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

Proof
Suppose $\RR$ is not a well-founded relation.

Hence there exists $T \subseteq S$ such that $T$ has no minimal element under $\RR$.

Let $a_0 \in T$.

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

So:
 * $\exists a_1 \in T: \paren {a_1 \mathrel \RR a_0} \text { and } a_1 \ne a_0$

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

So:
 * $\exists a_2 \in T: \paren {a_2 \mathrel \RR a_1} \text { and } a_2 \ne a_1$

Let $a_{k + 1}$ be an arbitrary element for which $a_{k + 1} \mathrel \RR a_k$ and $a_{k + 1} \ne a_k$.

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 $T$:
 * $\exists a_{k + 1} \in T: \paren {a_{k + 1} \mathrel \RR a_k} \text { and } \paren {a_{k + 1} \ne a_k}$

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

So, by the Axiom of Dependent Choice, it follows that:
 * $\forall n \in \N: \exists a_n \in T: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } a_{n + 1} \ne a_n$

Thus we have been able to construct an infinite sequence $\sequence {a_n}$ in $T$ such that:
 * $\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } a_{n + 1} \ne a_n$

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

then $\RR$ is a well-founded relation.