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

(Of course P1 is an instant consequence of P4).

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 $t_2 \in T$, and that (proof by contradiction) $t_1 \ne t_2$.

Define $A = P \setminus \{t_2\}$.

Thus $t_1 \in A \ne P$, and: $x \in A \Rightarrow s(x) \in A$.

Thus, by induction P5, $A = P$. A contradiction.

Thus $P \setminus s(P)$ cannot contain two different elements.