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\begin{document}
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\title[On the number of countable models of the DMT theory]{\large On the number of countable models of constant and unary predicates expansions of the dense meet-tree theory}


\author[A. B. DAULETIYAROVA]{{\bf A. B. Dauletiyarova}\protect\orcidicon{0000-0003-0051-870X}}

\author[V. V. VERBOVSKIY]{{\bf V. V. Verbovskiy}\protect\orcidicon{0000-0001-5177-8523}}

\address{Aigerim Baisultanovna Dauletiyarova  % Êîíòàêòíûå äàííûå âñåõ àâòîðîâ è ìåñòî ðàáîòû óêàçûâàþòñÿ òîëüêî íà àíãëèéñêîì ÿçûêå
\newline\hphantom{iii} SDU University
\newline\hphantom{iii} Abylai Khan street, 1/1
\newline\hphantom{iii} 040900, Kaskelen, Kazakhstan}%
\email{\textcolor{blue}{d\underline{\,\,\,}aigera95@mail.ru}}%

\address{Viktor Valerievich Verbovskiy  % Êîíòàêòíûå äàííûå âñåõ àâòîðîâ è ìåñòî ðàáîòû óêàçûâàþòñÿ òîëüêî íà àíãëèéñêîì ÿçûêå
\newline\hphantom{iii} Institute of Mathematics and Mathematical Modeling
\newline\hphantom{iii} Shevchenko street, 28
\newline\hphantom{iii} 050010, Almaty, Kazakhstan}%
\email{\textcolor{blue}{verbovskiy@math.kz}}%

%\thanks{\sc Ivanov, I.I.,
%On some problems of commutative algebra}
%\thanks{\copyright \ 2024 Ivanov I.I.}

\thanks{\rm The work is supported by CS of MSHE of RK (grant BR20281002 ``Fundamental
research in mathematics and mathematical modeling'')}
%\thanks{\it Received April, 12, 2024, Published December, 31, 2024}%


\semrtop \vspace{1cm}
\maketitle {\small

%\vspace{-12pt}
%\centerline{{\it Communicated by} {\sc P.P. Petrov}
%({\it filled by Editorial board})
%}

\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} 
In the paper, we investigate Ehrenfeucht theories, that is, theories which have finitely many countable models but which are not
countably categorical. 
More precisely, we count all possible numbers of countable models of the theory DMT of dense meet-trees expanded by several sequences of constants including decreas\-ing ones and by unary predicates with finite realizations. Also, we study the realizations of models over a certain set of formulas based on the Rudin-Keisler preorders on models. 
\medskip

\noindent{\bf Keywords:} Constant expansion, Ehrenfeucht theory, the number of countable models, 
the number of limit models, the number of prime models, small theory, Rudin-Keisler preorder.
\end{quote}
}

\bigskip
\selectlanguage{english}

\section{Introduction}

Since Andrzej Ehrenfeucht had constructed his example of a theory with three non-isomorphic countable models this field of Model Theory, called Ehrenfeucht theories, is one of most important in Model Theory.
One of the main problems here is to build new Ehrenfeucht theories, which are not based 
on a dense linear ordering. One of the first such examples was constructed by M.~Peretyat'kin in \cite{Per}. 
Alistair Lachlan posed the problem if there exists a stable Ehrenfeucht theory.
Sergey Sudoplatov solved this problem by having constructed such a theory \cite{S2009}.
Some other papers related to the topic are 
%\cite{ABKZ2018, BSV, EKS2017, Kul2016, KS2017, Kul2019, KS2020, Kul2020, KS2023, %KS2023(1), Men2021, Millar, S2004, S2007, S2018, ST, TMI}.
\cite{ABKZ2018}--\cite{KS2023(1)}, \cite{Millar}, \cite{S2004}, \cite{S2007}, 
\cite{S2018}--\cite{TMI}.
However, in this paper, we focus on the example by M.~Peretyat'kin, where he considered
a partially densely ordered set, the so-called Dense Meet Tree \cite{Men2021}, DMT for short, expanded by a countable sequence of constants.
While M.~Peretyat'kin considered only one increasing sequence of constants,
we consider several sequences, including decreasing ones, as well as expansions
by unary predicates with finite realizations. 
We give all possible numbers of countable models of such expansions of DMT (Theorem~\ref{mr}).

So, we give the axioms of DMT. A {\em dense meet-tree} ${\mathcal{M}}=\langle M;\leq, \sqcap \rangle$ 
is a lower semilattice (that is, for each elements $a$ and $b$ there exists their greatest lower bound, which we denote by $a\sqcap b$ and call the {\em meet} of $a$ and $b$) without the least and greatest elements such that:

(a) for each pair of incomparable elements, their join does not exist;

(b) for each pair of distinct comparable elements, there is an
    element between them;

(c) for each element $a$ there exist infinitely many pairwise
    incomparable  elements  greater  than  $a$,  whose  infimum  is
    equal to~$a$.

Note that $x$ and $y$ are incomparable if $x \nleq y$ and $y \nleq x$. We denote it by $x \parallel y$.

We study the realizations of models over a certain set of formulas based on the Rudin-Keisler preorders on models. 
The next set of definitions is taken from \cite{S2009}.


A model $\mathcal{M}$ is {\em prime over a type} $p$ if there is a tuple of elements $\bar{a}$ in $\mathcal{M}$ such that $\bar{a}$ is a realization of $p$ and $\mathcal{M}$ is prime over $\bar{a}$.  
We denote a prime model over a type $p$ by $\mathcal{M}_{p}$.
A model $\mathcal{M}$ is {\em almost prime} if it is prime over a realization of some type. 
If a model is not almost prime, we call it a {\em limit model}.



\begin{definition}\rm
Let $p$ and $q$ be types in $S(T)$. We say that
the type $p$ \emph{is dominated by the type $q$}, or $p$ \emph{does not exceed $q$ under the Rudin-Keisler preorder} (written $p \leq_{\mathrm{RK}} q$), if $\mathcal{M}_{q} \models p$, that
is, $\mathcal{M}_{p}$ is an elementary submodel of $\mathcal{M}_{q}$ (written $\mathcal{M}_{p} \preceq \mathcal{M}_{q}$).

Besides, we say that a model $\mathcal{M}_{p}$ \emph{is dominated by a model $\mathcal{M}_{q}$},
or \emph{$\mathcal{M}_{p}$ does not exceed $\mathcal{M}_{q}$ under the Rudin-Keisler preorder}, and
write $\mathcal{M}_{p} \leq_{\mathrm{RK}} \mathcal{M}_{q}$.
\end{definition}

\begin{definition}\rm
Types $p$ and $q$ are said to be \emph{domination-equivalent}, \emph{realization}-\emph{equivalent},
\emph{Rudin-Keisler equivalent}, or RK-\emph{equivalent} (written $p \sim _{\textrm{RK}} q$) if $p \leq_{\mathrm{RK}} q$ and $q \leq_{\mathrm{RK}} p$. 

Models $\mathcal{M}_{p}$ and $\mathcal{M}_{q}$ are said to be \emph{domination-equivalent, Rudin-Keisler equivalent}, or RK-\emph{equivalent} (written $\mathcal{M}_{p} \sim _{\textrm{RK}} \mathcal{M}_{q}$).
\end{definition}




\section{Constant expansions of the DMT theory}

In this section, we expand $T_{\text{dmt}}$  with the signature $\mathcal{L}_{\text{dmt}}=\{ \leq, \sqcap \}$ 
by countable sequences of constants and
%and countably many disjoint nonempty unary predicates with finite realizations, 
find all possible values $I(T, \omega)$ for these expansions $T \supseteq T_{\text{dmt}}$.

Let us extend $\mathcal{L}_{\text{dmt}}$ to $\mathcal{L}_0$ by constants $c_{k}^{(0)}$, $k \in \omega$,
and extend the theory $T_{\text{dmt}}$ to $T_{0}$, so that constants $c_{k}^{(0)}$, $k \in \omega$, form a strictly increasing sequence. In this case, the signature of  $T_{0}$ is $\mathcal{L}_{0} = \mathcal{L}_{\text{dmt}} \cup \{ c_{k}^{(0)} \mid k \in \omega \}$ and $T_{0} = T_{\text{dmt}} \cup \{ c_{k}^{(0)} < c_{k+1}^{(0)} \mid k \in \omega \}$. 
Note that the theory $T_{0}$ was constructed by Peretyat'kin in \cite{Per}, 
where he proved that it is Ehrenfeucht, namely, 
$T_{0}$ has exactly three countable models: the prime model, the saturated model, 
and the prime model over the realization of the powerful type $p_{0}(x)$, 
isolated by the set of formulas $\{ c_{k}^{(0)}<x \mid k \in \omega\}$. In Figure \ref{fig1}, we illustrate all possible realizations of the type $p_{0}$ (a.) and represent the Hasse diagram of Rudin-Keisler preorder $\leq_{\textrm{RK}}$ on the set of countable models (up to isomorphisms) of $T_{0}$ (b.).

\begin{figure}[!htb]
\centering
\begin{picture}(310, 95)

\put(30,20){\vector(0,1){50}} %first realizations
\put(10,40){\scriptsize{$c_{k}^{(0)}$}}
\put(25,82){$\varnothing$}

\put(100,20){\vector(0,1){50}} %second realizations
\put(80,40){\scriptsize{$c_{k}^{(0)}$}}
\put(100,85){\circle*{4}}
\put(105,83){\small{$x$}}
\put(98,0){\scriptsize{a.}}

