Equivalence of Formulations of Peano's Axioms

Theorem
The two formulations of Peano's Axioms:

S1

 * P1: $$\exists 0 \in \N$$: There exists a number $$0$$.


 * P2: $$\forall n \in \N: \exists n'$$: For every $$n$$, there exists another number $$n'$$, known as the successor of $$n$$.


 * P3: $$\neg \left({\exists n \in \N: n' = 0}\right)$$: No number has $$0$$ as its successor.


 * P4: $$\forall m, n \in \N: n' = m' \implies n = m$$: If two numbers have the same successor, they are the same number. Or: different numbers have different successors.


 * P5: $$\forall A \subseteq \N: \left({0 \in A \and \left({n \in A \implies n' \in A}\right)} \right) \implies A = \N$$: A subset of $$\N$$ with $$0$$ in it, such that it has the successor of every number in it, is the same set as $$\N$$. This is known as the principle of induction.

and:

S2

 * P1: $$\N \ne \varnothing$$: The set of natural numbers is not empty.


 * P2: $$\exists s: \N \to \N$$: there exists a mapping $$s$$ from $$\N$$ to itself. (This mapping can be called the successor mapping.)


 * P3: $$\forall m, n \in \N: s \left({m}\right) = s \left({n}\right) \implies m = n$$: the successor mapping is injective.


 * P4: $$\operatorname{Im} \left({s}\right) \subset \N$$: the image of $$s$$ is a proper subset of $$\N$$. (That is, $$s$$ is not surjective.)


 * P5: $$\forall A \subseteq \N: \left({x \in A: \neg \left({\exists y \in \N: x = s \left({y}\right)}\right) \and \left({z \in A \implies s \left({z}\right) \in A}\right)}\right) \implies A = \N$$: For any subset $$A$$ of $$\N$$ which has an element of $$\N$$ with no successor elements, such that it has the successor of every number in it, is the same set as $$\N$$.

are equivalent.

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


 * From S1: P1 we have that $$0 \in \N$$.

Thus S2: P1 is fulfilled, as $$\N \ne \varnothing$$.


 * From S1: P2, we define $$s: \N \to \N$$ as:
 * $$\forall n \in \N: s \left({n}\right) = n'$$.

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

Thus S2: P2 is fulfilled.


 * From S1: P3, we have that $$\neg \left({\exists n \in \N: n' = 0}\right)$$.

Thus from S2: P2 it follows that $$\neg \left({\exists n \in \N: s \left({n}\right) = 0}\right)$$.

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

Thus S2: P4 is fulfilled.


 * From S1: P4, we have that $$\forall m, n \in \N: n' = m' \implies n = m$$.

Thus from S2: P2 it follows that $$\forall m, n \in \N: s \left({n}\right) = s \left({m}\right) \implies n = m$$.

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


 * From S1: P5, by identifying $$0$$ with the element in $$\N$$ which is the successor of no other element, we see that:
 * $$\neg \left({\exists y \in \N: 0 = s \left({y}\right)}\right)$$

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

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

Thus S2: P5 is fulfilled.

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

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

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

From S2: P4 we establish that $$\exists x \in \operatorname{Im} \left({s}\right) \subset \N$$:
 * $$\exists x \in \N: \neg \left({\exists y \in \N: x = s \left({y}\right)}\right)$$

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

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

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

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

and so S1: P3 is fulfilled.

Also, we see that $$0 \in \N$$ and so S1: P1 is fulfilled.

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

that is, S1: P5 is fulfilled.

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