Element of Natural Number is Natural Number

Theorem
Let $n$ be a natural number.

Let $m \in n$.

Then $m$ is also a natural number.

Proof
The proof proceeds by induction.

For all $n \in \N$, let $\map P n$ be the proposition:
 * for all $m \in n$: $m$ is a natural number.

Basis for the Induction
$\map P 0$ is the case:
 * for all $m \in 0$: $m$ is a natural number.

This is true vacuously, as $\O$ has no elements by definition of empty class.

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k^+}$ is true.

So this is the induction hypothesis:
 * for all $m \in k$: $m$ is a natural number.

from which it is to be shown that:
 * for all $m \in k^+$: $m$ is a natural number.

Induction Step
This is the induction step:

Let $m \in k^+$.

Then either:
 * $m = k$

or:
 * $m \in k$

In the first case, $m$ is the natural number $k$.

In the second case, it follows by the induction hypothesis that $m$ is a natural number.

In both cases $m$ is a natural number.

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: \forall m \in n$: $m$ is a natural number.