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.
$\blacksquare$