# 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.

$\blacksquare$

## Sources

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- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts*: Introduction $\S 4$ (implicit)