only have the tree $T_7=T_{E_6}=(1,2).$

-   $E_1=\left\{ M_1=[3],M_2=[2],M_3=[2,2] \right\}$ and we have $\min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2),l(M_1))=\min(1,3,1)=1$. Thus we only have the tree, already found in $S_1^1([3,2,4])$, $T_1=T_{E_1}=(1,1).$

-   $E_7=\left\{ M_1=[3],M_2=[1],M_3=[3] \right\}$ and we have $\min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2),l(M_1))=\min(2,2,1)=1$. Hence we have the tree $T_8=T_{E_7}=(1,2)$.

-   $E_2=\left\{ M_1=[3],M_2=[2],M_3=[4] \right\}$ and we have $\min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2),l(M_1))=\min(1,3,1)=1$. Thus we only have the tree, already found in $S_1^1([3,2,4])$, $T_2=T_{E_2}=(1,1).$

Hence $S_2^1([3,2,4])=\left\{ T_1,T_2,T_7,T_8\right\}.$

Now we compute $S_2^2([3,2,4])$. $k_1=2$ is the only value such that $I(k_1) \neq \emptyset$, and $I(2)$ contains two elements. In fact, if we consider $B_1$ and $B_3$ and $M_1=[2]$, we have:

-   $ E_3=\left\{ M_1=[2],M_2=[1],M_3=[2,2] \right\} \textrm{ and } l(M_1)+c[1]-k_1=1+3-2=2 \leq \min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2)))=\min(2,2)=2$. Thus we only have the tree, already found in $S_1^1([3,2,4])$, $T_4=T_{E_3}=(2,2).$

-   $ E_5=\left\{ M_1=[2],M_2=[1],M_3=[3] \right\} \textrm{ and } l(M_1)+c[1]-k_1=1+3-2=2 \leq \min(\max(p_2,l(M_2),\textrm{Comp}(M_1,M_2)))=\min(2,2)=2$. Thus we only have the tree, already found in $S_1^2([3,2,4])$, $T_6=T_{E_5}=(2,2).$

Thus $S_2^1([3,2,4]) \cup S_2^2([3,2,4])=\left\{ T_1,T_2,T_4,T_6,T_7,T_8\right\}$. 0.3in

at(current bounding box.south)[$T_7$]{};

0.3in

at(current bounding box.south)[$T_8$]{};

Hence $ \textrm{Cond}([3,2,4])=\left\{ T_1,T_2,T_3,T_4,T_5,T_6,T_7,T_8\right\}$.

If we compute the set $\overline{\textrm{Cond}([3,2,4])}$, with the technique explained at the end of the previous section we will find that: $$\overline{\textrm{Cond}([3,2,4])}=\textrm{Cond}([3,2,4]) \bigcup \left\{ T_9, T_{10} \right\},$$ where

-   $T_9=M(T)_E=\left( \begin{matrix} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 &0 & 0 \end{matrix}\right)$ 