\put(170,20){\vector(0,1){50}} %third realizations
\put(150,40){\scriptsize{$c_{k}^{(0)}$}}
\put(170,85){\circle{4}}
\put(175,83){\small{$x$}}

\put(270,20){\circle*{3}} %Hasse diagram, countable models over an empty set
\put(270,20){\circle{8}}
\put(280,17){0}
\put(270,85){\circle*{3}} %Hasse diagram, countable models over the type p_{0}
\put(270,85){\circle{8}}
\put(280,82){1}
\put(247,82){$p_{0}$}
\put(270,20){\line(0,1){65}} 
\put(268,0){\scriptsize{b.}}
\end{picture}
\caption{\scriptsize{From left to right: a prime model, where there is no element greater than all $c_{k}^{(0)}$, a prime model, where among the elements of large $c_{k}^{(0)}$ there is the smallest element, and a limit model where among the elements of large $c_{k}^{(0)}$ there is no the smallest element (a.); RK($T_{0}$) is a linear order of two elements. In Hasse diagrams, we draw the realization of prime models over a certain set of formulas using the character $\bullet$ based on the Rudin-Keisler preorder $\leq_{\text{RK}}$ on the models. The circled circle $\bigcirc$ above the $\bullet$ means the possibility of realization of limit models over these sets of formulas. It should be noted here that the total number of characters $\bullet$ is the total number of prime models, and the numbers next to $\odot$ mean the number of limit models (b.).}}
\label{fig1}
\end{figure}

Now we construct a theory $T_{1}$. To do this, we expand  $\mathcal{L}_{0}$ with a strictly decreasing sequence of constants $c_{k}^{(1)}$, $k \in \omega$. %so that $c_{k}^{(0)} < c_{k}^{(1)}, k \in \omega$.
We put $\mathcal{L}_{1} = \mathcal{L}_{0} \cup \{ c_{k}^{(1)} \mid k \in \omega \}$ and $T_{1}=T_{0} \cup \{ c_{k}^{(1)}>c_{k+1}^{(1)} \mid k \in \omega \} \cup \{ c_{k}^{(0)}<{c}_{k}^{(1)} \mid k \in \omega \}$. 

It is straightforward to show that there exist exactly two non-principal $1$-types over an empty set in $T_1$:
$$
p(x)  =  p_{0}(x) \cup p_{1}(x)  = \{ c^{(0)}_{k} < x \mid k \in \omega\} \cup 
\{ x < c^{(1)}_{k} \mid k \in \omega \}
$$
and
$$
\bar{\bar{p}}(x)  =  p_{0}(x) \cup \bar{\bar{p}}_1(x)  = \{ c^{(0)}_{k} < x \mid k \in \omega\} \cup 
\{ x \parallel c^{(1)}_{k} \mid k \in \omega \}
$$
Here, $p_{1}(x)  = \{ x < c^{(1)}_{k} \mid k \in \omega \}$ and $\bar{\bar{p}}_1(x) = \{ x \parallel c^{(1)}_{k} \mid k \in \omega \}$.

We can write three completions of $p(x)\cup p(y)$. They are defined by the following sets of formulas:
\setlength\arraycolsep{2pt}
\begin{eqnarray*}
q_{1}(x, y) & = & p(x) \cup p(y)  \cup \{ x < y \}; \\
q_{2}(x, y) & = & p(x) \cup p(y)  \cup \{ y < x \}; \\
q_{3}(x, y) & = & p(x) \cup p(y)  \cup \{ x = y \}. %\\
%q_{4}(x, y) & = & p(x) \cup p(y)  \cup \{ x \parallel y \}.
\end{eqnarray*}

\noindent The reason for the existence of exactly three types involving two variables is the following: 
due to the quantifier elimination result and obvious logical equivalences of negation of these formulas, 
there are just four formulas in two variables: 
$x < y$, $y < x$, $x = y$, and $x \parallel y$.
Note that here the types $q_{1}(x, y)$ and $q_{2}(x, y)$ are the same up to a permutation of variables, the type $q_{3}(x, y)$ is logically equivalent to  $p(x)$, and, as it is easy to check that $p(x) \cup p(y)  \cup \{ x \parallel y \}$ is inconsistent. 
Indeed, let $(a,b) \models p(x) \cup p(y)  \cup \{ x \parallel y \}$. Then both $a$ and $b$ are less than $c^{(1)}_1$.
By Axiom~1 of $T_{\rm dmt}$ the elements $a$ and $b$ are comparable, for a contradiction.

Similarly, there are two completions of $p(x)\cup \bar{\bar{p}}(y)$. They are defined by the following sets of formulas:
\setlength\arraycolsep{2pt}
\begin{eqnarray*}
q_{4}(x, y) & = & p(x) \cup \bar{\bar{p}}(y)  \cup \{ x < y \}; \\
q_{5}(x, y) & = & p(x) \cup \bar{\bar{p}}(y)  \cup \{ x \parallel y \}.
\end{eqnarray*}

\begin{remark}
It holds that $p(x) \sim_{\text{RK}} q_{4}(x, y)$. Indeed, let $a_{1} \models p(x)$. By Axiom 3 of $T_{\rm dmt}$ there exist infinitely many pairwise incomparable elements $b_{1}$, $b_{2}$, \dots, $b_{n}$, $\dots$, such that $b_i\sqcap b_j = a_1$ for any $1\le i<j<\omega$. Assume that $b_{i}$ and $b_j$ are less than $c_{k}^{(1)}$ for some $i$, $j$, and $k \in \omega$. Then $b_i$ and $b_j$ are comparable by Axiom 1 of $T_{\rm dmt}$, for a contradiction. So, there is some $b_{i}$ such that $b_{i} > a_{1}$ and $b_{i} \parallel c_{k}^{(1)}$. 
Then $(a_1, b_i) \models q_{4}(x, y)$. Hence, $q_4(x,y) \le_{RK} p(x)$.

Since $p(x)\subseteq 
q_4(x,y)$, so $p(x)\le_{RK}  q_4(x,y)$.
Thus, $p(x)\sim_{RK}  q_4(x,y)$.
\end{remark}

\begin{remark}
It holds that $q_1 \sim_{\text{RK}} q_{5}$. Indeed, let $(a_1, a_{2}) \models q_1(x,y)$. 
As above, there exists $b$ such that $a_1 < b$ and $b\parallel a_2$. 
Then $(a_2, b) \models q_5$ and $q_5\le_{RK}  q_1$.

Conversely, let $(a_2,b)\models q_5$ and let $a_1 = a_2 \cap b$. 
Then $(a_1, b)\models q_1$, so $q_1\le_{RK}  q_5$.
Thus, $q_1\sim_{RK}  q_5$.
\end{remark}

We have the following pairwise non-isomorphic countable models of $T_{1}$:

$\bullet$ a prime model;% with empty set of realizations of  $p(x)$; 

$\bullet$ a prime model over a realization of $p(x)$, with a unique realization of this type;

$\bullet$ a prime model over the realization of $q_{1}(x, y)$ forming a closed interval $[a, b]$ with $c_{k}^{(0)}< a \leq b < c_{k}^{(1)}$;

$\bullet$ three limit models over the type $q_{1}(x, y)$, in which the sets of realizations of $p(x)$ are  $ [a, b), (a, b], (a, b)$, correspondingly. 

In Figure \ref{fig2}, we illustrate all possible realizations of the type $p(x)$ and represent the Hasse diagram of Rudin-Keisler preorders $\leq_{\textrm{RK}}$ on the set of countable models (up to isomorphisms) of the theory $T_{1}$, respectively.

\begin{figure}[!htb]
\begin{picture}(330,150)

\put(30, 20){\vector(0,1){40}}
\put(10, 35){\scriptsize{$c_{n}^{(0)}$}}

\put(30, 140){\vector(0,-1){40}}
\put(10, 120){\scriptsize{${c}_{n}^{(1)}$}}


\put(25, 77){$\varnothing$}

%\put(50, 55){\tiny{$I_{p}(T, p)=1$}}

%type p(x_1)


\put(70, 20){\vector(0,1){40}}
\put(50, 35){\scriptsize{$c_{k}^{(0)}$}}

\put(70, 140){\vector(0,-1){40}}
\put(50, 120){\scriptsize{${c}_{k}^{(1)}$}}


\put(70, 80){\circle*{3}}
\put(73, 78){\scriptsize{$x$}}

%\put(150, 55){\tiny{$I_{p}(T, p)=1$}}

%type q(x_1,x_2)



\put(110, 20){\vector(0,1){40}}
\put(90, 35){\scriptsize{$c_{k}^{(0)}$}}

\put(110, 140){\vector(0,-1){40}}
\put(90, 120){\scriptsize{${c}_{k}^{(1)}$}}

\put(110, 74){\circle*{3}}
\put(110, 86){\circle*{3}}
\put(110, 75.6){\line(0,1){10}}
\put(112, 72){\scriptsize{$x$}}
\put(112, 84){\scriptsize{$y$}}

