Figure 6: “Bottom up parsing” of $aabb \in \mathcal{L}(\mathcal{G}_1) \cap \mathcal{L}(\mathcal{G}_2)$
$x_4 \in (C \sqcap D)^I$ must hold, which obviously contradicts $x_4 \in (\neg(C \sqcap D))^I$. The example is therefore unsatisfiable. Considering Figure 6, it can be seen that the role box performs a “bottom up parsing” of the word $aabb$ – the two derivation trees shown in the figure can be immediately discovered as role compositions in the graph.
We can now prove the main result of this section by showing how to reduce the intersection problem of concatenation grammars to the satisfiability problem of $\mathcal{ALC}_{\mathcal{R}A\ominus}$:
Theorem 1 The satisfiability problem of $\mathcal{ALC}_{\mathcal{R}A\ominus}$ is undecidable. $\square$
Proof 4 We give an example for a pair $(E, \mathfrak{R})$ for which no algorithm exists that is capable of checking its satisfiability.
Let $\mathcal{G}_1 = (\mathcal{V}_1, \Sigma_1, \mathcal{P}_1, S_1)$ and $\mathcal{G}_2 = (\mathcal{V}_2, \Sigma_2, \mathcal{P}_2, S_2)$ be two arbitrary concatenation grammars. Without loss of generality we assume $\mathcal{V}_1 \cap \mathcal{V}_2 = \emptyset$.
For $i \in {1, 2}$, we define $\mathfrak{R}i ={def} {B \circ C \subseteq A \mid A \to B C \in \mathcal{P}i}$. Let $\Sigma ={def} \Sigma_1 \cup \Sigma_2$ and $\mathfrak{R} =_{def} \mathfrak{R}_1 \cup \mathfrak{R}_2$. Let $R? \notin \text{roles}(\mathfrak{R})$, and let
be the completion of $\mathfrak{R}$.
Then, $(E, \mathfrak{R}')$ is satisfiable iff $\mathcal{L}(\mathcal{G}_1) \cap \mathcal{L}(\mathcal{G}_2) = \emptyset$, where