User:Dfeuer/Membership is Asymmetric on Natural Numbers

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Theorem

Let $m$ and $n$ be natural numbers.


Then $\lnot(m \in n \land n \in m)$.


Proof

Suppose for the sake of contradiction that $m \in n$ and $n \in m$.

By User:Dfeuer/Natural Number is Transitive, $m$ is a transitive class.

Thus $m \in m$, contradicting User:Dfeuer/Natural Number does not Contain Itself.

$\blacksquare$