since we can always consistently rename the variables in one of the grammars and get $\mathcal{V}_1 \cap \mathcal{V}_2 = \emptyset$.
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 $\mathfrak{R} =_{def} \mathfrak{R}_1 \cup \mathfrak{R}_2$.
Then, for $i \in {1, 2}$, $w \in \mathcal{L}(\mathcal{G}_i)$ iff $\langle x_0, x_n \rangle \in S_i^\mathcal{I}$ in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top, \mathfrak{R})$. Obviously, $w \in \mathcal{L}(\mathcal{G}_1) \cap \mathcal{L}(\mathcal{G}_2)$ iff $\langle x_0, x_n \rangle \in S_1^\mathcal{I} \cap S_2^\mathcal{I}$ in all models $\mathcal{I}$ of $(\exists w_1...\exists w_n.\top, \mathfrak{R})$. $\square$
Proof 3 An easy consequence of the previous Lemma and of the requirement that $\mathcal{V}_1 \cap \mathcal{V}_2 = \emptyset$ (the derivation trees do not become "mixed", i.e. each grammar solely uses its own productions). $\square$
As an application of this Lemma, let us consider the two grammars
$\mathcal{G}_1 = ({\mathcal{S}_1}, {\boldsymbol{a}, \boldsymbol{b}}, \mathcal{P}_1, S_1)$, where $\mathcal{P}_1 = {\boldsymbol{S}_1 \to \boldsymbol{a}\boldsymbol{b} \mid a\boldsymbol{S}_1\boldsymbol{b}}$,
$\mathcal{G}_2 = ({\mathcal{S}_2}, {\boldsymbol{a}, \boldsymbol{b}}, \mathcal{P}_2, S_2)$, where $\mathcal{P}_2 = {\boldsymbol{S}_2 \to \boldsymbol{a}\boldsymbol{a}\boldsymbol{b} \mid aa\boldsymbol{S}_2\boldsymbol{b}\boldsymbol{b}}$.
Obviously, $\mathcal{L}(\mathcal{G}_1) = {\boldsymbol{a}^n\boldsymbol{b}^n \mid n \ge 1}$ and $\mathcal{L}(\mathcal{G}_2) = {\boldsymbol{a}^{2n}\boldsymbol{b}^{2n} \mid n \ge 1}$. Transformed into concatenation grammars we get
$\mathcal{G}'_1 = ({\mathcal{S}_1, A}, {\boldsymbol{a}, \boldsymbol{b}}, \mathcal{P}'_1, S_1)$, where $\mathcal{P}'_1 = {\boldsymbol{S}_1 \to \boldsymbol{a}\boldsymbol{b} \mid aA, A \to S_1\boldsymbol{b}}$, and
$\mathcal{G}'_2 = ({\mathcal{S}_2, B, C, D, E, F}, {\boldsymbol{a}, \boldsymbol{b}}, \mathcal{P}'_2, S_2)$, where $\mathcal{P}'_2 = {\begin{aligned}[t] &\boldsymbol{S}_2 \to aB, B \to aC, C \to bb, \ &S_2 \to aD, D \to aE, E \to S_2F, F \to bb \end{aligned}}$
The corresponding role box is
\begin{align*} \mathfrak{R} &= \left\{ \begin{array}{@{}l@{}} a \circ b \sqsubseteq S_1, a \circ A \sqsubseteq S_1, S_1 \circ b \sqsubseteq A \\ a \circ B \sqsubseteq S_2, a \circ C \sqsubseteq B, b \circ b \sqsubseteq C, \\ a \circ D \sqsubseteq S_2, a \circ E \sqsubseteq D, S_2 \circ F \sqsubseteq E, b \circ b \sqsubseteq F \end{array} \right\} \\ &\cup \\ &\left\{ \begin{array}{@{}l@{}} a^0 b^0 c^0 d^0 e^0 f^0 \\ a^1 b^1 c^1 d^1 e^1 f^1 \\ a^2 b^2 c^2 d^2 e^2 f^2 \\ a^3 b^3 c^3 d^3 e^3 f^3 \end{array} \right\}. \end{align*}
The “first part” of this role box corresponds to $\mathcal{P}'_1$, and the “sec- ond part” to $\mathcal{P}'_2$. The symbols of the grammars correspond to roles now. Please consider $(\forall S_1.C \sqcap \forall S_2.D \sqcap \exists a.\exists a.\exists b.\exists b.\neg(C \sqcap D), \mathfrak{R})$. Any model of $(\forall S_1.C \sqcap \forall S_2.D \sqcap \exists a.\exists a.\exists b.\exists b.\neg(C \sqcap D), \mathfrak{R})$ would also be a model of $(\exists a.\exists a.\exists b.\exists b.\top, \mathfrak{R})$, and must therefore contain $\langle x_0, x_4\rangle \in S_1^\mathcal{I} \cap S_2^\mathcal{I}$, because $w = aabb \in \mathcal{L}(\mathcal{G}'_1) \cap \mathcal{L}(\mathcal{G}'_2)$, due to Lemma 3. Since $x_0 \in (\forall S_1.C \sqcap \forall S_2.D)^\mathcal{I}$ also