Peano axioms without explicit reference to zero Consider $(N, S)$ with S injective and not surjective and suppose the induction principle holds, where the zero in inductive principle is an element which is not in $S(N)$.
Can I prove that $0$ is unique?
In other words can we delete explicit reference of zero in Peano's axioms?
Update
My attempt of proof.
If $0' \ne 0$ and $0'$ not in $S(N)$ define $A=N-\{0'\}$ so we have:
1) $0 \in A$;
2) if $n \in A$ then $S(n) \ne 0'$, beacuse $0' \notin S(N)$, so $S(n) \in A$.
Hence by induction we have $A = N$ which is a contradiction.
Is my proof correct?
I think this question is the same as the intro of the following:
Prove that the system $(P, S, 0)$ satisfy Peano Axioms.
 A: I tried to formalize what (I think) you are trying to do.
As I understand it, you are trying to remove the explicit reference to $0$ in the Peano Axioms, and instead say that there is just something that is outside the range of $s$. So, I figure you want to replace the usual axiom:
$$\forall x \ s(x) \neq 0$$
with something like:
$$\exists z \forall x \ s(x) \neq z$$
And, to remove the explicit reference to $0$ from the typical inductive schema:
$$\forall w_1 ... w_n ((\varphi(0) \land \forall x (\varphi(x) \to \varphi(s(x)))) \to \forall x \ \varphi(x))$$ 
you would get something like:
$$\forall w_1 ... w_n ((\exists z (\forall x \ s(x) \neq z \land \varphi(z)) \land \forall x (\varphi(x) \to \varphi(s(x)))) \to \forall x \ \varphi(x))$$ 
But the question now is:  is there only one such $z$ that is not covered by the $s$ function, i.e. is it true that:
$$\exists z \forall y (\forall x \ s(x) \neq y \to y = z)$$
Well, below is a formal proof, using an instance of the new inductive scheme with:
$$\varphi(y) := \forall x \ s(x) \neq y \to y = w$$

So yes, the $z$ is indeed unique.
