Finite Totally Ordered Set is Well-Ordered

Theorem
Every finite totally ordered set is well-ordered.

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

From Condition for Well-Foundedness, $$\left({S, \preceq}\right)$$ is well-founded iff there is no infinite sequence $$\left \langle {a_n}\right \rangle$$ of elements of $$S$$ such that $$\forall n \in \N: a_{n+1} \prec a_n$$.

If it were the case that $$\left({S, \preceq}\right)$$ had such an infinite sequence, then at least some of the elements would be repeated in that sequence.

So there would be, for example:
 * $$s_i \preceq s_j \preceq s_k \preceq s_i$$

and $$\preceq$$ would therefore not be transitive and so not a totally ordered set.

So $$\left({S, \preceq}\right)$$ is well-founded.

The result follows from the definition of well-ordered set.