# 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.

The following definitions of the concept of Peano's Axioms are equivalent:

### Formulation 1

 $(\text P 3)$ $:$ $\ds \forall m, n \in P:$ $\ds \map s m = \map s n \implies m = n$ $s$ is injective $(\text P 4)$ $:$ $\ds \forall n \in P:$ $\ds \map s n \ne 0$ $0$ is not in the image of $s$ $(\text P 5)$ $:$ $\ds \forall A \subseteq P:$ $\ds \paren {0 \in A \land \paren {\forall z \in A: \map s z \in A} } \implies A = P$ Principle of Mathematical Induction: Any subset $A$ of $P$, containing $0$ and closed under $s$, is equal to $P$

### Formulation 2

 $(\text P 3)$ $:$ $\ds \forall m, n \in P:$ $\ds \map s m = \map s n \implies m = n$ $s$ is injective $(\text P 4)$ $:$ $\ds \Img s \ne P$ $s$ is not surjective $(\text P 5)$ $:$ $\ds \forall A \subseteq P:$ $\ds \paren {\paren {\exists x \in A: \neg \exists y \in P: x = \map s y} \land \paren {\forall z \in A: \map s z \in A} }$ Principle of Mathematical Induction: $\ds \implies A = P$ Any subset $A$ of $P$, containing an element not in the image of $s$ and closed under $s$, is equal to $P$

## Proof

### 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 $(\text P 5)$.

$\Box$

### 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.

$\blacksquare$