C$ is contained in a finite strong subset of $M_n$ and for such subsets the quantifier-free type determines the type by Remark \[r:qftp\]. Hence there are only countably many types over a countable model.

In fact, it is easy to see directly without counting types that $T_n$ is superstable (see Remark \[r:superstable\]).

The has weak elimination of imaginaries.

Let $a$ be a vertex and $A$ any set. Then we can choose $\operatorname{Cb}(stp(a/A))$ as the projection of $a$ on $A$. This is a finite set.

The following immediate corollary will be very useful:

\[c:forking\] The vertex $a$ is independent from $A$ over $C$ if $\operatorname{proj}(a/AC)\subseteq\operatorname{acl}(C)$. In particular, $a$ is independent from $A$ over $\emptyset$ if and only if $a$ is not connected to any vertex of $\operatorname{acl}(A)$.

\[r:superstable\] As in [@TZU] we could have defined a notion of independence on models of $T_n$ by saying $${
 A\mathop{\mathpalette{
 \setbox0=\hbox{$x$}\kern\wd0\hbox to 0pt{\hss$
 \mid$\hss}\lower.9\ht0\hbox to 0pt{\hss$\smile$\hss}\kern\wd0
}{}}_{C}B
}$$ if and only if $\operatorname{proj}(a/BC)\subseteq\operatorname{acl}(C)$ for all $a\in\operatorname{acl}(A)$. It is easy to see that this notion of independece satisfies the characterizing properties of forking in stable theories (see [@TZ] Ch. 8) and hence agrees with the usual one. Note that the existence of nonforking extensions follows 