where $E=\left\{ M_1,M_2 \right\}$. From the previous remark we have $ d(1)-1=\max(l(M_1),p)$ e $d(2)-1=\max(l(M_2),p)$. We have the following cases:\
$\bullet\ \textrm{Case}\  d(1)-1=l(M_1)$ and $d(2)-1=l(M_2)$.

We have $p \leq \min(l(M_1),l(M_2))$. Furthermore we have $p \leq \textrm{Comp}(M_1,M_2)$ because $T$ is well defined. Because $T_{E}=(p) \in \textrm{Cond}(\textbf{c})$ we have: $$c[1]=\sum_{k=1}^{d(1)-1}{M_1[k]} =\sum_{k=1}^{l(M_1)}{M_1[k]}  \textrm{ and }  c[2]=\sum_{k=1}^{d(2)-1}{M_2[k]} =\sum_{k=1}^{l(M_2)}{M_2[k]},$$ and from it we deduce, if $c[1] \neq 1$ and $c[2] \neq 1 $, that $M_1 \in \textrm{Cond}(c[1])$ and $M_2 \in \textrm{Cond}(c[2])$. If $c[1]=1$ then we can deduce $M_1=[1] \in \textrm{Cond}(0)$ (the same for $c[2]$).

The multiplicity sequence $M=[1]$ is the only one such that the sum of its entries up to to its length is not equal to the conductor of its associated Arf numerical semigroup (that is $\mathbb{N}$ with conductor $0$). In order to have more compact notation in the following, we define the set $\textrm{Cond}^{*}(c)$ for all $c \in \mathbb{N} \setminus \left\{ 0 \right\}$, where

-   $\textrm{Cond}^{*}(c)=\textrm{Cond}(c)  \textrm{ for all } c \neq 1;$

-   $\textrm{Cond}^{*}(1)=\textrm{Cond}(0).$

So in this case $T$ belongs to the following set: $$\begin{aligned}
  S^1(\textbf{c})=\left\{ T_{E}=(k): E=\left\{M_1,M_2\right\}; M_i \in\textrm{Cond}^{*}(c[i]) \textrm{  and } \right. \\  \left. 1 \leq  k \leq \min(\textrm{Comp}(M_1,M_2),l(M_1),l(M_2)) \right\}.  \end{aligned}$$ On the other hand, we can easily show that 