# Does theory have the smallest model

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 decide if $T$ has the smallest model $M$. With smallest model I mean such model, that for any $M'$ of theory $T$ there exist injective homomorphism from $M$ to $M'$. Could somebody please help me with this?

No. Let $C_\infty$ be an infinite chain, and let $C_n$ be a finite loop of length $n$. Then $C_\infty$ is a model of your theory, and so is $\coprod_{n\geq 3} C_n$, the disjoint union of a loop of length $n$ for all $n\in \mathbb{N}$ with $n\geq 3$. But obviously neither of these models embed into the other.
However, any completion of this theory has a smallest model. In my answer to your previous question, I described all completions of this theory and all of their countable models. In any completion, the model with the fewest possible infinite chains ($0$ if the completion allows arbitrarily large finite loops, $1$ if not) will embed into every other model.