%\put(225, 55){\tiny{$I_{p}(T, q)=1$, $I_{l}(T, q)=3$}}

\put(150, 20){\vector(0,1){40}}
\put(130, 35){\scriptsize{$c_{k}^{(0)}$}}

\put(150, 140){\vector(0,-1){40}}
\put(130, 120){\scriptsize{${c}_{k}^{(1)}$}}


\put(150, 74){\circle*{3}}
\put(150, 86){\circle{3}}
\put(150, 75.6){\line(0,1){9.2}}
\put(152, 72){\scriptsize{$x$}}
\put(152, 84){\scriptsize{$y$}}

%%%%%%%%%%%%%%%%%%

\put(190, 20){\vector(0,1){40}}
\put(170, 35){\scriptsize{$c_{k}^{(0)}$}}

\put(190, 140){\vector(0,-1){40}}
\put(170, 120){\scriptsize{${c}_{k}^{(1)}$}}

\put(190, 74){\circle{3}}
\put(190, 86){\circle*{3}}
\put(190, 75.6){\line(0,1){10}}
\put(192, 72){\scriptsize{$x$}}
\put(192, 84){\scriptsize{$y$}}

%%%%%%%%%%%%%%%%%%

\put(230, 20){\vector(0,1){40}}
\put(210, 35){\scriptsize{$c_{k}^{(0)}$}}

\put(230, 140){\vector(0,-1){40}}
\put(210, 120){\scriptsize{${c}_{k}^{(1)}$}}

\put(230, 74){\circle{3}}
\put(230, 86){\circle{3}}
\put(230, 75.6){\line(0,1){9.1}}
\put(232, 72){\scriptsize{$x$}}
\put(232, 84){\scriptsize{$y$}}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\put(300,25){\circle*{3}}
\put(300,25){\circle{8}}
\put(310,22){0}

\put(300,25){\line(0,1){110}}

\put(300,80){\circle*{3}}
\put(300,80){\circle{8}}
\put(310,77){0}
\put(285,78){$p$}

\put(300,135){\circle*{3}}
\put(300,135){\circle{8}}
\put(310,132){3}
\put(285,133){$q_1$}

\put(130,0){\scriptsize{a.}}
\put(300,0){\scriptsize{b.}}
\end{picture}
\label{fig2}
\caption{\scriptsize{Realizations of $p(x)$ (a.); and Hasse diagram of RK($T_{1}$) (b.)}}
\end{figure}

So, the next is clear.
\begin{theorem}
The theory $T_1$ has exactly 6 countable models up to isomorphism.
\end{theorem}

Starting with the theory $T_{2}$, the situation looks a little different. First, we define the signature $\mathcal{L}_{2}$ for  $T_{2}$ as $\mathcal{L}_{1} \cup \{ c_{k}^{(2)} \mid k \in \omega \}$, where $c_{k}^{(2)}, k \in \omega$ is also a strictly decreasing sequence of constants on the tree, additionally $c_{k}^{(0)}, k \in \omega $ is comparable with $c_{k}^{(1)}, k \in \omega $ and $c_{k}^{(2)}, k \in \omega $, but $c_{k}^{(1)}, k \in \omega $ and $c_{k}^{(2)}, k \in \omega $ are incomparable. Therefore, the theory $T_{2}$ has the following form 
\setlength\arraycolsep{2pt}
\begin{eqnarray*}
T_{2} & = & T_{1}  \cup  \{ c_{k}^{(2)}>c_{k+1}^{(2)} \mid k \in \omega \} %   \cup {} \nonumber\\ & &   {} \quad 
\cup  \{ c_{k}^{(0)}<{c}_{k}^{(2)} \mid k \in \omega \}  \cup  \{ c_{k}^{(1)} \parallel {c}_{k}^{(2)} \mid k \in \omega \}. 
\end{eqnarray*}

\begin{lemma}
The meet of incomparable elements of the strictly decreasing sequences of constants $(c_{k}^{(1)})_{k \in \omega}$, $(c_{k}^{(2)})_{k \in \omega}$ does not depend on their representative constants, that is,
$c_l^{(1)} \sqcap c_m^{(2)} = c_i^{(1)}\sqcap c_j^{(2)}$ for any pairs
$(l, m)$ and $(i, j)$.
\end{lemma}

\begin{proof}
Since $c^{(k)}_i > c^{(k)}_{i+1}$ for both $k$, we obtain that
$$c_{i}^{(1)} \sqcap c_{i}^{(2)} \ge 
c_{i+1}^{(1)} \sqcap c_{i+1}^{(2)}
$$
Assume that $d_{12} \bydef c_{i}^{(1)} \sqcap c_{i}^{(2)} >
c_{i+1}^{(1)} \sqcap c_{i+1}^{(2)}$.
So, $\neg (d_{12} \le c^{(1)}_{i+1} \land d_{12} \le c^{(2)}_{i+1})$
holds. Both $d_{12}$ and $c^{(k)}_{i+1}$ are less than $c^{(k)}_{i}$,
so they are comparable, and $\neg(d_{12} \le c^{(k)}_{i+1})$ is equivalent $d_{12} > c^{(k)}_{i+1}$, for both $k$. 
Then 
$$\neg (d_{12} \le c^{(1)}_{i+1} \land d_{12} \le c^{(2)}_{i+1})
\Leftrightarrow
(d_{12} > c^{(1)}_{i+1} \lor d_{12} > c^{(2)}_{i+1})$$
If $d_{12}>c^{(1)}_{i+1}$ then by transitivity we obtain that
$c^{(2)}_{i} > c^{(1)}_{i+1}$, because $c^{(2)}_{i}>d_{12}$, for a contradiction. The case $d_{12}>c^{(2)}_{i+1}$ is similar.

By mathematical induction, we obtain that 
$c_{i}^{(1)} \sqcap c_{i}^{(2)} = 
c_{j}^{(1)} \sqcap c_{j}^{(2)}
$ for all $i$ and $j<\omega$.
Now, let $i<j$. Then
$$
c_{i}^{(1)} \sqcap c_{i}^{(2)} \ge 
c_{i}^{(1)} \sqcap c_{j}^{(2)} \ge
c_{j}^{(1)} \sqcap c_{j}^{(2)} = 
c_{i}^{(1)} \sqcap c_{i}^{(2)}
$$
that proves the lemma.
\end{proof}

Let $d_{1, 2}= c_{k}^{(1)} \sqcap c_{k}^{(2)}$, $k \in \omega$. 
The element $d_{1, 2}$ exists in any model of $T_{2}$. 
So, we redefine our types as follows

\begin{eqnarray*}
    p_{0}(x_{0}) & = & \{ c_{k}^{(0)} < x_{0} \mid k \in \omega  \} \cup \{ x_{0} < d_{1, 2} \}; \\
    p_{1}(x_{0}) & = & \{ x_{0} < c_{k}^{(1)} \mid k \in \omega  \} \cup \{  d_{1, 2} < x_{0}\}; \\
    p_{2}(x_{0}) & = & \{ x_{0} < c_{k}^{(2)} \mid k \in \omega  \} \cup \{ d_{1, 2} < x_{0} \}.
\end{eqnarray*}

\bigskip 

\noindent Note that $p_{0}(x_{0}) \cup p_{1}(x_{1}) \cup p_{2}(x_{2})$ defines a complete type. Then we can consider each type separately, by analogy with the theory $T_{0}$. Therefore, according to the orthogonality of the types $p_{0}$, $p_{1}$, and $p_{2}$, two prime models and one limit model arise over the realization of each type $p_{i}(x_{i})$ for any $i \in \{0, 1, 2 \}$. 
In total, $I(T_{2}, \omega)=3^{3}=27$, where $2^3=8$ of them are prime models (over some set).

In Figure \ref{fig3}, we illustrate all possible realizations of the types $p_{0}$, $p_{1}$ and $p_{2}$ (a.) and represent the Hasse diagram of Rudin-Keisler preorder $\leq_{\textrm{RK}}$ on the set of countable models (up to isomorphisms) of $T_{2}$ (b.).

\begin{figure}[!htb]
\begin{picture}(310,170)

\put(80, 20){\vector(0,1){30}}
\put(60, 25){\scriptsize{$c_{k}^{(0)}$}}

\put(22, 145){\vector(1,-1){30}}
\put(23, 120){\scriptsize{${c}_{k}^{(1)}$}}

\put(139, 145){\vector(-1,-1){30}}
\put(120, 120){\scriptsize{${c}_{k}^{(2)}$}}

\put(78, 62){$\ast$}
\put(58, 102){$\ast$}
\put(98, 102){$\ast$}

\multiput(63,102)(2,-2){10}%
{\circle*{1}}
\multiput(98,102)(-2,-2){10}%
{\circle*{1}}
\multiput(80.5,84)(0,-2){10}%
{\circle*{1}}

