diagrams. The only correct diagram $\delta$ connects $\gstar{x, -b(g(x))}$ and $\gstar{+b(y), y}$. We have to solve the problem $\mathcal{P}(\delta) = \{b(g(x)) \doteq b(y)\}$ producing $\theta = \{y \mapsto g(x)\}$ and we finally obtain $\exec(\Sigma_{ex}) = \theta\gstar{x, y} = \gstar{x, g(x)}$ since $\Sigma_{ex}$ has only two diagrams. One can check that the other diagram fails.

\[def:strongnormal\] A constellation $\Sigma$ is strongly normalising w.r.t. a set of colours $\mathcal{A}\subseteq\mathcal{C}$ if and only if $\exec(\Sigma)$ is a finite constellation. When $\mathcal{A}=\mathcal{C}$, we simply say that $\Sigma$ is strongly normalising.

\[thm:associativity\] Let $\Sigma$ be a constellation, and $\mathcal{A,B}$ be two disjoint sets of colours, i.e. $\mathcal{A}\subseteq\mathcal{C}$, $\mathcal{B}\subseteq\mathcal{C}$, and $\mathcal{A}\cap\mathcal{B}=\emptyset$. Then: $$\exec^*_\mathcal{B}(\exec^*_\mathcal{A}(\Sigma)) = \exec^*_\mathcal{A\cup B}(\Sigma) = \exec^*_\mathcal{A}(\exec^*_\mathcal{B}(\Sigma)),\quad \mathrm{(*\in\{\emptyset,tree\}})$$

We establish the first isomorphism, which is enough by symmetry. First note that the disjointness of the sets of colours implies that the set of edges in $\ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$ is the disjoint union of the sets of edges in $\ugraph{\mathcal{A}}{\Sigma}$ and those in $\ugraph{\mathcal{B}}{\Sigma}$. As a consequence, any diagram on $\ugraph{\mathcal{A}}{\Sigma}$ (resp. $\ugraph{\mathcal{B}}{\Sigma}$) can be thought of as a diagram on $\ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$.

Let $\delta : D_\delta \rightarrow \ugraph{\mathcal{B}}{\exec^*_\mathcal{A}(\Sigma)}$ be a correct saturated $\mathcal{B}$-diagram of $\exec^*_\mathcal{A}(\Sigma)$. For each $s\in\abs{\exec^*_\mathcal{A}(\Sigma)}$, the star $\exec^*_\mathcal{A}(\Sigma)(s)$, which we will write $\sigma_s$, corresponds to 