Show two models of the set of axioms for the successor are isomorphic Let $\Sigma$ be the set of the following axioms.
$\forall x \forall y (S(x)= S(y) \rightarrow x = y) $ 
$\forall y (y \neq 0 \rightarrow \exists x (S(x)=y)) $ 
$\forall x (S(x) \neq 0) $ 
$\forall x (S^n(x)\neq x) \quad , n \in \mathbb{N}$

Let $\mathcal{C}=\langle C; s; 0\rangle$ be a model of $\Sigma$. Take into account the following definitions:

*

*$a,b\in C$ are $s-$independent if for all $k\in\mathbb{N}$ neither $a=s^k(b)$ nor $b=s^k(a)$.

*$D\subseteq C$ is $s-$independent if for all $a,b\in D$, $a$ and $b$ are $s-$independent.

*$D\subseteq C$ is called an $s-$base if is is a maximal independent subset of $C$ such that for all $x\in C$ there is $k\in\mathbb{N}$, $d\in D$ such that $x=s^k(d)$ or $d=s^k(x)$ (i.e. $x$ is not $s-$independent of $d$).


I am now asked to show that if  $\mathcal{C}=\langle C; s; 0\rangle$ and $\mathcal{C}^{'}=\langle C^{'}; s^{'}; 0^{'}\rangle$ are models of $\Sigma$, $D\subseteq C$ and $D^{'}\subseteq C^{'}$ are $s-$basis then EVERY bijection $\pi: D\rightarrow D^{'}$ induces and isomorphism $\mathcal{C}\simeq\mathcal{C}^{'}$.

Note that my problem is not showing that any two models of the same cardinality are isomorphic, this is done in Enderton's A Mathematical Introduction to Logic and it was the approach I had seen for this problem.
What I'm struggling with is the way I'm asked to show this. I don't believe it is right, if the models $\mathcal{C}$ and $\mathcal{C}^{'}$ are isomorphic then the image of $0$ must be $0^{'}$ but this imposes a condition on the bijections $\pi$ that actually induce the isomorphism, right? In the sense that if there is a bijection from $D$ to $D^{'}$ such that $\pi(0)=d^{'}$ and $d^{'}\neq 0^{'}$, then this won't induce an isomorphism as it is stated in the question.
 A: There should be an additional requirement in the definition.
$D$ is $s$-independent if and only if:

*

*For all $a,b\in D$ and all  $k\in\mathbb{N}$, $s^k(a) \neq b$ and $s^k(b) \neq a$, and

*For all $c\in D$ and all  $k\in \mathbb{N}$, $s^k(0)\neq c$.

Note also that in the definition of $s$−base, it is enough to say that $D$ is a maximal independent subset of $C$. The condition that there is no element $s$-independent from  every element of $D$ follows from this.

The context for this question is that $\Sigma$ is a strongly minimal theory: for any model $M$, every formula with one free variable and parameters from $M$ defines a finite or cofinite subset of $M$. In every strongly minimal theory, we can define a notion of independence: a subset $D$ of a model is independent if and only if for all $d\in D$, $d\notin \mathrm{acl}(D\setminus \{d\})$. In your particular theory  $\Sigma$, the general definition of independence involving $\mathrm{acl}$ specializes to the one I gave above.
Now a basis for a model $M$ is a maximal independent set, and one can prove that if $M$ and $N$ are models of the same strongly minimal theory, then any  bijection between a basis for $M$ and a basis for $N$ extends to an isomorphism $M\cong N$. Thus models are determined up to isomorphism by their dimension: the cardinality of a basis. (And further one can prove that any two bases for the same model  have the same cardinality, so the dimension is a well-defined invariant).
All this generalizes the situation in the most familiar example of a strongly minimal theory: the theory of infinite vector spaces over a fixed field $k$. In this theory, independence is linear independence and dimension is the ordinary dimension of the vector space.
The abstraction to general strongly minimal theories lies at the heart of Morley's Categoricity Theorem and its vast generalization to Shelah's Classification Theory.
A: You are correct that it is necessary that the isomorphism take $0$ to $0'$.  If $C$ is countable it consists of one copy of the naturals and zero or more copies of the integers.  If $C$ and $C'$ are isomorphic they must have the same number of copies of the integers and you can make any element of one copy correspond to any element of any copy in the other set.  There is nothing wrong with requiring that the defined elements be matched up.  Group isomorphisms require that the identities be matched, for example.
