\sum_{k=1}^{p}{M_1[k]}=\sum_{k=1}^{l(M_1)}{M_1[k]}+ \sum_{k=l(M_1)+1}^{p}{M_1[k]}=\sum_{k=1}^{l(M_1)}{M_1[k]}+p-l(M_1)$$ $$c[2]= \sum_{k=1}^{p}{M_2[k]}=\sum_{k=1}^{l(M_2)}{M_2[k]}+ \sum_{k=l(M_2)+1}^{p}{M_2[k]}=\sum_{k=1}^{l(M_2)}{M_2[k]}+p-l(M_2),$$ If we denote by $k(1)=c[1]-(p-l(M_1))$ and by $k(2)=c[2]-(p-l(M_2))$, we have $M_1 \in \textrm{Cond}^*(k(1))$ and $M_2 \in \textrm{Cond}^*(k(2))$.

Furthermore notice that $k(1)<c[1]$ and $k(2)<c[2]$. Now, for all $k_1<c[1]$ and $k_2<c[2]$ we define the set: $$\begin{aligned}
 I(k_1,k_2)=\left\{ T_{E}=(p): E=\left\{M_1,M_2\right\}, M_i \in \textrm{Cond}^*(k_i), \textrm{ for } i=1,2 \right.\\ \left.  p=l(M_1)+c[1]-k_1=l(M_2)+c[2]-k_2 \leq \textrm{Comp}(M_1,M_2) \right\} \end{aligned}$$ so $T$ belongs to $$S^3(\textbf{c})=\bigcup_{k_i < c[i]}{I(k_1,k_2)} \subseteq \textrm{Cond}(\textbf{c})$$ Even in this case we can show that $ S^3(\textbf{c}) \subseteq  \textrm{Cond}(\textbf{c})$.

We have studied all the possible cases so we proved that $$S^1(\textbf{c}) \cup S_1^2(\textbf{c}) \cup S_2^2(\textbf{c}) \cup S^3(\textbf{c})=\textrm{Cond}(\textbf{c}).$$ All the previous set can be computed by using the procedure given in the case $r=1$ so we have found a procedure to compute $\textrm{Cond}(\textbf{c})$ when $r=2$.

Let us compute $ \textrm{Cond}([4,5])$.

First of all we compute $S^1([4,5])$. We need $\textrm{Cond}^*(4)$ and $\textrm{Cond}^*(5)$. They are: $$\textrm{Cond}^*(4)=\left\{[4],[2,2]\right\} \textrm{ and } \textrm{Cond}^*(5)=\left\{ [5], [3,2] \right\}.$$ Hence when we compute $S^1([4,5])$ we find:

-   $E_1=\left\{M_1=[4] , M_2=[5]\right\}$. Thus $\textrm{Comp}(M_1,M_2)=5$ and $\min(l(M_1),l(M_2))=1$. Then we have only the tree $T_1=T_{E_1}=(1)$.

-   $E_2=\left\{M_1=[4] , M_2=[3,2]\right\}$. Thus $\textrm{Comp}(M_1,M_2)=3$ and $\min(l(M_1),l(M_2))=1$. Then we have only the tree $T_2=T_{E_2}=(1)$.

-   $E_3=\left\{M_1=[2,2] , M_2=[5]\right\}$. Thus $\textrm{Comp}(M_1,M_2)=2$ and $\min(l(M_1),l(M_2))=1$. Then we have only the tree $T_3=T_{E_3}=(1)$.

