Image of Successor Mapping forms Peano Structure

Theorem
Let $\left({P, s, 0}\right)$ be a Peano structure.

Let $P'$ be the set $s \left[{P}\right]$, that is:


 * $P' = \left\{{s \left({n}\right): n \in P}\right\}$

Let $s'$ be the restriction of $s$ to $P'$.

Then $\left({P', s', s \left({0}\right)}\right)$ is also a Peano structure.

Proof
We need to check that all of Peano's axioms hold for $\left({P', s', s \left({0}\right)}\right)$.

Now from Restriction of Injection is Injection, because $s$ is an injection then so is $s'$.

So Axiom $(P3)$ holds.

Suppose that for some $n \in P'$, we would have:


 * $s' \left({n}\right) = s \left({0}\right)$

Then by definition of $s'$ as the restriction of $s$:


 * $s \left({n}\right) = s \left({0}\right)$

However, since $0 \notin P'$, it follows that $n \ne 0$.

This contradicts the fact that $s$ is an injection.

Hence for all $n \in P'$:


 * $s' \left({n}\right) \ne s \left({0}\right)$

So Axiom $(P4)$ holds as well.

Let $A \subseteq P'$ be such that $s \left({0}\right) \in A$ and:


 * $\forall z \in A: s' \left({z}\right) \in A$

Let $A_0$ be defined as:


 * $A_0 = A \cup \left\{{0}\right\}$

Then $0 \in A_0$, and since:


 * $\forall z \in A: s' \left({z}\right) \in A$
 * $s \left({0}\right) \in A$

it follows that:


 * $\forall z \in A_0: s \left({z}\right) \in A_0$

Therefore, since $\left({P, s, 0}\right)$ is a Peano structure:


 * $A_0 = P$

Therefore, since $s \left({n}\right) \ne 0$ for all $n \in P$:


 * $\forall n \in P: s \left({n}\right) \in A$

Hence, by definition of $P'$:


 * $P' \subseteq A$

By definition of set equality:


 * $A = P'$

meaning that Axiom $(P5)$ holds as well.

Having verified all axioms, it follows that $\left({P', s', s \left({0}\right)}\right)$ is a Peano structure.