\put(80.5, 84){\circle*{3}}
\put(85, 84){\scriptsize{$d_{1, 2}$}}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\put(220,45){\line(1,1){30}}
\put(220,45){\line(-1,1){30}}
\put(250,75){\line(-1,1){30}}
\put(190,75){\line(1,1){30}}
\put(220,45){\circle*{3}} \put(220,45){\circle{8}}
\put(218,28){\small{0}}
\put(190,75){\circle*{3}}
\put(190,75){\circle{8}} \put(173,73){\small{$p_{0}$}}
\put(198,72){\small{1}}
\put(250,75){\circle*{3}}
\put(250,75){\circle{8}} 
\put(258,73){\small{$p_{2}$}}
\put(248,60){\small{1}}
\put(220,105){\circle*{3}}
\put(220,105){\circle{8}} 
\put(207,111){\small{$q_{1}$}}
\put(227,102){\small{3}}
\put(220,45){\line(0,1){30}}
\put(190,75){\line(0,1){30}}
\put(250,75){\line(0,1){30}}
\put(220,105){\line(0,1){30}}
\put(220,75){\line(1,1){30}}
\put(220,75){\line(-1,1){30}}
\put(250,105){\line(-1,1){30}}
\put(190,105){\line(1,1){30}}
\put(220,75){\circle*{3}} \put(220,75){\circle{8}}
\put(218,84){\small{$p_{1}$}}
\put(228,72){\small{1}}
\put(190,105){\circle*{3}}
\put(190,105){\circle{8}} 
\put(174,103){\small{$q_{0}$}}
\put(198,102){\small{3}}
\put(250,105){\circle*{3}}
\put(250,105){\circle{8}} 
\put(245,115){\small{$q_{2}$}}
\put(258,102){\small{3}}
\put(220,135){\circle*{3}}
\put(220,135){\circle{8}} 
\put(217,144){\small{$r$}}
\put(228,133){\small{7}}

\put(78,0){\scriptsize{a.}}
\put(218,0){\scriptsize{b.}}
\end{picture}
\caption{\scriptsize{Here, $d_{1, 2}$ is the meet of strictly decreasing sequences of  constants $(c_{k}^{(1)})_{k \in \omega}$ and $(c_{k}^{(2)})_{k \in \omega}$, and the character * means one of the $\{ \varnothing, \bullet,  \circ \}$ (a.); RK($T_{2}$) is a Hasse diagram with boolean of $\{p_{0}, p_{1}, p_{2}\}$, here the type $q_{0} = p_{0} \cup p_{1}$, $q_{1} = p_{0} \cup p_{2}$, $q_{2} = p_{1} \cup p_{2}$, $r= p_{0} \cup p_{1} \cup p_{2}$ (b.).}} 
\label{fig3}
\end{figure}

But in the theory 
\begin{eqnarray*}
T_{3}  & = & T_{2}  \cup  \{ c_{k}^{(3)}>c_{k+1}^{(3)} \mid k \in \omega \}  \cup  
\{ c_{k}^{(1)} \parallel {c}_{k}^{(3)} \mid k \in \omega \} \cup 
\\
&  & \hphantom{T_{2}}\cup \{ c_{k}^{(0)}<{c}_{k\hphantom{+1}}^{(3)} \mid k \in \omega \}
 \cup \{ c_{k}^{(2)} \parallel {c}_{k}^{(3)} \mid k \in \omega \}
\end{eqnarray*} 
with the signature $\mathcal{L}_{3} = \mathcal{L}_{2} \cup \{ c_{k}^{(3)} \mid k \in \omega \}$ there are three meets: 
$$d_{1, 2}=c_{k}^{(1)} \sqcap c_{k}^{(2)},\ d_{1, 3}=c_{k}^{(1)} \sqcap c_{k}^{(3)}, \mbox{ and } d_{2, 3}=c_{k}^{(2)} \sqcap c_{k}^{(3)}
\mbox{ for any } k \in \omega.$$ 
Note that any two of $d_{1, 2}$, $d_{1, 3}$, $d_{2, 3}$ are comparable. 
Since $d_{1, 2}<c_{k}^{(1)}$ and $d_{1, 3}<c_{k}^{(1)}$, then $d_{1, 2}$ and $d_{1, 3}$ are comparable. A similar situation is repeated for pairs $d_{1, 2}$ and $d_{2, 3}$; $d_{1, 3}$ and $d_{2, 3}$. 

Let $d_{1, 2} \leq d_{2, 3}$. Since $d_{1, 2} < c_{k}^{(1)}$ and $d_{1, 2} \leq d_{2, 3} < c_{k}^{(3)}$ for any $k \in \omega$, we have $d_{1, 2} \leq d_{1, 3}$. If $d_{2, 3} \leq d_{1, 3}$, then $d_{2, 3} \leq d_{1, 2}$, which contradicts $d_{1, 2} \leq d_{2, 3}$, because $d_{2, 3} < c_{k}^{(2)}$ and $d_{2, 3} \leq d_{1, 3} < c_{k}^{(1)}$. Hence, $d_{1, 3} < d_{2, 3}$. By the fact $d_{1, 3} < c_{k}^{(1)}$ and $d_{1, 3} \leq d_{2, 3} < c_{k}^{(2)}$, we have $d_{1, 3} \leq d_{1, 2}$. 
Thus, there are two cases:

{\it Case 1.} Assume that $d_{1, 2} = d_{1, 3} = d_{2, 3}$. 
Then $T_3\cup \{d_{1, 2} = d_{1, 3} = d_{2, 3}\}$ is complete.
Then the types of this theory are defined by the following sets of formulas for every $i \in \{1, 2, 3\}$: 
\begin{eqnarray*}
    p_{0}(x_{0}) & = & \{ c_{k}^{(0)} < x_{0} \mid k \in \omega  \} \cup \{ x_{0} < d_{1, 2} \};\\ %\cup \{ d_{1, 2}=d_{1, 3}=d_{2, 3} \} ; \\
    p_{i}(x_{0}) & = & \{ x_{0} < c_{k}^{(i)} \mid k \in \omega  \} \cup \{  d_{1, 2} < x_{0}\}.% \cup \{ d_{1, 2}=d_{1, 3}=d_{2, 3} \}.
\end{eqnarray*}

Since the union of types $p_{0}(x_{0}) \cup p_{1}(x_{1}) \cup p_{2}(x_{2}) \cup p_{3}(x_{3})$ defines a complete type and these types are orthogonal, we obtain $3^{4}=81$ countably pairwise non-isomorphic models over the realizations of these types, where $2^{4}=16$ of them are prime models.

{\it Case 2.} Assume now that $d_{1, 2} = d_{1, 3} < d_{2, 3}$. As in the previous case, $T_3\cup \{d_{1, 2} = d_{1, 3} < d_{2, 3}\}$ is complete and
we define the types of this theory:
\begin{eqnarray*}
    p_{0}(x_{0}) & = & \{ c_{k}^{(0)} < x_{0} \mid k \in \omega  \} \cup \{ x_{0} < d_{1, 2} \};\\% \cup \{ d_{1, 2}=d_{1, 3}<d_{2, 3} \}; \\
    p_{1}(x_{0}) & = & \{ x_{0} < c_{k}^{(1)} \mid k \in \omega  \} \cup \{  d_{1, 2} < x_{0}\}\\% \cup \{ d_{1, 2}=d_{1, 3}<d_{2, 3} \}.
		p_{i}(x_{0}) & = & \{ x_{0} < c_{k}^{(i)} \mid k \in \omega  \} \cup \{  d_{2, 3} < x_{0}\}, \mbox{ for } i = 2, 3.% \cup \{ d_{1, 2}=d_{1, 3}<d_{2, 3} \}.
\end{eqnarray*}

Therefore, we obtain the $3^{4}$ of countable models over
the realizations of these types as in the first case. 

Note that, in the each cases $d_{1, 2} = d_{2, 3} < d_{1, 3}$ and $d_{1, 3} = d_{2, 3} < d_{1, 2}$, we also obtain $3^4$ countable models.

Below we illustrate the above-described cases for elements $d_{1, 2}$, $d_{1, 3}$ and $d_{2, 3}$ in Figure \ref{fig4}.

\begin{figure}[!htb]
\begin{picture}(325,160)

\put(0, 115){$\ast$}
\put(20, 115){$\ast$}
\put(40, 115){$\ast$}
\put(20, 50){$\ast$}

\put(23, 75){\circle*{3}}
\put(27, 75){\scriptsize{$d_{1, 2}=d_{1, 3}=d_{2, 3}$}}

\multiput(3,115)(3,-6){7}%
{\circle*{1}}
\multiput(23,115)(0,-6){7}%
{\circle*{1}}
\multiput(42,115)(-3,-6){7}%
{\circle*{1}}
\multiput(23,75)(0,-3){7}%
{\circle*{1}}


\put(3, 158){\vector(0,-1){30}}
\put(-12, 148){\scriptsize{$c_{k}^{(1)}$}}
\put(23, 158){\vector(0,-1){30}}
\put(8, 148){\scriptsize{$c_{k}^{(2)}$}}
\put(43, 158){\vector(0,-1){30}}
\put(28, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(23, 10){\vector(0,1){30}}
\put(8, 20){\scriptsize{$c_{k}^{(0)}$}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\put(90, 115){$\ast$}
\put(110, 115){$\ast$}
\put(130, 115){$\ast$}
\put(110, 50){$\ast$}


\put(113, 95){\circle*{3}}
\put(117, 95){\scriptsize{$d_{2, 3}$}}
\put(113, 75){\circle*{3}}
\put(117, 75){\scriptsize{$d_{1, 2}=d_{1, 3}$}}

\multiput(93,115)(3,-6){7}%
{\circle*{1}}
\multiput(113,115)(0,-6){7}%
{\circle*{1}}
\multiput(131,115)(-3,-3){7}%
{\circle*{1}}
\multiput(113,75)(0,-3){7}%
{\circle*{1}}


