• If $|w| = 2$, $w = w_1w_2$, and $V \xrightarrow{+} w$, then there must be a production of the form $V \to w_1w_2 \in \mathcal{P}$. Note that there cannot be productions of the form $V \to w_1B$, $V \to Aw_2$, $V \to AB$, since $\mathcal{G}$ is a concatenation grammar – we would additionally need productions of the form $A \to w_1\ldots$, or even productions of the form $A \to \epsilon$. If $\mathcal{I}$ is a model of $\mathfrak{R}$ and $\langle x_0, x_1 \rangle \in w_1^\mathcal{I}$, $\langle x_1, x_2 \rangle \in w_2^\mathcal{I}$, then, due to $w_1 \circ w_2 \subseteq V \in \mathfrak{R}$ we have $\langle x_0, x_2 \rangle \in V^\mathcal{I}$ in every model $\mathcal{I}$.
• Let $w = w_1 \dots w_n$, $n \ge 3$. Let $V \xrightarrow{+} w$. Since $\mathcal{G}$ is a concatenation grammar, there must be a production of the form $V \to XY \in \mathcal{P}$, and the following cases can occur:
$X \in \mathcal{V}$, $Y \in \mathcal{\Sigma}$: then, there is a derivation $X \xrightarrow{+} w_1 \dots w_{n-1}$, and $Y = w_n$. Due to the induction hypothesis we have $\langle x_0, x_{n-1} \rangle \in X^{\mathcal{I}}$ in every model $\mathcal{I}$. Since we consider a model of $(\exists w_1...\exists w_{n-1}.\exists w_n.\top, \mathfrak{R})$, with $\langle x_{n-1}, x_n \rangle \in w_n^{\mathcal{I}}$, we have $\langle x_0, x_n \rangle \in V^{\mathcal{I}}$, because $\mathcal{I}$ is a model of $\mathfrak{R}$ with $X \circ w_n \subseteq V \in \mathfrak{R}$.
$X \in \mathcal{\Sigma}$, $Y \in \mathcal{V}$: same argumentation.
$X \in \mathcal{V}$, $Y \in \mathcal{V}$: let $w = uv$ be the partition of $w$ corresponding to the derivations $X \xrightarrow{+} u$, $Y \xrightarrow{+} v$. Let $u = w_1\dots w_i$, $v = w_{i+1}\dots w_n$. Due to the induction hypothesis we have $\langle x_0, x_i \rangle \in X^{\mathcal{I}}$ and $\langle x_{i+1}, x_n \rangle \in Y^{\mathcal{I}}$, since both $u$ and $v$ have a length smaller than $n$. We have $X \circ Y \subseteq V \in \mathfrak{R}$. This shows that $\langle x_0, x_n \rangle \in V^{\mathcal{I}}$.
Summing up we have shown that $\langle x_0, x_n \rangle \in V^{\mathcal{I}}$ holds in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top, \mathfrak{R})$, if $V \xrightarrow{+} w$.
“⇒” If $\langle x_0, x_n \rangle \in V^\mathcal{I}$ holds in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top, \mathfrak{R})$, then the presence of $\langle x_0, x_n \rangle \in V^\mathcal{I}$ is a logical consequence of $(\exists w_1...\exists w_n.\top, \mathfrak{R})$. Therefore, $\langle x_0, x_n \rangle \in V^\mathcal{I}$ is enforced by the role axioms in $\mathfrak{R}$. One can easily construct a derivation tree for $w$, showing that $V \xrightarrow{+} w$, by inspecting one of these models. More formally this could be shown using induction as well, and the proof would be very similar to the previous one.
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Since we are trying to reduce the intersection problem of concatenation grammars to the satisfiability problem of $\mathcal{ALC}_{\mathbb{R}\mathcal{A}}^\ominus$, we have to deal with two grammars. Please note that concatenation grammars are not closed under intersection (i.e. for two grammars $\mathcal{G}1$ and $\mathcal{G}2$ there is in general no concatenation grammar $\mathcal{G}{1,2}$ such that $\mathcal{L}(\mathcal{G}{1,2}) = \mathcal{L}(\mathcal{G}_1) \cap \mathcal{L}(\mathcal{G}_2)$). In order to deal with this problem we have two put two concatenation grammars into one role box:
Lemma 3 Let $\mathcal{G}_1 = (\nu_1, \Sigma_1, P_1, S_1)$ and $\mathcal{G}_2 = (\nu_2, \Sigma_2, P_2, S_2)$ be two arbitrary concatenation grammars. Without loss of generality can assume $\nu_1 \cap \nu_2 = \emptyset$,