Natural Numbers cannot be Elements of Each Other

Theorem

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

Then it cannot be the case that both $m \in n$ and $n \in m$.

Proof

Aiming for a contradiction, suppose both $m \in n$ and $n \in m$.

We have $m \in n$

From Natural Number is Transitive Set:

$m \subseteq n$

by definition of transitive.

Thus:

$n \in m \subseteq n$

and so:

$n \in n$

But from Natural Number is Ordinary Set:

$n \notin n$

The result follows by Proof by Contradiction.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.3$