Image of Successor Mapping forms Peano Structure

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

Let $P\,'$ be the set $P \setminus \left\{{0}\right\}$, that is:
 * $P\,' = \left\{{x \in P: x \ne 0}\right\}$

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

Then the structure:
 * $\mathcal P\,' = \left({P\,', s \left({0}\right), s'}\right)$

is also a Peano structure.

Proof
We need to check that all of Peano's axioms hold for $\mathcal P\,'$.

In $\mathcal P = \left({P, 0, s}\right)$ there exists the element $0$.

Although this element is not in $\mathcal P\,'$, its successor is.

So $P\,' \ne \varnothing$, and P1 holds.

We have that $s'$ is the restriction of $s$ to $P\,'$.

As $\neg \left({\exists x \in P: s \left({x}\right) = 0}\right)$, it is clear that $0$ is not the image of any $x \in P$ under $s$.

Therefore it can not be the image of any $x \in P\,'$ under $s'$.

So $\operatorname{Im} \left({s'}\right) \subseteq P\,'$ and so:
 * $\exists s': P\,' \to P\,'$

and P2 holds for $\mathcal P\,'$.

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

So P3 holds for $\mathcal P\,'$.

Because $s$ is an injection, $s \left({0}\right)$ is the successor of only one element of $P$, that is, $0$.

But as $0 \notin P\,'$, that means $s \left({0}\right) \notin s \left({P\,'}\right) = s' \left({P\,'}\right)$.

But $s \left({0}\right) \in P\,'$ as $s \left({0}\right) \ne 0$, and so $P\,' \ne s' \left({P\,'}\right)$.

So $s'$ is not a surjection and so P4 holds for $\mathcal P\,'$.

Now consider $A \subseteq P$ such that:
 * $\exists x \in A: \neg \left({\exists y \in P: x = s \left({y}\right)}\right)$;
 * $z \in A \implies s \left({x}\right) \in A$.

As $\mathcal P$ is a Peano structure it follows that $A = P$.

Now consider $A' = A \setminus \left\{{0}\right\}$.

As $0 \in A$ we have that $s \left({0}\right) \in A$ and as $s \left({0}\right) \ne 0$ it follows that $s \left({0}\right) \in A'$.

The only element of $A$ not in $A'$ is $0$ and so for any other element $z$, if $z \in A'$ then $s \left({z}\right) \in A'$.

So $A'$ is also of the form:
 * $\exists x \in A': \neg \left({\exists y \in P\,': x = s \left({y}\right)}\right)$;
 * $z \in A' \implies s \left({x}\right) \in A'$.

But $A' = A \setminus \left\{{0}\right\}$, and $A = P$.

So it follows that $A' = P \setminus \left\{{0}\right\} = P\,'$.

So P5 holds for $\mathcal P\,'$.

Hence the result.