Find some complete theory $U \supseteq T$

Question

I have a language $L=\{P\}$ with equation, where $P$ is binary predicate symbol. Language's formulas are:

$\varphi \equiv \forall x \forall y (\neg P(x,x) \land (P(x,y) \to P(y,x)))$,

$\psi \equiv \forall x \exists y \exists z(y\ne z \land P(x, y) \land P(x,z) \land \forall v (P(x,v) \to (v=y \lor v =z )))$

and for every $n \in N^+$

$\xi_n \equiv \exists x_1 \exists x_2 ... \exists x_n (\bigwedge \limits _{i=1}^{n-1}P(x_i, x_{i+1}) \land \bigwedge \limits _{i=1 }^{n}\bigwedge \limits _{j=i+1}^{n} x_i \ne x_j)$

Theory $T=\{\varphi,\psi\} \cup \{\xi_n | n \in N^+\}$ is being considered.

In my question models with countable field only are considered(I mean for every implementation $M$ injection from $M$ to $N$ should exist)

I need to find some complete theory $U \supseteq T$ under language $L$ and unambiguously show if some formula $\varphi'$ of language $L$ is such that $\varphi' \in U$. Could somebody please help me with this?

Is it possible to prove that all models of theory $U$ are mutually isomorphic?

Answer 1

Let

$$\sigma_n(x_0,\ldots,x_{n-1})\equiv\bigwedge_{i=0}^{n-1}P(x_i,x_{i+1})\land\bigwedge_{i=1}^{n-1}\bigwedge_{j=i+1}^nx_i\ne x_j\;,$$

where all arithmetic in the subscripts is carried out modulo $n$. Let $\eta_n$ be

$$\exists x_0\,\exists x_1\ldots\exists x_{n-1}\left(\sigma_n(\vec x)\land\forall y_0\,\ldots\forall y_{n-1}\left(\sigma_n(\vec y)\to\bigwedge_{i=0}^{n-1}\tau_{n,i}(\vec x,\vec y)\right)\right)\;,$$

where $\vec x=\langle x_0,\ldots,x_{n-1}\rangle$ and similarly for $\vec y$, and $\tau_{n,i}(\vec x,\vec y)$ is

$$\bigwedge_{k=0}^{n-1}x_k=y_{i+k}\;,$$

where again the arithmetic in the subscripts is modulo $n$. Add $\eta_n$ to $T$ for each $n\ge 3$; the resulting theory is the theory of $2$-regular graphs having exactly one $n$-cycle for each $n\ge 3$. This theory has infinitely many non-isomorphic countable models, since a graph satisfying it can have any countable number of bi-infinite chains.

However, it’s $\kappa$-categorical in every uncountable cardinal $\kappa$ (why?) and therefore complete by the Łoś-Vaught test.