Well-Ordering Minimal Elements are Unique
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Theorem
Let $\struct {S,\preceq}$ be a well-ordered set.
Then every non-empty subset of $S$ has a unique minimal element.
Proof
The proof consists of a uniqueness and an existence part.
Let $S'$ be a non-empty subset of $S$.
Existence
As $S$ is well-ordered by $\preceq$, we have by definition the existence of a minimal elements $s$ of $S'$ (with respect to $\preceq$).
$\Box$
Uniqueness
Let $s, s'$ be minimal elements of $S'$.
By Well-Ordering is Total Ordering, we have at least one of $s \preceq s'$, $s'\preceq s$.
Since both $s$ and $s'$ are minimal elements, we conclude that necessarily $s = s'$.
$\blacksquare$