T_1 \notin\textrm{Cond}((c[1],\ldots,c[t])); $

-   $ T_2 \notin \textrm{Cond}((c[t],\ldots,c[r])). $

Let us consider the following integers (which are linked to the conductor): $$d(i)=\min\left\{  j \in \mathbb{N}: M_i[j]=1 \textrm{ and the $i$-th branch in $T$ is not glued to other branches at level $j$} \right\},$$ for $  i=1,\ldots,r .$ $$d_1(i)=\min\left\{  j \in \mathbb{N}: M_i[j]=1 \textrm{  and the $i$-th branch in $T_1$ isn't glued to other branches at level $j$}  \right\},$$ for $  i=1,\ldots,t .$ $$d_2(i)=\min\left\{  j \in \mathbb{N}: M_i[j]=1 \textrm{ and the $i$-th branch in $T_2$ isn't glued to other branches at level $j$}  \right\},$$ for $  i=t,\ldots,r .$

We clearly have $d_1(l)=d(l)$ for all $l=1,\ldots,t-1$ and $d_2(m)=d(m)$ for all $m=t+1,\ldots,r$. Furthermore $d_1(t)\leq d(t)$ e $d_2(t) \leq d(t)$. In fact we have noticed that $d(t)-1=\max(l(M_t),p_{t-1},p_t)$, while $d_1(t)-1=\max(l(M_t),p_{t-1})$ and $d_2(t)-1=\max(l(M_t),p_t)$.

From $T \in \textrm{Cond}(\textbf{c})$ we deduce that $$\sum_{k=1}^{d(i)-1}{M_i[k]}=c[i]  \textrm{ for }  i=1,\ldots,r.$$ We denote respectively by $(c_1[1],\ldots,c_1[t])$ and by $(c_2[t],\ldots,c_2[r]))$ the conductors of $T_1$ and of $T_2$.

We have: $$c_1[l]=\sum_{k=1}^{d_1(l)-1}{M_l[k]}=\sum_{k=1}^{d(l)-1}{M_l[k]}=c[l] \textrm{   for  } l=1,\ldots,t-1$$ $$\textrm{  and } c_2[m]=\sum_{k=1}^{d_2(m)-1}{M_m[k]}=\sum_{k=1}^{d(m)-1}{M_m[k]}=c[m],  \textrm{ for } m=t+1,\ldots,r$$ and this implies, because $ T_1 \notin \textrm{Cond}((c[1],\ldots,c[t])) $ and $ T_2 \notin \textrm{Cond}((c[t],\ldots,c[r]))$, that $$c_1[t]=\sum_{k=1}^{d_1(t)-1}{M_t[k]}\neq c[t] \textrm{   and  }  c_2[t]=\sum_{k=1}^{d_2(t)-1}{M_t[k]} \neq c[t],$$ and therefore we have $d_1(t)<d(t)$ and $d_2(t)<d(t)$.

From 