Equivalence of Formulations of Peano's Axioms

Theorem
Let $P$ be a set.

Let $s: P \to P$ be a mapping.

Let $0 \in P$ be a distinguished element.

Formulation 1 implies Formulation 2
For $(\text P 3)$, there is nothing to prove.

Next, axiom $(\text P 4)$ of Formulation 2.

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


 * $\Img s = \set {n \in P: \exists m \in P: \map s m = n}$

From this definition, it follows that $0 \notin \Img s$, and hence:


 * $\Img s \ne P$

Lastly, axiom $(\text P 5)$ 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 $(\text P 3)$ of Formulation 1, we know this element is $0$.

Therefore, the premises:


 * $0 \in A$

and:


 * $\exists x \in A: \neg \exists y \in P: x = \map s y$

are logically equivalent, and hence so are both forms of axiom $(\text P 5)$.

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 $(\text P 3)$ and $(\text P 4)$ of Formulation 1 are satisfied.

Lastly, axiom $(\text P 5)$ of Formulation 1.

Since $0$ satisfies the premise:


 * $\neg \exists y \in P: 0 = \map s y$

we conclude that axiom $(\text P 5)$ of Formulation 1 follows a fortiori from axiom $(\text P 5)$ of Formulation 2.