Successor Mapping of Peano Structure has no Fixed Point

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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 $\map s n \ne n$.


Aiming for a contradiction, suppose $m \notin T$.

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.

$\blacksquare$

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