Answering Fuzzy Conjunctive Queries Over Finitely Valued Fuzzy Ontologies

Abstract

Fuzzy Description Logics (DLs) provide a means for representing vague knowledge about an application domain. In this paper, we study fuzzy extensions of conjunctive queries (CQs) over the DL ${\mathcal {SROIQ}}$ based on finite chains of degrees of truth. To answer such queries, we extend a well-known technique that reduces the fuzzy ontology to a classical one, and use classical DL reasoners as a black box. We improve the complexity of previous reduction techniques for finitely valued fuzzy DLs, which allows us to prove tight complexity results for answering certain kinds of fuzzy CQs. We conclude with an experimental evaluation of a prototype implementation, showing the feasibility of our approach.

Appendices

Appendix

Proof of Theorem 12

Table 7 depicts the reduction rules for transforming a ${\mathscr {C}}$-${\mathcal {SROIN}}$ ontology $\mathcal {O}$ into a classical ${\mathcal {SROIN}}$ ontology $\mathcal {O} _\mathsf {c}$. In this table, we use the notation

$$\begin{aligned} {\ominus ^-}(d) := \max \{d'\in {\mathscr {C}} \mid \ominus d'\geqslant d\}. \end{aligned}$$

Likewise, we define ${\otimes ^-}(d)$ as the set of all pairs $(d_1,d_2)\in {\mathscr {C}} ^2$ that satisfy $d_1\otimes d_2\geqslant d$ and are minimal w.r.t. the component-wise ordering on ${\mathscr {C}} ^2$. This means that all elements of ${\otimes ^-}(d)$ are incomparable, i.e., for all $(d_1,d_2),(d_1',d_2')\in {\otimes ^-}(d)$ we have either $d_1>d_1'$ and $d_2<d_2'$ or vice versa. The set ${\oplus ^-}(d)$ is defined analogously. For the implication, we need a slightly different definition, characterizing all pairs of elements whose implication does not exceed a specified value d. More precisely, we define ${\Rightarrow ^-}(d)$ as the set of all $(d_1,d_2)\in {\mathscr {C}} ^2$ satisfying $d_1\Rightarrow d_2<d$, and minimize here w.r.t. the first component and maximize w.r.t. the second component since $\Rightarrow$ is antitone in the first argument and monotone in the second argument.

Table 7 Mapping of concepts, roles, and axioms to classical ${\mathcal {SROIN}}$

Note that all expressions of the form $>d$ in Table 7 are well defined since we have $d<1$ in all such cases. In particular, it holds that ${\ominus ^-}(d)<1$ whenever $d>0$, and $d_2<1$ for all $(d_1,d_2)\in {\Rightarrow ^-}(d)$. We can now prove Theorem 12. In the following, let $\mathcal {O}$ be an arbitrary (not necessarily normalized) ontology in ${\mathscr {C}}$-${\mathcal {SROIN}}$, and $\mathcal {O} _\mathsf {c}$ be its reduced form according to Table 7.

1.1 Soundness

Consider first a classical model $\mathcal {J}$ of $\mathcal {O} _\mathsf {c}$. We construct a fuzzy interpretation $\mathcal {I}$, with the goal of showing that $\mathcal {I}$ is a model of $\mathcal {O}$, as follows (for all $x,y\in \Delta ^{\mathcal {J}}$, $a\in \mathsf{N_I}$, $A\in \mathsf{N_C}$, $r\in \mathsf{N_R}$, and $d\in {\mathscr {C}}$):

$$\begin{aligned}&\Delta ^{\mathcal {I}} :=\Delta ^{\mathcal {J}} \\&a^\mathcal {I}:=a^\mathcal {J} \\&A^{\mathcal {I}}\left( x\right) :=\max \big \{ {d}\mid x\in A_{\geqslant {d}}^{\mathcal {J}}\big \} \\&r^{\mathcal {I}}\left( x,y\right) :=\max \big \{ {d}\mid (x,y)\in r_{\geqslant {d}}^{\mathcal {J}}\big \} \end{aligned}$$

To show that $\mathcal {I}$ is a model of $\mathcal {O}$, we first prove the following proposition:

