Consider a first order language with a single binary relation $R$ and consider the theory of a set $A$, Th($A$), where $A$ is the set containing a single axiom describing that there are exactly three elements: $A = \{ \exists x \exists y \exists z (x \neq y \land x \neq z \land y \neq z \land \forall w(w = x \lor w = y \lor w = z))\}$.
I'm trying to show that i. $T$ has countably infinite many structures up to isomorphism and ii. to find a decision procedure for $T$.
Here are my thoughts: i. I understand how to find a class of countably many structures, e.g., the class of structures whose domains consist of $i$, $i+1$, $i+2$ for any $i \in \mathbb{N}$. We see in my example any two models are isomorphic, but I'm unsure of how to show that there can't be a larger class or isomorphic models of any sort.
ii. Since $A$ contains a single axiom and is therefore enumerable, then its consequences, are enumerable and hence $T$ is enumerable. We may write out all strings representing proofs and by Los Vaught Test (we showed in part i. that $T$ is $\aleph_0$-categorical), we know $T$ is complete so for every formula we must reach at some point a proof of either it or its negation.
I'd appreciate any insight as to whether I am on the right track and hints of any sort would be much appreciated!
There are two key observations here.
The first is purely combinatorial:
The second observation is:
Since this will let us count the models of $T$ up to isomorphism, by counting the models of $T$ with domain $\{a, b, c\}$ up to isomorphism.
So suppose $(\{x, y, z\}; R)$ is a model of $T$. Pick a bijection $f:\{x,y,z\}\rightarrow\{a, b, c\}$; do you see a way to use $f$ to define a relation $S$ on $\{a, b, c\}$ such that $f$ is an isomorphism from $(\{x, y, z\}; R)$ to $(\{a, b, c\}; S)$?
HINT: you're going to define $S$ in terms of $R$ and $h$. Suppose for example that $R(x, y)$ holds; using $h$, does this suggest some fact about the $S$ we want to build?