Peano Structure is Unique

Theorem
Let $\struct {P, s, 0}$ and $\struct {P', s', 0'}$ 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:


 * $A := \set {n \in P: \map f n = \map g n}$

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

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

Hence $\map s n \in A$.

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:


 * $A' := \set {n \in P: \forall m \in P: \map f n = \map f m \implies n = m}$

It follows from Axiom $(P4)$ that $\map f n = 0'$ implies $n = 0$.

Hence $0 \in A'$.

Suppose now that $n \in A'$, but $\map s n \notin A'$.

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


 * $\map f {\map s n} = \map f {\map s m}$

That is, such that:


 * $\map {s'} {\map f n} = \map {s'} {\map f m}$

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

Hence $\map f n = \map f m$, contradicting the assumption that $n \in A'$.

Therefore $\map s n \in A'$, so that $A' = P$ by Axiom $(P5)$.

That is, $f$ is an injection.

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


 * $A'' := \set {n' \in P': \exists n \in P: \map f n = n'}$

Since $\map f 0 = 0'$, we have $0' \in A''$.

Suppose now that $n' \in A''$, and let $\map f n = n'$.

Then:

meaning $\map {s'} {n'} \in A''$.

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

That is, $f$ is surjective.

It follows that $f$ is the sought bijection.