Successor Mapping of Peano Structure has no Fixed Point

Theorem
Let $\mathcal P = \left({P, 0, s}\right)$ be a Peano structure.

Then:
 * $\forall n \in P: s \left({n}\right) \ne n$

That is, the successor mapping has no fixed points.

Proof
Let $T$ be the set:


 * $T = \left\{{n \in P: s \left({n}\right) \ne n}\right\}$

From Axiom $(P4)$, a fortiori:


 * $s \left({0}\right) \ne 0$

Hence $0 \in T$.

Let $n \in T$.

Let $m = s \left({n}\right)$.

Then $n \ne m$, as $s \left({n}\right) \ne n$.

Suppose $m \notin T$, that is:


 * $s \left({m}\right) = m$

Then that would mean:


 * $s \left({n}\right) = s \left({m}\right) = m$

But from Axiom $(P3)$, $s$ is injective.

From this contradiction, it follows that $m \in T$.

Hence we have that:


 * $n \in T \implies s \left({n}\right) \in T$

Recall also that $0 \in T$.

By Axiom $(P5)$, we conclude that $T = P$.

From the definition of $T$, the result follows.