# Inductive Construction of Natural Numbers fulfils Peano's Axioms

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## Theorem

Let $P$ denote the set of natural numbers by definition as an inductive set.

Then $P$ fulfils Peano's axioms.

## Proof

By definition of inductive set:

$\O \in P$

By definition of the natural numbers, $\O$ is identified with $0$ (zero).

Thus Peano's Axiom $\text P 1$: $0 \in P$ holds.

$\Box$

Let $x$ be a natural number.

By definition, $x$ is an element of every inductive set.

Thus if $x \in P$ it follows that $x^+ \in P$.

Thus Peano's Axiom $\text P 2$: $n \in P \implies \map s n \in P$ holds.

$\Box$

$\Box$

For all $n$, $n$ is an element of $n^+$.

But $0$ is identified with the empty set $\O$.

By definition, $\O$ has no elements.

Therefore it is not possible for $\O$ to equal $n^+$ for any $n$.

Thus Peano's Axiom $\text P 4$: $0 \notin \Img s$ holds.

$\Box$

Peano's Axiom $\text P 5$: Principle of Mathematical Induction holds immediately by definition of inductive set.

$\blacksquare$