User:Dfeuer/Double Induction Principle/Naturals

Theorem
Let $\mathcal R$ be a relation such that:


 * For each natural number $n$, $n \mathrel{\mathcal R} 0$
 * For any natural numbers $m$ and $n$:
 * $(n \mathrel{\mathcal R} m) \land m \mathrel{\mathcal R} n \implies m \mathrel{\mathcal R} (n^+)$

Then $ n \mathrel{\mathcal R} m $ for any naturals $m$ and $n$.

Proof
Let $P$ be the class of all rationals $n$ such that $0 \mathrel {\mathcal R} n$.

$0 \in P$.

Suppose $n \in P$.

Then $0 \mathrel {\mathcal R} n$ and $n \mathrel {\mathcal R} 0$, so $0 \mathrel {\mathcal R} n^+$.

Thus for each natural $n$: $n \in P \implies n^+ \in P$.

Thus $P$ contains all the naturals. That is, for each natural $n$, $0 \mathrel {\mathcal R} n$.

Let $A$ be the class of all naturals $n$ such that for all naturals $m$, $m \mathrel{\mathcal R} n$ and $n \mathrel {\mathcal R} m$.

$0 \in A$.

Suppose that $n \in A$.

Let $B$ be the class of all naturals $m$ such that $m \mathrel{\mathcal R} n^+$ and $n^+ \mathrel {\mathcal R} m$.

We have shown that $0 \in B$.

Suppose that $m \in B$.

Then by the definition of $B$: $m \mathrel{\mathcal R} n^+$ and $n^+ \mathrel {\mathcal R} m$.

Thus $n^+ \mathrel{\mathcal R} m^+$ because $\mathcal R$ is double inductive.

By the definition of $A$, $m^+ \mathrel{\mathcal R} n$ and $n \mathrel{\mathcal R} m^+$.

Because $\mathcal R$ is doubly inductive, $m^+ \mathrel{\mathcal R} n^+$.

As we already have $n^+ \mathrel{\mathcal R} m^+$, we see that $m^+ \in B$.

Thus $B$ is inductive, so it contains every natural.

Thus $A$ is inductive, so it contains every natural.