No Membership Loops

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For any proper classes or sets $A_1, A_2, \ldots, A_n$:

$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$



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Either $A_1, A_2, \ldots, A_n$ are all sets, or there exists a proper class $A_m$ such that $1 \le m \le n$.

Suppose there exists a proper class $A_m$.

Then, by the definition of a proper class, $\neg A_m \in A_{m+1}$, since it is not a member of any class.

The result then follows directly.

Otherwise it follows that all $A_1, A_2, \ldots, A_n$ are sets.

Then, by the fact that Epsilon Relation is Strictly Well-Founded and a Strictly Well-Founded Relation has no Relational Loops, it follows that:

$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$


Also see