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

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

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 $\exists x \notin \operatorname{Im} \left({s}\right) \subset P$:
 * $\exists x \in P: \neg \left({\exists y \in P: x = s \left({y}\right)}\right)$

... 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 $0$ is no successor, it follows that:
 * $\forall A \subseteq P: \left({0 \in A \land \left({n \in A \implies n' \in A}\right)} \right) \implies A = P$

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.