\put(93, 158){\vector(0,-1){30}}
\put(78, 148){\scriptsize{$c_{k}^{(1)}$}}
\put(113, 158){\vector(0,-1){30}}
\put(98, 148){\scriptsize{$c_{k}^{(2)}$}}
\put(133, 158){\vector(0,-1){30}}
\put(118, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(113, 10){\vector(0,1){30}}
\put(98, 20){\scriptsize{$c_{k}^{(0)}$}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\put(180, 115){$\ast$}
\put(200, 115){$\ast$}
\put(220, 115){$\ast$}
\put(200, 50){$\ast$}


\put(203, 95){\circle*{3}}
\put(185, 95){\scriptsize{$d_{1, 3}$}}
\put(203, 75){\circle*{3}}
\put(207, 75){\scriptsize{$d_{1, 2}=d_{2, 3}$}}

\multiput(184,115)(3,-3){7}%
{\circle*{1}}
\multiput(203,115)(0,-6){7}%
{\circle*{1}}
\multiput(222,115)(-3,-6){7}%
{\circle*{1}}
\multiput(203,75)(0,-3){7}%
{\circle*{1}}


\put(183, 158){\vector(0,-1){30}}
\put(168, 148){\scriptsize{$c_{k}^{(1)}$}}
\put(203, 158){\vector(0,-1){30}}
\put(188, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(223, 158){\vector(0,-1){30}}
\put(208, 148){\scriptsize{$c_{k}^{(2)}$}}
\put(203, 10){\vector(0,1){30}}
\put(188, 20){\scriptsize{$c_{k}^{(0)}$}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\put(270, 115){$\ast$}
\put(290, 115){$\ast$}
\put(310, 115){$\ast$}
\put(290, 50){$\ast$}


\put(293, 95){\circle*{3}}
\put(275, 95){\scriptsize{$d_{1, 2}$}}
\put(293, 75){\circle*{3}}
\put(297, 75){\scriptsize{$d_{1, 3}=d_{2, 3}$}}

\multiput(274,115)(3,-3){7}%
{\circle*{1}}
\multiput(293,115)(0,-6){7}%
{\circle*{1}}
\multiput(313,115)(-3,-6){7}%
{\circle*{1}}
\multiput(293,75)(0,-3){7}%
{\circle*{1}}


\put(273, 158){\vector(0,-1){30}}
\put(258, 148){\scriptsize{$c_{k}^{(1)}$}}
\put(293, 158){\vector(0,-1){30}}
\put(278, 148){\scriptsize{$c_{k}^{(2)}$}}
\put(313, 158){\vector(0,-1){30}}
\put(298, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(293, 10){\vector(0,1){30}}
\put(278, 20){\scriptsize{$c_{k}^{(0)}$}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\put(78,0){\scriptsize{a.}}
%\put(218,0){\scriptsize{b.}}



\end{picture}
\caption{\scriptsize{}}
\label{fig4}
\end{figure}

Let us proceed to consider the case of arbitrary $n \geq 2$. 
The signature $\mathcal{L}_{n} = \mathcal{L}_{n-1} \cup \{ c_{k}^{(n)} \mid k \in \omega \}$ 
and the theory $T_{n}$ is given as 
\begin{multline*}
T_n  = T_{n-1} \cup \{ c_{k}^{(n)} > c_{k+1}^{(n)} \mid k \in \omega \} \cup \{ c_{k}^{(0)} < c_{k}^{(n)} \mid k \in \omega \} \cup \\
 \cup  \{ c_{i}^{(n)} \parallel c_{j}^{(m)} \mid m<n, \forall i, j \in \omega \}.
\end{multline*}
 In this theory, we have $C(n,2)$ meets: $d_{1, 2}$, $d_{1, 3}$, $\dots$, $d_{n-1, n}$.

%d_{i_{n-1} i_{n}}$. 

Here it is enough to consider the case when $d_{1, 2}=d_{1, 3}= \dots =d_{n-1, n}$, since other cases give the same number of models. The list of complete types is defined as follows
(here, $i \in \{1, 2, \dots, n\}$): 
\begin{eqnarray*}
    p_{0}(x) & = & \{ c_{k}^{(0)} < x \mid k \in \omega  \} \cup \{ x < d_{1, 2} \}; \\ % \cup \{ d_{i_{1} i_{2}}=d_{i_{1} i_{3}}= \dots =d_{i_{n-1} i_{n}} \} ; \\
    p_{i}(x) & = & \{ x < c_{k}^{(i)} \mid k \in \omega  \} \cup \{  d_{1, 2} < x\}. % \cup \{ d_{i_{1} i_{2}}=d_{i_{1} i_{3}}= \dots =d_{i_{n-1} i_{n}} \}.
\end{eqnarray*}

 Thus, we have the following:

 \begin{theorem}
Let $T_{n}$, where $n\geq 2$, be a countable constant expansion of the dense meet-tree theory
$T_{\mathrm{dmt}}$ with an increasing sequence of constants $(c_{k}^{(0)})_{k \in \omega}$ and 
$n$ decreasing sequences of constants $(c_{k}^{(1)})_{k \in \omega}, \dots, (c_{k}^{(n)})_{k \in \omega}$, so that $c_{k}^{(0)} < c_{k}^{(1)}, \dots, c_{k}^{(0)} < c_{k}^{(n)}$, $k \in \omega$ and $c_{k}^{(j)} \parallel c_{k}^{(t)}$ for each $1 \leq j \neq t \leq n$. Then $T$ has exactly $3^{n+1}$ countable models, where $2^{n+1}$ of them are prime models.
 \end{theorem}

Note that the number of limit models 
is equal to subtraction of the number of
prime models from the total number of countable models, i.e. $3^{n+1}-2^{n+1}$. 

Since the complete type $p_{0}(x_{0}) \cup p_{1}(x_{1}), \dots p_{n}(x_{n})$ is powerful, then the Rudin-Keisler preorder $\leq_{\text{RK}}$ is a Boolean of the $(n+1)$-element set.

If there are several sequences that are increasing, but pairwise incomparable, and each one has several sequences that are decreasing from above, then this can be considered as a disjunctive union. Then the following theorem will be true.

\begin{theorem}\label{contantsss}
Let $T$ be a countable constant expansion of the dense meet-tree theory
$T_{\text{dmt}}$ with 
$n$ increasing sequences of constants $(c_{k}^{(i, 0)})_{k \in \omega}$ 
such that 
$$
\sup\limits_{k\in \omega} c_{k}^{(i, 0)} \ne \sup\limits_{k\in \omega} c_{k}^{(j, 0)} \mbox { for all } i\ne j.
$$
Let for each $i$ there are $\tau (i)$  decreasing sequences %of constants 
$(c_{k}^{(1, i)})_{k \in \omega}$, \dots, $(c_{k}^{(\tau(i), i)})_{k \in \omega}$, such that $c_{k}^{(i, 0)} < c_{k}^{(1, i)}, \dots, c_{k}^{(i, 0)} < c_{k}^{(\tau(i), i)}$ for all $k \in \omega$, and $c_{k}^{(j, i)} \parallel c_{k}^{(t, i)}$ for each $1 \leq j \neq t \leq \tau(i)$. 
We define $d(m)$ as $|\{i \mid \tau(i)=m \}|$ and we put $I = \{ i : 1\le i \le n \land d(i)\ne 1\}$.
Then $T$ has exactly 
$$6^{d(1)} \cdot \prod\limits_{i\in I} 3^{\tau(i)+1}$$ 
countable models, where 
$$3^{d(1)} \cdot \prod\limits_{i\in I} 2^{\tau(i)+1}$$ 
of them are prime models.
\end{theorem}

Note that $T$ from Theorem~\ref{contantsss} has as many countable models as some finite disjoint union of theories of the form $T_{0}, T_{1}$, and $T_{n}$, for $n\geq 2$.    

\section{On unary predicates expansions of DMT}

Let $T_{n}$ be the complete constant expansion of $T_{\text{dmt}}$ as we have considered above. 
We replace $L_{n}$ with a new language $\mathcal{L}^{P}_{n}=\mathcal{L}_{0} \cup \{ P_{k} \mid k \in \omega \}$, where we interpret the predicates $P_{k}$ as the  set of constants $\{ c_{k}^{(1)}, c_{k}^{(2)}, \dots c_{k}^{(n)}\}$, for each $k \in \omega$. 
Let $T_{n}^{P}=\text{Th}\{(\mathcal{M}, \mathcal{L}_{n}^{P}) \mid 
(\mathcal{M}, {L}_{n}) \models T_n\}$. 
We aim to count the number of countable models of $T_{n}^{P}$. As $T_{n}$ is not complete, so is $T_{n}^{P}$, that is why we consider various completions of $T_{n}$. 


The difference between models of $T_{n}$ and $T_{n}^{P}$ is that in countable models of $T_{n}^{P}$ there are automorphisms which make some permutation of $P_{k}(\mathcal{M})$. So, we have the following schemes of axioms:

$1)_{k}$ $ (\exists^{!n} x) P_{k}(x);$

$2)_{k}$ $\forall x (P_{k}(x) \rightarrow \exists ! y (P_{k+1} (y) \wedge y<x));$

$3)_{k}$ $\forall x \forall y (P_{k} (x) \wedge P_{k}(y) \wedge x \neq y \rightarrow x \parallel y);$

We work simultaneously in models of $T_{n}$ and $T_{n}^{P}$. Since each model of $T_{n}$ after replacing the constants with the predicates becomes a model of $T_{n}^{P}$, and vice versa, each model of $T_{n}^{P}$ after a suitable replacement of the predicates with the constants becomes a model of $T_{n}$. 

