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.