a sequence $n_i$ and elements $b_{i}\in A_{n_i}$ by induction. Let $n_{0} = n$ and $b_{0} = a$. Given $b_{i} \in A_{n_{i}}$, by hypothesis there are $m>n_{i}$ and $b \in A_{m}$ such that whenever $b' {\mathbin{R}}^{A_m}b$ it follows that ${\varphi}^{m}_{n_{i}}(b')=b_{i}$. Set $n_{i+1} = m$ and $b_{i+1} = b$. Thus ${\varphi}^{n_{i+1}}_{n_{i}}(b_{i+1}) = b_{i}$ for each $i$, so there exists $x \in \mathbb A $ such that ${\varphi}_{n_{i}} (x) = b_{i}$, for each $i \in {\mathbb N}$. In particular ${\varphi}_{n}(x) = a$. Let $y {\mathbin{R}}^{ \mathbb A }x$; if towards contradiction $y \neq x$ then there is $i \in {\mathbb N}$ such that ${\varphi}_{n_{i}}(y) \neq {\varphi}_{n_{i}}(x) = b_{i}$. But ${\varphi}_{n_{i+1}}(y) {\mathbin{R}}^{A_{n_{i+1}}}{\varphi}_{n_{i+1}}(x)= b_{i+1}$, so $ {\varphi}_{n_{i}}(y)= {\varphi}^{n_{i+1}}_{n_{i}} {\varphi}_{n_{i+1}}(y)= b_{i}$ by construction of $b_{i+1}$, a contradiction.

If $ {\varphi}: \mathbb A \to A$ is an epimorphism onto a finite $ {\mathcal{L}}_R$-structure $A$ and $a\in A$, we let $${\llbracket a \rrbracket}_{ {\varphi}}=p[ {\varphi}^{-1}(a)], \qquad {\llbracket A \rrbracket}_{ {\varphi}}= \setnew{ {\llbracket a \rrbracket}_{ {\varphi}}}{a\in A} .$$

If the quotient map $p: \mathbb A \to {{\raisebox{.1em}{$ \mathbb A \!$}\left/\raisebox{-.1em}{$R^{ \mathbb A }$}\right.}}$ is irreducible, then $ {\llbracket A \rrbracket}_{ {\varphi}}$ is a regular quasi-partition of $ {{\raisebox{.1em}{$ \mathbb A \!$}\left/\raisebox{-.1em}{$R^{ \mathbb A }$}\right.}} $ by , and the 