If $\mathcal{M}=(M, \mathcal{L}_{n})$ is a model of $T_{n}$, we denote 
the corresponding model of $T_{n}^{P}$ by $\mathcal{M}_{P}=(M, \mathcal{L}_{n}^{P})$. 
Since the universes of $\mathcal{M}$ and $\mathcal{M}^{P}$ are the same, $p_{k}(\mathcal{M})$ being a type of $T_{n}$, defines a subset of $M$, which is also
a subset of $\mathcal{M}^{P}$. 

The first case is $c_{1}^{(i)} \sqcap c_{1}^{(j)} = c_{1}^{(s)} \sqcap c_{1}^{(t)}$ for all $i \neq j$ and $s \neq t$. We denote this theory by $T_{n,=}^{P}$. 

In $T_{n}$ we have 3 possible realizations of each type $p_{k}$, where $k \in \{1, \dots n \} $. Because of automorphisms we can permute realizations of $p_{k}(\mathcal{M})$ and $p_{m}(\mathcal{M})$. Then the number of non-isomorphic countable models of $T_{n,=}^{P}$ is equal to the number of combinations of a three-element set with repetitions. Hence we have $3\cdot C(3+n-1, n)=3\cdot\frac{(n+2)(n+1)}{2}$ countable models.  
Note that the number of non-isomorphic prime countable models of $T_{n,=}^{P}$ is equal to the number of combinations of a two-element set with repetitions, so the number is $2\cdot C(2+n-1, n) = 2\cdot(n+1)$. 

Now we consider $T_3$ and its possible completions different from the con\-si\-der\-ed above.
We consider the case that $d_{1, 2} = d_{1, 3} < d_{2, 3}$, 
where $d_{i, j} = c^{(i)}_0\sqcap c^{(j)}_0$ (see Figure \ref{fig4}). Then there is an automorphism of $\mathcal{M}^P$
which moves $p_2(M)$ to $p_3(M)$ and $p_3(M)$ to $p_2(M)$.
So, we have 3 kinds of realizations for each of $p_0$ and $p_1$,
and by the above formula $C(3+n-1, n)$ realizations for $p_2$ and $p_3$, where $n=2$.
Thus, the number of countable models is $3^2\cdot 6$
and the number of countable prime models is $2^2\cdot 3$, where the number $3$ 
is the result of the formula $C(2+n-1, n) = n+1$ for $n=2$.  
Since other completions of $T_3$ give the same $T^P_3$, we are done.

We describe Hasse diagrams for the completions $T^P_{3,=}$ and $T^P_{3,<}$ of $T_3^P$ by 
$d_{1, 2}=d_{1, 3}=d_{2, 3}$ and by $d_{1, 2}=d_{1, 3}<d_{2, 3}$, respectively. 

Let's describe the realizations of types in this theory. For this, we set $d_{1, 2}$ as a unique element which satisfies the following formula
$$\exists x_{1} \exists x_{2} (P_{k}(x_{1}) \wedge P_{k}(x_{2})\land x_1 \parallel x_2 \wedge y=x_{1} \sqcap x_{2}).$$ 

\noindent As we showed this above, $d_{1, 2}$ does not depend on the choice of $k$.

First we consider $T^P_{3,=}$. 
We desrcibe all types up to Rudin-Keisler preorder:
\begin{eqnarray*}
p_{1, 0}(x_{0}) & = & \{ c_{k}^{(0)} < x_{0} \mid k \in \omega  \} \cup \{ x_{0} < d_{1, 2} \} \\
p_{0, 1}(x_{1}) & = & \{d_{1, 2}<x_{1} \} \cup \{ \exists t (P_{k}(t) \wedge x_{1}<t) \mid k \in \omega\} \\
p_{0, 2}(x_{1}, x_{2}) & = & p_{0,1}(x_1)\cup p_{0,1}(x_2)\cup \{x_{1} \parallel x_{2} \} 
 \\ 
p_{0, 3}(x_{1}, x_2, x_3) & = & p_{0,2}(x_1, x_2)\cup p_{0,2}(x_1, x_3) \cup p_{0,2}(x_2, x_3) \\
p_{1,1}(x_0, x_1) & = & p_{1, 0}(x_0)\cup p_{0,1}(x_1)\\
p_{1,2}(x_0, x_1, x_2) & = & p_{1,1}(x_0,x_1) \cup p_{0,2}(x_1, x_2)\\
p_{1,3}(x_0, x_1, x_2, x_3) & = & p_{1,1}(x_0,x_1) \cup p_{0,3}(x_1, x_2, x_3)
\end{eqnarray*}
Hasse diagram of Rudin-Keisler preorder for the types of $T^P_{3,=}$
is given in the Figure~\ref{fig5}.

Now we consider $T^P_{3,<}$. 
As in the previous case, $T^P_{3,<}\cup \{d_{1, 2} = d_{1, 3} < d_{2, 3}\}$ is complete and
we desrcibe all types up to Rudin-Keisler preorder:
\begin{eqnarray*}
p_{1, 0, 0}(x_{0}) & = & \{ c_{k}^{(0)} < x_{0} \mid k \in \omega  \} \cup \{ x_{0} < d_{1, 2} \}  \\
p_{0, 1, 0}(x_{1}) & = & \{d_{1, 2}<x_{1} \} \cup \{ \exists t (P_{k}(t) \wedge x_{1}<t) \mid k \in \omega\} \\
p_{0, 0, 1}(x_{1}') & = & \{d_{1, 2}<x_{1}' \} \cup \{ \exists t (P_{k}(t) \wedge x_{1}'<t) \mid k \in \omega\} 
 \\ 
p_{0, 2, 0}(x_{1}, x_{2}) & = &  p_{0, 1, 0} (x_{1}) \cup p_{0, 1, 0} (x_{2}) \cup \{x_{1} \parallel x_{2} \}\\
p_{0, 1, 1}(x_{1}, x_{1}') & = & p_{0, 1, 0}(x_{1}) \cup p_{0, 0, 1}(x_{1}') \cup \{ x_{1} \parallel x_{1}' \}\\
p_{0, 2, 1}(x_{1}, x_{2}, x_{1}') & = & p_{0, 2, 0}(x_{1}, x_{2}) \cup p_{0, 1, 1}(x_{1}, x_{1}')\\
p_{1, 1, 0}(x_{0}, x_{1}) & = & p_{1, 0, 0}(x_{0}) \cup p_{0, 1, 0} (x_{1}) \\
p_{1, 2, 0}(x_{0}, x_{1}, x_{2}) & = & p_{1, 1, 0}(x_{0}, x_{1}) \cup p_{0, 2, 0} (x_{1}, x_{2}) \\
p_{1, 0, 1}(x_{0}, x_{1}') & = & p_{1, 0, 0}(x_{0}) \cup p_{0, 0 , 1} (x_{1}') \\
p_{1, 1, 1}(x_{0}, x_{1}, x_{1}') & = & p_{1, 1, 0}(x_{0}, x_{1}) \cup p_{1, 0, 1} (x_{0}, x_{1}') \cup p_{0, 1, 1}(x_{1}, x_{1}') \\
p_{1, 2, 1}(x_{0}, x_{1}, x_{2}, x_{1}') & = & p_{1, 2, 0}(x_{0}, x_{1}, x_{2}) \cup p_{1, 1, 1} (x_{0}, x_{1}, x_{1}') \cup p_{0, 2, 1}(x_{1}, x_{2}, x_{1}') 
\end{eqnarray*}

Hasse diagram of Rudin-Keisler preorder for the types of $T^{P}_{3, <}$
is given in the Figure~\ref{fig5}.


\begin{figure}[!htb]
\begin{picture}(330,150)

\put(15, 20){\circle*{3}}
\put(15, 20){\circle{7}}
\put(17, 25){\tiny{0}}
%\put(132,9){\scriptsize{$p_{0, 0}$}}
%\put(132,9){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0000}}}$}}

\put(15, 20){\line(0,1){30}}
\put(15, 20){\line(3,1){90}}

\put(105, 50){\circle*{3}}
\put(105, 50){\circle{7}}
\put(97,52){\tiny{1}}
\put(112,49){\scriptsize{$p_{1, 0}$}}
%\put(182,59){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0000}}}$}}

\put(15, 50){\circle*{3}}
\put(15, 50){\circle{7}}
\put(17,55){\tiny{1}}
\put(-5,49){\scriptsize{$p_{0,1}$}}
%\put(80,59){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0001}}}$}}


\put(15, 50){\line(0,1){30}}
\put(15, 50){\line(3,1){90}}
\put(105, 50){\line(0,1){30}}

\put(105, 80){\circle*{3}}
\put(105, 80){\circle{7}}
\put(97,82){\tiny{3}}
\put(112,79){\scriptsize{$p_{1,1}$}}
%\put(182,99){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0001}}}$}}

\put(15, 80){\circle*{3}}
\put(15, 80){\circle{7}}
\put(17,85){\tiny{2}}
\put(-5,79){\scriptsize{$p_{0, 2}$}}
%\put(80,99){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0011}}}$}}


\put(15, 80){\line(0,1){30}}
\put(15, 80){\line(3,1){90}}
\put(105, 80){\line(0,1){30}}

\put(105, 110){\circle*{3}}
\put(105, 110){\circle{7}}
\put(97,112){\tiny{5}}
\put(112,109){\scriptsize{$p_{1, 2}$}}
%\put(182,139){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0011}}}$}}

