Non-Successor Element of Peano Structure is Unique

Theorem
Let $$P$$ be a set which fulfils the Peano Axiom schema:


 * P1: $$P \ne \varnothing$$


 * P2: $$\exists s: P \to P$$


 * P3: $$\forall m, n \in P: s \left({m}\right) = s \left({n}\right) \implies m = n$$


 * P4: $$\operatorname{Im} \left({s}\right) \ne P$$


 * P5: $$\forall A \subseteq P: \left({x \in A: \neg \left({\exists y \in P: x = s \left({y}\right)}\right) \and \left({z \in A \implies s \left({z}\right) \in A}\right)}\right) \implies A = P$$

Then:
 * $$P \setminus s \left({P}\right)$$ is a singleton set

where:
 * $$\setminus$$ denotes set difference;
 * $$s \left({P}\right)$$ denotes the image of the mapping $$s$$.

Proof
Let $$T = P \setminus s \left({P}\right)$$.

From P4 we know that $$T \ne \varnothing$$.

Now suppose that $$t_1 \in T$$ and $$t_2 \in T$$.

Let us form $$A \in P$$ such that $$t_1 \in A$$ and $$z \in A \implies s \left({z}\right) \in A$$, but such that $$t_2 \notin A$$.

As $$t_1 \in P \setminus s \left({P}\right)$$, it follows that $$\neg \left({\exists y \in P: t_1 = s \left({y}\right)}\right)$$.

Thus $$A$$ is of the form:
 * $$\left({t_1 \in A: \neg \left({\exists y \in P: t_1 = s \left({y}\right)}\right) \and \left({z \in A \implies s \left({z}\right) \in A}\right)}\right)$$

By axiom P5, it follows that $$A = P$$.

That is, $$P \setminus A = \varnothing$$.

But $$t_2 \notin A$$ and so $$t_2 \in P \setminus A$$.

This contradiction, demonstrates that, given the existence of $$t_1 \in P \setminus s \left({P}\right)$$, there can be no $$t_2 \in P \setminus s \left({P}\right): t_1 \ne t_2$$.

Hence $$t_1$$ is unique, and $$P \setminus s \left({P}\right)$$ is singleton.