# Complete $n$-types for the theories of $( \mathbb Z , s )$ and $( \mathbb Z , s , < )$

This is exercise 4.5.2 from Marker's Model Theory: An Introduction (p.163), quoted verbatim:

Let $T$ be the theory of $(\mathbb Z,s)$ where $s(x) = x+1$. Determine the types in $S_n(T)$ for each $n$. Which types are isolated? Do the same for $(\mathbb Z, <, s)$.

Here $S_n (T)$ denotes the set of all complete $n$-types $p$ such that $p \cup T$ is satisfiable.

Consider fist Th$({\mathbb Z},s)$. There is only one type in $S_1(T)$, because translations are automorphisms and we can translate any $n\in{\mathbb Z}$ to any $m\in{\mathbb Z}$. The same argument also shows that for each $n\in{\mathbb Z}$ the formula $s^nx=y$ isolates a complete type in $S_2(T)$. In $S_2(T)$ there is one non isolated type $p(x)=\{s^nx\neq y:n\in{\mathbb Z}\}$.

In $S_3(T)$ the isolated types are those that contains the formulas $s^mx=s^ny=z$. There are two sorts of non isolated types (if I count well).

1. contains a formula of the form $s^nx= y$ for some $n\in{\mathbb Z}$ and says $z\neq s^mx$ for every $m\in{\mathbb Z}$ (and similar types obtained permuting the variables);

2. $s^nx\neq y$ and $s^nz\neq y$ and $s^nz\neq x$ for every $n\in{\mathbb Z}$

You may guess $S_n(T)$.

As for the model $({\mathbb Z},<,s)$, the isolated/algebraic types remain the same, non isolated types split into different types. For instance, in $S_2(T)$, the type $p(x)$ above splits into:

a. $p_1(x)=\{s^nx< y:n\in{\mathbb Z}\}$;

b. $p_2(x)=\{s^ny< x:n\in{\mathbb Z}\}$.

(This is the idea, I hope the counting above is right, the details are tedious.)