Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 2

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Suppose that there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
 * $\forall n \in \N: a_{n + 1} \prec a_n$

Then $\struct {S, \preceq}$ is well-founded.

Proof
Suppose $\struct {S, \preceq}$ is not well-founded.

Let $T \subseteq S$ have no minimal element.

Let $a_0 \in T$.

We have that $a_0$ is not minimal in $T$.

So:
 * $\exists a_1 \in T: a_1 \prec a_0$

Similarly, $a_1$ is not minimal in $T$.

So:
 * $\exists a_2 \in T: a_2 \prec a_1$

Let $a_{k + 1}$ be an arbitrary element for which $a_{k + 1} \prec a_k$.

In order to allow this to be possible in the infinite case, it is necessary to invoke the Axiom of Dependent Choice as follows:

Let $a_k \in T$.

Then as $a_k$ is not minimal in $T$:
 * $\exists a_{k + 1} \in T: a_{k + 1} \prec a_k$

Hence by definition $\prec$ is a right-total relation.

So, by the Axiom of Dependent Choice, it follows that:
 * $\forall n \in \N: \exists a_n \in T: a_{n + 1} \prec a_n$

Thus we have been able to construct an infinite sequence $\sequence {a_n}$ in $T$ such that:
 * $\forall n \in \N: a_{n + 1} \prec a_n$.

It follows by the Rule of Transposition that if there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
 * $\forall n \in \N: a_{n + 1} \prec a_n$

then $\struct {S, \preceq}$ is well-founded.