Strictly Well-Founded Relation has no Relational Loops

Theorem
Let $\prec Fr A$ and let $x _{1}, x _{2},...x_{n} \in A$

Then $\neg ( x _{1} \prec x _{2} \land x _{2} \prec x _{3} ... \land x _{n} \prec x _{1} )$

Thus there are no relational "loops" within A.

Proof
Since $x _{1}, x _{2},...x_{n} \in A$, there exists a $y$ such that $y = \{ x _{1}, x _{2},...x_{n} \}$

Now, suppose $( x _{1} \prec x _{2} \land x _{2} \prec x _{3} ... \land x _{n} \prec x _{1} )$. 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 $x _{1}, x _{2},...x_{n}$, 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.