\left({S, \preceq}\right)$.

Then:
 * $(1): \quad m$ is the smallest element of $\left({S, \preceq}\right)$.
 * $(2): \quad m$ is the only minimal element of $\left({S, \preceq}\right)$.

Proof
By definition of minimal element:
 * $\forall y \in S: y \preceq m \implies m = y$

As $\left({S, \preceq}\right)$ is a totally ordered set, by definition $\preceq$ is connected.

That is:
 * $\forall x, y \in S: y \preceq x \lor x \preceq y$

It follows that:
 * $\forall y \in S: m = y \lor m \preceq y$

But as $m = y \implies m \preceq y$ by definition of $\preceq$, it follows that:
 * $\forall y \in S: m \preceq y$

which is precisely the definition of smallest element.

Hence $(1)$ holds.

Suppose $m_1$ and $m_2$ are both minimal elements of $S$.

By $(1)$ it follows that both are smallest elements.

It follows from Smallest Element is Unique that $m_1 = m_2$.

That is, $(2)$ holds.

Also see

 * Maximal Element in Toset is Unique and Greatest