that are easier to study, because the twisted one can be always obtained by permutation of an untwisted tree.

The aim of this and the following section is to find a procedure that let us to determine all the local Arf semigroup $S \subseteq\mathbb{N}^r$ with a given conductor $\textbf{c} \in \mathbb{N}^r$. For the Remark \[remark0\], we can focus only on the untwisted trees.

We denote by $\textrm{Cond}(\textbf{c})$ the set of all the untwisted multiplicity trees of Arf semigroups in $\mathbb{N}^r$ with conductor $\textbf{c}\in \mathbb{N}^r$ (in the case $r=1$ we have the multiplicity sequences and from the previous section we have a procedure to determine such a set).

We notice the following general fact.

Let $S$ be an Arf semigroup of $\mathbb{N}^r$, $T$ the corresponding multiplicity tree and $M_i$ for $i=1,\ldots,n $ the multiplicity sequences of its branches.

We introduce the following integers $$d(i)=\min\left\{  j \in \mathbb{N}: M_i[j]=1 \textrm{ and  the $i-$th branch is not glued to other branches at level $j$} \right\},$$ for $  i=1,\ldots,r .$

Then $\textbf{c}=(c[1],\ldots,c[n])$ is the conductor of $S$ where $$c[i]=\sum_{k=1}^{d(i)-1}{M_i[k]}  \textrm{ for }  i=1,\ldots,n.$$

**Proof**. Denote by $N(T)= \left\{ \textbf{n}_i^j \right\}$ the set of the nodes of $T$. We will call $e_i=(0,\ldots,0,1,0,\ldots,0)$ where the non 