Successor Mapping of Peano Structure has no Fixed Point

Theorem
Let $\mathcal P = \struct {P, s, 0}$ be a Peano structure.

Then:
 * $\forall n \in P: \map s n \ne n$

That is, the successor mapping has no fixed points.

Proof
Let $T$ be the set:


 * $T = \set {n \in P: \map s n \ne n}$

We will use Axiom $(P5)$ to prove that $T = P$.

Part 1: $0 \in T$
From Axiom $(P4)$, a fortiori:


 * $\map s 0 \ne 0$

Hence $0 \in T$.

Part 2: $n \in T \implies \map s n \in T$
Let $n \in T$.

Let $m = \map s n$.

Then $n \ne m$, as $\map s n \ne n$.

$m \notin T$.

That is:


 * $\map s m = m$

Then that would mean:


 * $\map s n = \map s m = m$

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

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

Hence we have that:


 * $n \in T \implies \map s n \in T$

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

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