Equivalence of Formulations of Peano's Axioms

Theorem
The two formulations of Peano's Axioms:

S1
Let $$P$$ be a set which fulfils the following properties:


 * P1: $$P \ne \varnothing$$


 * P2: $$\exists s: P \to P$$


 * P3: $$\forall m, n \in P: s \left({m}\right) = s \left({n}\right) \implies m = n$$


 * P4: $$\operatorname{Im} \left({s}\right) \ne P$$


 * P5: $$\forall A \subseteq P: \left({x \in A: \neg \left({\exists y \in P: x = s \left({y}\right)}\right) \and \left({z \in A \implies s \left({z}\right) \in A}\right)}\right) \implies A = P$$

and:

S2
Let $$P$$ be a set (see Peano's Axioms Uniquely Define Natural Numbers) which fulfils the following properties:


 * P1: $$\exists 0 \in P$$


 * P2: $$\forall n \in P: \exists n' \in P$$


 * P3: $$\neg \left({\exists n \in P: n' = 0}\right)$$


 * P4: $$\forall m, n \in P: n' = m' \implies n = m$$


 * P5: $$\forall A \subseteq P: \left({0 \in A \and \left({n \in A \implies n' \in A}\right)} \right) \implies A = P$$

are logically equivalent.

S1 implies S2
Let $$P$$ be a set that fulfils schema S1.

By identifying $$s \left({n}\right)$$ with $$n'$$, we see that:
 * S1: P2 implies S2: P2;
 * S1: P3 implies S2: P4.

From S1: P4 we establish that $$\exists x \in \operatorname{Im} \left({s}\right) \subset P$$:
 * $$\exists x \in P: \neg \left({\exists y \in P: x = s \left({y}\right)}\right)$$

... and so there is at least one element of $$P$$ which is the successor of no element of $$P$$.

But from Non-Successor Element of Peano Axiom Schema is Unique, we see that there must be exactly one such element.

Let us give a name to that element, and so say $$x = 0$$.

Thus:
 * $$P \setminus \operatorname{Im} \left({s}\right) = \left\{{0}\right\}$$

and so S2: P3 is fulfilled.

Also, we see that $$0 \in P$$ and so S2: P1 is fulfilled.

As $$0$$ has no successors, it follows that:
 * $$\forall A \subseteq P: \left({0 \in A \and \left({n \in A \implies n' \in A}\right)} \right) \implies A = P$$

that is, S2: P5 is fulfilled.

So S2: P1 - P5 are all fulfilled, and hence we see that S1 implies S2.

S2 implies S1
Let $$P$$ be a set that fulfils schema S2.


 * From S2: P1 we have that $$0 \in P$$.

Thus S1: P1 is fulfilled, as $$P \ne \varnothing$$.


 * From S2: P2, we define $$s: P \to P$$ as:
 * $$\forall n \in P: s \left({n}\right) = n'$$.

As (implicitly) $$n'$$ is unique for a given $$n$$, it follows that $$s: P \to P$$ is a mapping.

Thus S1: P2 is fulfilled.


 * From S2: P3, we have that $$\neg \left({\exists n \in P: n' = 0}\right)$$.

Thus from S1: P2 it follows that $$\neg \left({\exists n \in P: s \left({n}\right) = 0}\right)$$.

So $$0 \notin s \left({P}\right)$$, so $$s$$ is not surjective.

Thus S1: P4 is fulfilled.


 * From S2: P4, we have that $$\forall m, n \in P n' = m' \implies n = m$$.

Thus from S1: P2 it follows that $$\forall m, n \in P: s \left({n}\right) = s \left({m}\right) \implies n = m$$.

That is, $$s$$ is an injection.

Thus S1: P3 is fulfilled.


 * From S2: P5, by identifying $$0$$ with the element in $$P$$ which is the successor of no other element, we see that:
 * $$\neg \left({\exists y \in P: 0 = s \left({y}\right)}\right)$$

That is, there is no $$y$$ in $$P$$ which has $$0$$ as its successor.

So any subset $$A \subseteq P$$ with $$0$$ in it fulfilling S2: P5 also fulfils S1: P5.

Thus S1: P5 is fulfilled.

So S1: P1 - P5 are all fulfilled, and hence we see that S1 implies S2.