+\newcommand{\ns}{$\hspace{.1em}$}
\begin{xpl}
Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=2$), and that
\end{array}
\right.$.
-So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.010004000000000000000000011005 ...$
+So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.01\ns00\ns04\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns01\ns10\ns05 ...$
Indeed, the $p=2$ first digits are 01, as $|v^0-\check{v}^0|=1$,
and we use $p$ digits to code this difference ($\mathcal{P}$ being $\{1,2,11\}$, this difference can be equal to 10). We then take the $v^0=1$ first terms of $u$, each term being coded in $n=2$ digits, that is, 06. As we can iterate
at most $\max{(\mathcal{P})}$ times, we must complete this
Conversely, if $\Gamma_{\mathcal{P}}(f)$ is not strongly connected, then there are
2 vertices $e_1$ and $e_2$ such that there is no path between $e_1$ and $e_2$.
That is, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$
-and $n \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$
+and $n\in \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$
cannot be reached from any neighborhood of $e_1$, and thus $G_f$ is not transitive.
\end{proof}
$$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
\subset \mathcal{B}(x,\varepsilon),$$
and $y=G_f^{k_1}(e,(u,v))$. $\Gamma_{\mathcal{P}}(f)$ being strongly connected,
-there is at least a path from the Boolean state $y_1$ of $y$ and $e$.
+there is at least a path from the Boolean state $y_1$ of $y$ and $e$ \ANNOT{Phrase pas claire : "from ... " mais pas de "to ..."}.
Denote by $a_0, \hdots, a_{k_2}$ the edges of such a path.
Then the point:
$$(e,((u^0, ..., u^{v^{k_1-1}},a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
