Foundational Relation has no Relational Loops
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Theorem
Let $\prec$ be a foundational relation on $A$ 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 $y = \left\{{x_1, x_2, \ldots, x_n}\right\}$.
Then $y$ is a nonempty subset of $A$.
So, by the definition of a foundational relation:
- $\exists w \in y: \forall z \in y: \neg w \prec z$
Now, suppose $x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1$.
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.
$\blacksquare$
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