Peano Structure is Unique

Theorem
Let $\left({P, s, 0}\right)$ and $\left({P', s', 0'}\right)$ be Peano structures.

Then there is a unique bijection $f: P \to P'$ such that:

Proof
First to establish uniqueness of $f$.

Suppose that $f, g: P \to P'$ both satisfy the conditions.

Define $A \subseteq P$ as:


 * $A := \left\{{n \in P: f \left({n}\right) = g \left({n}\right)}\right\}$

Then the first condition implies that $0 \in A$.

Now suppose that $n \in A$. Then:

Hence $s \left({n}\right) \in A$.

Since $P$ is a Peano structure, it follows that $A = P$.

Hence by Equality of Mappings, $f = g$.

For the existence of such an $f$, we apply the Principle of Recursive Definition with $0' \in P'$ and $s': P' \to P'$.

Lastly, it is to be established that the $f$ so obtained is a bijection.

Let $A' \subseteq P$ be defined as:


 * $A' := \left\{{n \in P: \forall m \in P: f \left({n}\right) = f \left({m}\right) \implies n = m}\right\}$

It follows from Axiom $(P4)$ that $f \left({n}\right) = 0'$ implies $n = 0$.

Hence $0 \in A'$.

Suppose now that $n \in A'$, but $s \left({n}\right) \notin A'$.

Then, since $f \left({s \left({n}\right)}\right) \ne 0'$, there exists some $m \in P, m \ne n$ such that:


 * $f \left({s \left({n}\right)}\right) = f \left({s \left({m}\right)}\right)$

That is, such that:


 * $s' \left({f \left({n}\right)}\right) = s' \left({f \left({m}\right)}\right)$

But by Axiom $(P3)$, $s'$ is injective.

Hence $f \left({n}\right) = f \left({m}\right)$, contradicting the assumption that $n \in A'$.

Therefore $s \left({n}\right) \in A'$, so that $A' = P$ by Axiom $(P5)$.

That is, $f$ is an injection.

Next, define $A'' \subseteq P'$ as:


 * $A'' := \left\{{n' \in P': \exists n \in P: f \left({n}\right) = n'}\right\}$

Since $f \left({0}\right) = 0'$, we have $0' \in A''$.

Suppose now that $n' \in A''$, and let $f \left({n}\right) = n'$.

Then:

meaning $s' \left({n'}\right) \in A''$.

Hence by Axiom $(P5)$, $A'' = P'$.

That is, $f$ is surjective.

It follows that $f$ is the sought bijection.