$ S^1(\textbf{c}) \subseteq  \textrm{Cond}(\textbf{c})$ (using the inverse implications).\
$\bullet\ \textrm{Case}\  d(1)-1=l(M_1)$ and $d(2)-1 \neq l(M_2)$.

Hence $d(2)-1=p$ and we have $l(M_2)<p \leq \min(l(M_1),\textrm{Comp}(M_1,M_2)).$ Therefore: $$c[1]= \sum_{k=1}^{l(M_1)}{M_1[k]} \textrm{ , }  c[2]= \sum_{k=1}^{p}{M_2[k]}=\sum_{k=1}^{l(M_2)}{M_2[k]}+ \sum_{k=l(M_2)+1}^{p}{M_2[k]}=\sum_{k=1}^{l(M_2)}{M_2[k]}+p-l(M_2),$$ and from this we can deduce that $M_1 \in \textrm{Cond}^*(c[1])$ and, denoted by $k(2)=c[2]-(p-l(M_2))$, $M_2 \in \textrm{Cond}^*(k(2))$. Notice that $k(2)<c[2]$.

Now, for all $k<c[2]$ we define the set: $$\begin{aligned}
 I_1(k)=\left\{ T_{E}=(p): E=\left\{M_1,M_2\right\}, M_1 \in \textrm{Cond}^*(c[1]),M_2 \in \textrm{Cond}^*(k) \right.\\ \left. p= l(M_2)+c[2]-k \leq \min(l(M_1),\textrm{Comp}(M_1,M_2)) \right\}, \end{aligned}$$ therefore $T$ belongs to $$S_2^1(\textrm{c})=\bigcup_{k<c[2]} {I_1(k)}.$$ With the inverse implication we can easily show that $ S_2^1(\textbf{c}) \subseteq  \textrm{Cond}(\textbf{c})$.\
$\bullet\ \textrm{Case}\  d(1)-1\neq l(M_1)$ e $d(2)-1 = l(M_2)$.

We have $d(1)-1=p$ and $l(M_1)<p \leq \min(l(M_2),\textrm{Comp}(M_1,M_2)).$ Hence: $$c[1]= \sum_{k=1}^{p}{M_1[k]}=\sum_{k=1}^{l(M_1)}{M_1[k]}+ \sum_{k=l(M_1)+1}^{p}{M_1[k]}=\sum_{k=1}^{l(M_1)}{M_1[k]}+p-l(M_1)  \textrm{; }  c[2]= \sum_{k=1}^{l(M_2)}{M_2[k]},$$ and from this we obtain $M_2 \in \textrm{Cond}^*(c[2])$ and, denoted by $k(1)=c[1]-(p-l(M_1))$, $M_1 \in \textrm{Cond}^*(k(1))$. Notice that $k(1)<c[1]$.

For all $k<c[1]$ we define the set: $$\begin{aligned}
 I_2(k)=\left\{ T_{E}=(p): E=\left\{M_1,M_2\right\}, M_1 \in \textrm{Cond}^*(k),M_2 \in \textrm{Cond}^*(c[2]) \right.\\ \left. p= l(M_1)+c[1]-k \leq \min(l(M_2),\textrm{Comp}(M_1,M_2)) \right\}, \end{aligned}$$ therefore $T$ belongs to $$S_2^2(\textrm{c})=\bigcup_{k<c[1]} {I_2(k)}.$$ With the inverse implication we can easily show that $ S_2^2(\textbf{c}) \subseteq  \textrm{Cond}(\textbf{c})$.\
$\bullet\ \textrm{Case}\  d(1)-1\neq l(M_1)$ and $d(2)-1 \neq l(M_2)$.

Then $d(2)-1=p$, $d(1)-1=p$ and we have $\max(l(M_1),l(M_2))<p \leq \textrm{Comp}(M_1,M_2).$ It follows: $$c[1]= 