Proposition 19

Let C be a concept, r a role, $x,y\in \Delta ^\mathcal {I}$, and $d\in {\mathscr {C}} _>0$. Then, we have

$$\begin{aligned}&{C}^{\mathcal {I}}\left( x\right) \geqslant {d} \text { iff } x\in \rho \left( C,\geqslant {d}\right) ^\mathcal {J} \text { and}\\&{r}^{\mathcal {I}}\left( x,y\right) \geqslant {d} \text { iff } (x,y)\in \rho \left( r,\geqslant {d}\right) ^{\mathcal {J}}. \end{aligned}$$

Proof

For role names r, the claim holds by the construction of $r^\mathcal {I}$ and the fact that $r_{\geqslant d_{\mathrm{next}}}\sqsubseteq r_{\geqslant d}$ is contained in $\mathcal {O} _\mathsf {c}$ for every $d\in {\mathscr {C}} _{<1}$. For the universal role u, the equivalence trivially holds since we have that $(x,y)\in u^\mathcal {J}$ and $u^\mathcal {I} (x,y)=1$. Finally, for inverse roles it is an immediate consequence of the facts that $(r^{-})^\mathcal {I} (x,y)=$ $r^\mathcal {I} (y,x)$, and $(x,y)\in (r_{\geqslant d}^-)^\mathcal {J}$ iff $(y,x)\in r_{\geqslant d}^\mathcal {J}$.

The rest of the proposition is proved by induction on the structure of C. For concept names A, the equivalence holds by the definition of $A^\mathcal {I}$ and the fact that $A_{\geqslant d_{\mathrm{next}}}\sqsubseteq A_{\geqslant d}$ is in $\mathcal {O} _\mathsf {c}$ for every $d\in {\mathscr {C}} _{<1}$. For the induction step, we consider all possible concept constructors:

Top Concept: The proof for this case is an immediate consequence of the facts that $\top ^\mathcal {I} (x)=1$, $\rho (\top ,\geqslant d)=\top$, and $\top ^\mathcal {J} =\Delta ^\mathcal {J} =\Delta ^\mathcal {I}$.

Bottom Concept: For bottom, we have $\bot ^\mathcal {I} (x)=0$, $\rho (\bot ,\geqslant d)=\bot$, and $\bot ^\mathcal {J} =\emptyset$.

Concept Negation: We have that $x \in \rho (\lnot C,\geqslant d)^\mathcal {J}$ holds if and only if $x \notin \rho (C,>{\ominus ^-}(d))^\mathcal {J}$. By the induction hypothesis, this is equivalent to $C^\mathcal {I} (x)\leqslant \max \{d'\in {\mathscr {C}} \mid \ominus d'\geqslant d\}$. Since $\ominus$ is antitone, this is finally equivalent to $(\lnot C)^\mathcal {I} (x)=\ominus C^\mathcal {I} (x)\geqslant d$.

Concept Conjunction: If $C_1^\mathcal {I} (x)\otimes C_2^\mathcal {I} (x)\geqslant {d}$, then by the definition of ${\otimes ^-}(d)$ there is at least one pair $(d_1,d_2)\in {\otimes ^-}(d)$ such that $C_1^\mathcal {I} (x)\geqslant d_1$ and $C_2^\mathcal {I} (x)\geqslant d_2$. Since $d>0$, we also know that $d_1>0$ and $d_2>0$. Thus, by the induction hypothesis we have

$$\begin{aligned} x\in \big (\rho (C_1,\geqslant {d_1})\sqcap \rho (C_2,\geqslant {d_2})\big )^\mathcal {J} \subseteq \rho (C_1\sqcap C_2,\geqslant d)^\mathcal {J}. \end{aligned}$$

Conversely, suppose that $x\in (\rho (C_1,\geqslant d_1)\sqcap \rho (C_2,\geqslant d_2))^\mathcal {J}$ holds for some $(d_1,d_2)\in {\otimes ^-}(d)$. Then, by induction hypothesis, $C_1^\mathcal {I} (x)\geqslant {d_1}$ and $C_2^\mathcal {I} (x)\geqslant {d_2}$. Because of the monotonicity of $\otimes$ we then get that $C_1^\mathcal {I} (x)\otimes C_2^\mathcal {I} (x)\geqslant d_1\otimes d_2\geqslant d$, as we wanted to show.

