Non-Successor Element of Peano Structure is Unique

Theorem
Consider the Peano Axiom schema:


 * P1: $$\N \ne \varnothing$$: The set of natural numbers is not empty.


 * P2: $$\exists s: \N \to \N$$: there exists a mapping $$s$$ from $$\N$$ to itself. (This mapping can be called the successor mapping.)


 * P3: $$\forall m, n \in \N: s \left({m}\right) = s \left({n}\right) \implies m = n$$: the successor mapping is injective.


 * P4: $$\operatorname{Im} \left({s}\right) \subset \N$$: the image of $$s$$ is a proper subset of $$\N$$. (That is, $$s$$ is not surjective.)


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

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

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

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

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

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

Let us form $$A \in \N$$ 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 $$t_1 \in \N \setminus s \left({\N}\right)$$, it follows that $$\neg \left({\exists y \in \N: t_1 = s \left({y}\right)}\right)$$.

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

By axiom P5, it follows that $$A = \N$$.

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

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

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

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