\oplus G_2$ the graph obtained from $G$ by taking the disjoint union of $G_1$ and $G_2$ and identifying vertices with the same label in the boundaries of $G_1$ and $G_2$. In $G_1 \oplus G_2$ there is an edge between two labeled vertices if there is an edge between them in $G_1$ or in $G_2$.

Given $G = G_1 \oplus G_2$ and $G_2'$, we say that $G' = G_1 \oplus G_2'$ is the graph obtained from $G$ by *replacing* $G_2$ with $G_2'$. The following notion was introduced by Bodlaender *el al*. [@BFL+09].

\[defi:cano\] Let $\Pi$ be a parameterized problem and let $t \in \mathbb{N}$. Given $G_1,G_2 \in {\ensuremath{{\mathcal}{B}_t}\xspace}$, we say that $G_1 \equiv_{\Pi} G_2$ if $\Lambda(G_1) = \Lambda(G_2)$ and there exists a transposition constant ${\ensuremath{\Delta_{\Pi,t}}\xspace}(G_1,G_2) \in \mathbb{Z}$ such that for every $H \in {\ensuremath{{\mathcal}{B}_t}\xspace}$ and every $k \in \mathbb{Z}$, it holds that $(G_1 \oplus H, k) \in \Pi$ if and only if $(G_2 \oplus H, k+{\ensuremath{\Delta_{\Pi,t}}\xspace}(G_1,G_2)) \in \Pi$.

A *tree decomposition* of a graph $G$ is a couple $(T,{\mathcal}{X} = \{ B_x : x \in V(T) \})$, where $T$ is a tree and such that $\bigcup_{x \in V(T)} B_x = V(G)$, for every edge $\{u,v\} \in E (G)$ there exists $x 