-   $E_4=\left\{M_1=[2,2] , M_2=[3,2]\right\}$. We have $\textrm{Comp}(M_1,M_2)=2$ and $\min(l(M_1),l(M_2))=2$. So we have the trees $T_4=T_{E_4}=(1)$ and $T_5=T_{E_4}=(2)$ .

Hence $S^1([4,5])=\left\{ T_1,T_2,T_3,T_4,T_5\right\}$. 0.3in

\]at(current bounding box.south)[$T_1$]{};

0.3in

at(current bounding box.south)[$T_2$]{};

0.3in

at(current bounding box.south)[$T_3$]{};

0.3in

at(current bounding box.south)[$T_4$]{};

0.3in

\]at(current bounding box.south)[$T_5$]{};

Now we compute $S_1^2([4,5])$. The only value $k$ such that $I_1(k) \neq \emptyset $ is $k=4$ and we have:

-   $c[1]=4$, $k=4$. If we consider $M_1=[2,2] \in \textrm{Cond}^*(4)$ , $M_2=[4] \in \textrm{Cond}^*(4) $ and $E_5=\left\{ M_1,M_2 \right\}$ we have $$l(M_2)+c[2]-k=2 \leq \min(l(M_1),\textrm{Comp}(M_1,M_2))=\min(2,2)=2.$$ Hence we have the tree $T_6=T_{E_5}=(2).$

Therefore $S_1^2([4,5])=\left\{T_6\right\}$. Let us compute $S_2^2([4,5])$. The only value $k$ such that $I_2(k) \neq \emptyset$ is $k=3$ and we have:

-   $k=3$, $c[2]=5$. If we consider $M_1=[3] \in \textrm{Cond}^*(3)$ , $M_2=[3,2] \in \textrm{Cond}^*(5) $ and $E_6=\left\{ M_1,M_2 \right\}$ we have $$l(M_1)+c[1]-k=1+4-3=2 \leq \min(l(M_2), \textrm{Comp}(M_1,M_2)=\min(2,3)=2.$$ Hence we have the tree $T_7=T_{E_6}=(2).$

Therefore $S_2^2([4,5])=\left\{T_7\right\}$.

We finally compute $S^3([4,5])$.

The only values of $k_1$ and $k_2$ such that $I(k_1,k_2) \neq \emptyset$ are the following:

-   $k_1=2$, $k_2=3$. If we consider $M_1=[2] \in \textrm{Cond}^*(2)$, $M_2=[3] \in \textrm{Cond}^*(3) $ and $E_7=\left\{ M_1,M_2 \right\}$ we have $$l(M_1)+c[1]-k_1=1+4-2=3=1+5-3=l(M_2)+c[2]-k_2 \leq \textrm{Comp}(M_1,M_2)=3.$$ Thus we have the tree $T_8=T_{E_7}=(3).$

-   $k_1=3$, $k_2=4$. If we consider $M_1=[3] \in \textrm{Cond}^*(3)$ , $M_2=[4] \in 