Definition 11 (Concatenation Grammar) A context-free grammar $G = (\mathcal{V}, \Sigma, \mathcal{P}, S)$ is called a concatenation grammar iff $\mathcal{P} \subseteq \mathcal{V} \times ((\mathcal{V} \cup \Sigma) \times (\mathcal{V} \cup \Sigma))$. $\square$
We say that a language is a concatenation language iff it has a generating concatenation grammar. For example, the language ${a, b}$ is not a concatenation language. The language ${a^n b^n \mid n \ge 1}$ is a concatenation language, since it is generated by the grammar $({S, X}, {a, b}, {S \to a b, S \to a X, X \to S b}, S)$.
Lemma 1 The intersection problem for concatenation languages is undecidable. $\square$
Proof 1 (Thanks to Harald Ganzinger who has suggested this proof) Let $G_1 = (\mathcal{V}_1, \Sigma_1, \mathcal{P}_1, S_1)$ and $G_2 = (\mathcal{V}_2, \Sigma_2, \mathcal{P}_2, S_2)$ be two arbitrary context-free grammars in Chomsky Normal Form.⁴ Let $#\notin \mathcal{V}_1 \cup \mathcal{V}_2 \cup \Sigma_1 \cup \Sigma_2$ be a new terminal symbol, for $i \in {1, 2}: \Sigma'i ={def} \Sigma_i \cup {#}, \mathcal{P}'i ={def} {A \to B C \mid A \to B C \in \mathcal{P}_i} \cup {A \to a# \mid A \to a \in \mathcal{P}_i}$, and $G'_i = (\mathcal{V}_i, \Sigma'_i, \mathcal{P}'_i, S_i)$.
Then, $G'_1 \cap G'_2 = \emptyset$ iff $G_1 \cap G_2 = \emptyset$. Since the latter is an undecidable problem for context-free grammars (e.g. see [21]), the former is undecidable as well. $\square$
Given an arbitrary concatenation grammar, the key-observation is now that one can simply reverse the productions $\mathcal{P}$ of the grammar and get a role box $\mathfrak{R}$. If a word can be derived “top down” by the grammar using a derivation tree, then it is possible to “parse” this word in a bottom-up style using the role axioms. The following Lemma fixes the relationship between words that are derivable by a concatenation grammar and the models of the role box corresponding to this grammar:
Lemma 2 Let $\mathcal{G} = (\mathcal{V}, \Sigma, \mathcal{P}, S)$ be an arbitrary concatenation grammar. Let $w = w_1...w_n$ be a word, $w \in \Sigma^+$ with $|w| \ge 2$, and $\mathcal{I}$ be a model of $(\exists w_1...\exists w_n.\top,\mathfrak{R})$ with $\mathfrak{R}={def} {B \circ C \sqsubseteq A \mid A \to B C \in \mathcal{P}}$. Let $\langle x_0, x_1 \rangle \in w_1^\mathcal{I}, ... \langle x{n-1}, x_n \rangle \in w_n^\mathcal{I}$ be an arbitrary path in the model $\mathcal{I}$ corresponding to $w$.
Let $V \in V$ be an arbitrary non-terminal of $\mathcal{G}$. Then, $\langle x_0, x_n \rangle \in V^\mathcal{I}$ holds in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top,\mathfrak{R})$ iff there is a derivation of $w$ having $V$ as the root node: we write $V \xrightarrow{\cdot} w$. As a consequence, $\langle x_0, x_n \rangle \in S^\mathcal{I}$ in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top,\mathfrak{R})$ iff $w \in L(\mathcal{G})$. $\square$
Proof 2 "⇔" This can be shown using induction over the length of $w$.
⁴A context-free grammar $G = (\mathcal{V}, \Sigma, \mathcal{P}, S)$ is in Chomsky Normal Form, iff $\mathcal{P} \subseteq \mathcal{V} \times ((\mathcal{V} \times \mathcal{V}) \cup \Sigma))$ (see [21]).