\put(15, 110){\circle*{3}}
\put(15, 110){\circle{7}}
\put(17,102){\tiny{3}}
\put(-5,109){\scriptsize{$p_{0, 3}$}}
%\put(80,139){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0111}}}$}}


%\put(135, 110){\line(0,1){30}}

\put(15, 110){\line(3,1){90}}
\put(105, 110){\line(0,1){30}}

\put(105, 140){\circle*{3}}
\put(105, 140){\circle{7}}
\put(97,130){\tiny{7}}
\put(112,139){\scriptsize{$p_{1, 3}$}}
%\put(50,142){\tiny{$2n-3$}}
%\put(182,174){\tiny{$p_{\underset{n+1}{\underbrace{1, 001 \dots 11}}}$}}

%%%%%%%%%%%%

\put(200, 20){\circle*{3}}
\put(200, 20){\circle{7}}
\put(202, 26){\tiny{0}}
%\put(132,9){\scriptsize{$p_{0, 0}$}}
%\put(132,9){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0000}}}$}}

\put(200, 20){\line(0,1){30}}
\put(200, 20){\line(3,2){45}}
\put(200, 20){\line(3,1){90}}

\put(200, 50){\circle*{3}}
\put(200, 50){\circle{7}}
\put(205,45){\tiny{1}}
\put(173,49){\scriptsize{$p_{0, 1, 0}$}}
%\put(80,59){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0001}}}$}}

\put(245, 50){\circle*{3}}
\put(245, 50){\circle{7}}
\put(235,50){\tiny{1}}
\put(252,49){\scriptsize{$p_{0, 0, 1}$}}
%\put(182,59){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0000}}}$}}

\put(290, 50){\circle*{3}}
\put(290, 50){\circle{7}}
\put(280,50){\tiny{1}}
\put(297,49){\scriptsize{$p_{1,0, 0}$}}
%\put(80,59){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0001}}}$}}

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\put(200, 50){\line(1,1){30}}
\put(200, 50){\line(2,1){60}}
\put(245, 50){\line(-1,2){15}}
\put(245, 50){\line(3,2){45}}
\put(290, 50){\line(-1,1){30}}
\put(290, 50){\line(0,1){30}}

\put(200, 80){\circle*{3}}
\put(200, 80){\circle{7}}
\put(205,77){\tiny{2}}
\put(173,79){\scriptsize{$p_{0, 2, 0}$}}
%\put(80,99){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0011}}}$}}

\put(230, 80){\circle*{3}}
\put(230, 80){\circle{7}}
\put(228,85){\tiny{3}}
\put(235,79){\scriptsize{$p_{0, 1, 1}$}}
%\put(182,99){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0001}}}$}}

\put(260, 80){\circle*{3}}
\put(260, 80){\circle{7}}
\put(258,70){\tiny{3}}
\put(265,79){\scriptsize{$p_{1, 1, 0}$}}
%\put(182,99){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0001}}}$}}

\put(290, 80){\circle*{3}}
\put(290, 80){\circle{7}}
\put(283,85){\tiny{3}}
\put(297,79){\scriptsize{$p_{1, 0, 1}$}}
%\put(182,99){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0001}}}$}}


\put(200, 80){\line(0,1){30}}
\put(200, 80){\line(3,2){45}}
\put(230, 80){\line(-1,1){30}}
\put(230, 80){\line(2,1){60}}
\put(260, 80){\line(-1,2){15}}
\put(260, 80){\line(1,1){30}}
\put(290, 80){\line(0,1){30}}


\put(200, 110){\circle*{3}}
\put(200, 110){\circle{7}}
\put(206,107){\tiny{5}}
\put(173,109){\scriptsize{$p_{0, 2, 1}$}}
%\put(80,139){\tiny{$p_{\underset{n+1}{\underbrace{0, 00 \dots 0111}}}$}}

\put(245, 110){\circle*{3}}
\put(245, 110){\circle{7}}
\put(235,109){\tiny{5}}
\put(252,109){\scriptsize{$p_{1, 2, 0}$}}
%\put(182,139){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0011}}}$}}

\put(290, 110){\circle*{3}}
\put(290, 110){\circle{7}}
\put(280,109){\tiny{7}}
\put(297,109){\scriptsize{$p_{1, 1, 1}$}}
%\put(182,139){\tiny{$p_{\underset{n+1}{\underbrace{1, 00 \dots 0011}}}$}}


\put(200, 110){\line(3,1){90}}
\put(245, 110){\line(3,2){45}}
\put(290, 110){\line(0,1){30}}

\put(290, 140){\circle*{3}}
\put(290, 140){\circle{7}}
\put(282,130){\tiny{11}}
\put(297,139){\scriptsize{$p_{1, 2, 1}$}}
%\put(50,142){\tiny{$2n-3$}}
%\put(182,174){\tiny{$p_{\underset{n+1}{\underbrace{1, 001 \dots 11}}}$}}

\end{picture}
\caption{\scriptsize{Hasse diagrams for $T^P_{3,=}$ and $T^P_{3,<}$}}
\label{fig5}
\end{figure}


It is turn to consider possible completions on $T_4^P$ via completions of $T_4$, where
some of $d_{i, j}$ are not the same. There are 5 such completions of $T_4^P$ (see Figure \ref{fig6}, in this figure we omit the case when all $d_{i, j}$ are equal).

\begin{figure}[!htb]
\begin{picture}(330,160)

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\put(40, 115){$\ast$}
\put(60, 115){$\ast$}
\put(30, 50){$\ast$}


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\put(7, 94){\tiny{$d_{1, 2}=$}}
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\put(37, 75){\tiny{$d_{1, 4}=d_{2, 4}=d_{3, 4}$}}


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\put(63, 158){\vector(0,-1){30}}
\put(48, 148){\scriptsize{$c_{k}^{(4)}$}}
\put(33, 10){\vector(0,1){30}}
\put(18, 20){\scriptsize{$c_{k}^{(0)}$}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\put(130, 115){$\ast$}
\put(150, 115){$\ast$}
\put(120, 50){$\ast$}


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\put(87, 95){\tiny{$d_{1, 2}$}}
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\put(127, 75){\tiny{$d_{1, 3}=d_{2, 3}$}}
\put(127, 68){\tiny{$=d_{1, 4}=d_{2, 4}=d_{3, 4}$}}

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{\circle*{1}}
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{\circle*{1}}



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\put(118, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(153, 158){\vector(0,-1){30}}
\put(138, 148){\scriptsize{$c_{k}^{(4)}$}}
\put(123, 10){\vector(0,1){30}}
\put(108, 20){\scriptsize{$c_{k}^{(0)}$}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\put(203, 115){$\ast$}
\put(223, 115){$\ast$}
\put(243, 115){$\ast$}
\put(213, 50){$\ast$}


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\put(178, 95){\tiny{$d_{1, 2}$}}
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\put(166, 84){\tiny{$d_{1, 3}=d_{2, 3}$}}
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\put(220, 75){\tiny{$d_{1, 4}=d_{2, 4}=d_{3, 4}$}}

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{\circle*{1}}



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\put(211, 148){\scriptsize{$c_{k}^{(3)}$}}
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\put(231, 148){\scriptsize{$c_{k}^{(4)}$}}
\put(216, 10){\vector(0,1){30}}
\put(201, 20){\scriptsize{$c_{k}^{(0)}$}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\put(293, 115){$\ast$}
\put(313, 115){$\ast$}
\put(333, 115){$\ast$}
\put(303, 50){$\ast$}


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\put(270, 95){\tiny{$d_{1, 2}$}}
\put(326, 95){\circle*{3}}
\put(310, 95){\tiny{$d_{3, 4}$}}
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\put(308, 72){\tiny{$d_{1, 3}=d_{2, 3}$}}
\put(308, 65){\tiny{$=d_{1, 4}=d_{2, 4}$}}

\multiput(277,115)(1.5,-3){7}%
{\circle*{1}}
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{\circle*{1}}
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{\circle*{1}}
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{\circle*{1}}
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{\circle*{1}}
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{\circle*{1}}
\multiput(326,95)(-3,-3){6}%
{\circle*{1}}



\put(276, 158){\vector(0,-1){30}}
\put(261, 148){\scriptsize{$c_{k}^{(1)}$}}
\put(296, 158){\vector(0,-1){30}}
\put(281, 148){\scriptsize{$c_{k}^{(2)}$}}
\put(316, 158){\vector(0,-1){30}}
\put(301, 148){\scriptsize{$c_{k}^{(3)}$}}
\put(336, 158){\vector(0,-1){30}}
\put(321, 148){\scriptsize{$c_{k}^{(4)}$}}
\put(306, 10){\vector(0,1){30}}
\put(291, 20){\scriptsize{$c_{k}^{(0)}$}}

%\put(78,0){\scriptsize{a.}}
%\put(218,0){\scriptsize{b.}}

\end{picture}
\caption{\scriptsize{}}
\label{fig6}
\end{figure}



{ \it Case 1.} $d_{1, 2} = d_{1, 3}=d_{2, 3} > d_{1, 4} = d_{2, 4} = d_{3, 4}$.
There are 3 realizations for each of $p_0$ and $p_4$.
The number of realization of $p_1$, $p_2$, and $p_3$ is expressed by $C(3+n-1, n)$,
where $n=3$. So we have $3^2\cdot 10$ countable models.
The number of countable prime models is $2^2\cdot C(2+n-1, n) = 2^2\cdot 4$.

