Strictly Well-Founded Relation has no Relational Loops

Theorem
Let $\prec Fr A$ and let $x_1, x_2, \ldots, x_n \in A$.

Then:
 * $\neg \left({x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1}\right)$

That is, there are no relational loops within $A$.

Proof
Since $x_1, x_2, \ldots, x_n \in A$, there exists a $y$ such that $y = \left\{{x_1, x_2, \ldots, x_n}\right\}$.

Now, suppose $x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1$.

Then $y$ is a nonempty subset of $A$.

So:
 * $\exists w \in y: \forall z \in y: \neg w \prec z$

But since the elements of $y$ are $x_1, x_2, \ldots, x_n$, then this contradicts the previous statement, since:
 * $\forall w \in y: \exists z \in y: w \prec z$

Thus, a founded relation has no relational loops.