Natural Number m is Less than n iff m is an Element of n

Theorem
Let $\omega$ be the set of natural numbers defined as the von Neumann construction.

Let $m, n \in \omega$.

Then:
 * $m < n \iff m \in n$

That is, every natural number is the set of all smaller natural numbers.

Proof
By definition of the ordering on von Neumann construction:


 * $m \le n \iff m \subseteq n$

Necessary Condition
Let $m \in n$.

Then from Natural Number is Transitive Set:
 * $m \subseteq n$

But if $m = n$, we would have:
 * $m \in m$

which contradicts Natural Number is Ordinary Set.

So we have:
 * $m \le n$ and $m \ne n$

that is:
 * $m < n$

Sufficient Condition
Let $m < n$.

Then by Natural Number m is Less than n implies n is not Greater than Successor of n:
 * $m^+ \le n$

But we have $m \in m^+$ by construction.

Hence:
 * $m \in n$