{ \it Case 2.} $d_{1, 2}> d_{1, 3}=d_{2, 3} = d_{1, 4} = d_{2, 4} = d_{3, 4}$.
The number of realizations of $p_1$ and $p_2$ is expressed by $C(3+n-1, n)$,
where $n=2$, as well as the number of realizations of $p_3$ and $p_4$. So we have $3\cdot 6^2$ countable models.
The number of prime countable models is $2\cdot (C(2+n-1, n))^2 = 2\cdot 4^2$.

{ \it Case 3.} $d_{1, 2}> d_{1, 3}=d_{2, 3} > d_{1, 4} = d_{2, 4} = d_{3, 4}$.
There are 3 realizations for each of $p_0$, $p_3$, and $p_4$.
The number of realization of $p_1$ and $p_2$ is expressed by $C(3+n-1, n)$,
where $n=2$. So we have $3^3\cdot 6$ countable models.
The number of countable prime models is $2^3\cdot C(2+n-1, n) = 2^3\cdot 3$.

{ \it Case 4.} $d_{1, 2}> d_{1, 3}=d_{2, 3} = d_{1, 4} = d_{2, 4}$, $d_{3, 4}>d_{1, 4}$, and $d_{1, 2}\parallel d_{3, 4}$.
This is the most interesting case here. There are automorphisms of $\mathcal{M}^P$
which swap $p_1(\mathcal{M})$ and $p_2(\mathcal{M})$, 
$p_3(\mathcal{M})$ and $p_4(\mathcal{M})$, and
$p_1(\mathcal{M})\cup p_2(\mathcal{M})$ and $p_3(\mathcal{M})\cup p_4(\mathcal{M})$.

First, we calculate the number of countable prime models. It is $2\cdot 6 = 12$. 
The number 6 is obtained by the following calculation. For each of $p_i$, we have 2 possibilities: either the realization is empty, or not. We denote it by 0 and 1, correspondingly.
In order to code all possible variants, we use multisets.
Since we have 2 elements: 0 and 1, there are exactly 3 multisets of cardinality 2: $\{ 0, 0\}$, $\{ 0, 1\}$, and $\{ 1, 1\}$.
Now we calculate the number of combinations of three-element set with repetitions, when we take 2 elements. It is equal to $C(4,2) = 6$.
Below, we list all possible  choices for realization of $p_1$, $p_2$, $p_3$, and $p_4$:
$$
\{\{0,0\},\{0,0\}\}, \ 
\{\{0,0\},\{0,1\}\}, \ 
\{\{0,0\},\{1,1\}\}$$
$$
\{\{0,1\},\{0,1\}\}, \ 
\{\{0,1\},\{1,1\}\}, \ 
\{\{1,1\},\{1,1\}\}
$$

Now we calculate the number of all countable models. We denote by 2
the limit realization of a type. 
Since we have 3 elements: 0, 1, and 2, there are exactly 6 multisets of cardinality 2: $\{ 0, 0\}$, $\{ 0, 1\}$, $\{ 0, 2\}$, $\{ 1, 1\}$, $\{ 1, 2\}$, and $\{ 2, 2\}$.
Now we calculate the number of combinations of six-element set with repetitions, when we take 2 elements. It is equal to $C(7,2) = 21$.
Recall, that $p_0$ has 3 kinds of realizations.
Thus, the number of countable models is $3\cdot 21 = 63$.

We express the general way for calculating the number of countable models. Let $\mathcal{T}$ be a finite rooted tree. Let $G$ be the group of all automorphisms of $\mathcal{T}$ as a rooted tree.
Now we color each leaf of $\mathcal{T}$ into $k$ colors,
where $k\in \{2,3\}$. 
Two colorings of $\mathcal{T}$ are said to be equivalent if there
is a $g\in G$, which takes the first coloring to the second one.
Each completion of $T_n$ correspond to some tree with $n$ leaves, where  each vertex has at least two sons or it is a leaf.
Then the number of countable models of a completion of $T_n$ is equal to 
the number of non-equivalent colorings of the corresponding tree, where the number of colors is equal to 3 modulo the number of realizations of the type $p_0$. So, the total number of countable models is 3 times the number of non-equivalent colorings of the corresponding tree.
For counting the number of prime models of $T_n$ we take the number of colors to be 2.

We consider some example of $\mathcal{T}$. 
Assume that the root of $\mathcal{T}$ has $m_3$ sons, 
each son has $m_2$ sons, and each grandson of the root has $m_1$ sons.
Then there are $\gamma_1 = C(3+m_1-1, m_1)$ ways to color the sons of some grandson.
Now there are $\gamma_2 = C(\gamma_1+m_2-1, m_2)$ ways to color
the grandsons with their children.
So, there are $\gamma_3 = C(\gamma_2+m_3-1, m_3)$ ways to color 
the sons with their sons and grandson.
Also, we have 3 coloring of the root.
Totally, there are 
$$3\cdot C(C(C(3+m_1-1, m_1)+m_2-1, m_2)+m_3-1, m_3)$$
countable models of the corresponding theory.

Recall that the height of a vertex in a rooted tree is the length of the longest downward path to a leaf from that vertex.
Let $a$ and $b$ be two parents of two leaves, that is, of height 1. Then there exists an automorphism that swaps $a$ and $b$, swaps
sons of $a$ and sons of $b$ and fixes the rest of the tree if and only if they have the same number of sons. 
So, we say that a  vertex of the height 1 which has exactly $m$ sons is of the type $m$.

Let $c$ be a vertex of height 2. 
We say that $c$ is of the type 
$$((m_0, 0), (m_1, k_1), \dots, (m_s, k_s))$$
if it has exactly $m_0+m_1+\dots +m_s$ sons, moreover, it has exactly $m_0$ sons which are leaves, and exactly 
$m_i$ sons of the type $k_i$, for each $i$.

Let $d$ be a vertex of height $p+1$, and let $K_i$ be a type
of vertex $c_i$ of the height $p$. 
We say that $d$ is of the type 
$$((m_1, K_i), \dots, (m_s, K_s))$$
if it has exactly $m_1+\dots +m_s$ sons, moreover, it has exactly $m_i$ sons of the type $K_i$ for each $i$.

It is straightforward to prove that for any two vertices, there is an automorphism which swaps them and their descendants and fixes the other part of the tree if and only if the types of these elements are the same. 

Let $a$ be a vertex which is fixed by each automorphism of $\mathcal{T}$
and which has at least two sons of the same type.
Let $b_0$ be a leaf that is a descendant of $a$.
Let $b_1$ be the parent of $b_0$ of type $m_1^0$, that is, 
$b_1$ has exactly $m_1^0$ sons. 

Let $b_2$ be the parent of $b_1$ and have exactly $m_2^0$ sons of the same type with $b_1$. And so on, let $b_k = a$ be the parent of $b_{k-1}$
and let it have exactly $m_k^0$ sons of the same type as $b_{k-1}$ has.
Let 
$$\gamma_1^0 = C(3+m_1^0-1, m_1^0) \mbox{ and }
\gamma_{i+1}^0 = C(\gamma_{i}^0+m_i^0-1, m_i^0),$$
$$\delta_1^0 = C(2+m_1^0-1, m_1^0) \mbox{ and } \delta_{i+1}^0 = C(\delta_{i}^0+m_i^0-1, m_i^0).$$

Let $B_0 = G(b_0)$, where $G$ is the group of all automorphisms of the rooted tree $\mathcal{T}$. Then similarly to the example above
we can prove that there exist $\gamma_k^0$ colorings of $B_0$ into 3 colors and $\delta_k^0$ colorings of $B_0$ into 2 colors.
Given a set $B_0$, we denote $\gamma_k^0$ by $\Gamma_0$
and $\delta_k^0$ by $\Delta_0$.

From the above considerations, Theorem~\ref{mr} follows.
\begin{theorem}\label{mr}
Let $T$ be a completion of $T_n^P$ and let $\mathcal{T}$ be the corresponding rooted tree. Let $B_0$, \dots, $B_w$ be a partition of the set of leaves of $\mathcal{T}$, where each $B_i$ is the orbit of some leaf under the action of the group of automorphisms of $\mathcal{T}$.
Then the number of countable models of $T$ is equal to
$$
3 \cdot \prod_{i\le w} \Gamma_i
$$
and the number of countable prime models is equal to
$$
2 \cdot \prod_{i\le w} \Delta_i.
$$
\end{theorem}

Now we give the diagram for $T_{n, =}^{P}$ in Figure~\ref{fig7}, 
where $T_{n, =}^{P}$ is a such completion of $T_n^P$ where all meets of the descending sequences are the same.
We omit the Hasse diagram of the Rudin-Keisler preorders $\leq_{\text{RK}}$ for the other theories since the drawing these diagrams seems cumbersome. 


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\end{picture}
\caption{\scriptsize{Hasse diagram for $T^P_{n,=}$}}
\label{fig7}
\end{figure}






\bigskip
\begin{thebibliography}{1} % Ññûëêè íà ëèòåðàòóðó ïðèâîäÿòñÿ òîëüêî íà àíãëèéñêîì ÿçûêå (äàæå åñëè ó èñòî÷íèêà íåò àíãëîÿçû÷íîé âåðñèè).

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\end{thebibliography}
\end{document}

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