Equivalence of Formulations of Peano's Axioms

Theorem
Let $P$ be a set, $s: P \to P$ be a mapping and $0 \in P$ be a distinguished element.

The following formulations of Peano's Axioms are logically equivalent:

Formulation 1 implies Formulation 2
For $(P3)$, there is nothing to prove.

Next, axiom $(P4)$ of Formulation 2.

Recall the definition of the image of a mapping as applicable to $s$:


 * $\operatorname{Im} \left({s}\right) = \left\{{n \in P: \exists m \in P: s \left({m}\right) = n}\right\}$.

From this definition, it follows that $0 \notin \operatorname{Im} \left({s}\right)$, and hence:


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

Lastly, axiom $(P5)$ of Formulation 2.

By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.

By axiom $(P3)$ of Formulation 1, we know this element is $0$.

Therefore, the premises:


 * $0 \in A$

and:


 * $\exists x \in A: \neg \exists y \in P: x = s \left({y}\right)$

are logically equivalent, and hence so are both forms of axiom $(P5)$.

Formulation 2 implies Formulation 1
By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.

We will identify this element with the distinguished element $0$.

This implies that axioms $(P3)$ and $(P4)$ of Formulation 1 are satisfied.

Lastly, axiom $(P5)$ of Formulation 1.

Since $0$ satisfies the premise:


 * $\neg \exists y \in P: 0 = s \left({y}\right)$

we conclude that axiom $(P5)$ of Formulation 1 follows a fortiori from axiom $(P5)$ of Formulation 2.