The proof for disjunction is similar to the proof for conjunction.

Existential Restriction: Suppose that $(\exists r. C)^\mathcal {I} (x)\geqslant {d}$. Since ${\mathscr {C}}$ is finite, there must exist some $y\in \Delta ^\mathcal {I}$ with $r^\mathcal {I} (x,y)\otimes C^\mathcal {I} (y)\geqslant {d}$. We have $r^\mathcal {I} (x,y)\geqslant {d_1}$ and $C^\mathcal {I} (y)\geqslant {d_2}$ for some $({d_1},{d_2})\in {\otimes ^-}(d)$. By induction, we get $(x,y)\in \rho (r,\geqslant {d_1})^\mathcal {J}$ and $y\in \rho (C,\geqslant {d_2})^\mathcal {J}$. Therefore,

$$\begin{aligned} x \in \big (\exists \rho (r,\geqslant {d_1}).\rho (C,\geqslant {d_2}) \big )^\mathcal {J} \subseteq \rho (\exists r .C, \geqslant {d})^\mathcal {J}. \end{aligned}$$

Conversely, suppose that $x\in (\exists \rho (r,\geqslant {d_1}).\rho (C,\geqslant {d_2}))^\mathcal {J}$ for some pair $({d_1},{d_2})\in {\otimes ^-}(d)$. Thus, there exists y with $(x,y)\in \rho (r,\geqslant {d_1})^\mathcal {J}$ and $y \in \rho (C,\geqslant {d_2})^\mathcal {J}$. By induction, we have that $r^\mathcal {I} (x,y)\geqslant d_1$ and $C^\mathcal {I} (y)\geqslant d_2$, and therefore

$$\begin{aligned} (\exists r.C)^\mathcal {I} (x) \geqslant r^\mathcal {I} (x,y)\otimes C^\mathcal {I} (y) \geqslant d_1\otimes d_2 \geqslant d. \end{aligned}$$

Universal Restriction: If $(\forall r.C)^\mathcal {I} (x)\geqslant d$, then for all $y\in \Delta ^\mathcal {I}$ we have $\big (r^\mathcal {I} (x,y)\Rightarrow C^\mathcal {I} (y)\big )\geqslant {d}$. Consider any $y\in \Delta ^\mathcal {I}$ and $(d_1,d_2)\in {\Rightarrow ^-}(d)$ with $r^\mathcal {I} (x,y)\geqslant d_1$. If $C^\mathcal {I} (y)\leqslant d_2$, then we immediately get that

$$\begin{aligned} r^\mathcal {I} (x,y)\Rightarrow C^\mathcal {I} (y) \leqslant d_1\Rightarrow d_2 < d, \end{aligned}$$

which contradicts our assumption. Thus, we must have $C^\mathcal {I} (y)>d_2$, and hence $x\in \big (\forall \rho (r,\geqslant d_1).\rho (C,>d_2)\big )^\mathcal {J}$ by the induction hypothesis (as $d_1>0$). As this argument applies to all pairs $(d_1,d_2)\in {\Rightarrow ^-}(d)$, we obtain $x\in \rho (\forall r.C,\geqslant d)^\mathcal {J}$, as required.

For the opposite direction, assume that $(\forall r.C)^\mathcal {I} (x)<d$. Then there must be a $y\in \Delta ^\mathcal {I}$ such that $r^\mathcal {I} (x,y)\Rightarrow C^\mathcal {I} (y)<d$. By the definition of ${\Rightarrow ^-}(d)$, we can find a pair $(d_1,d_2)\in {\Rightarrow ^-}(d)$ with $r^\mathcal {I} (x,y)\geqslant d_1$ and $C^\mathcal {I} (y)\leqslant d_2$. Since $d_1>0$, the induction hypothesis yields that $(x,y)\in \rho (r,\geqslant d_1)^\mathcal {J}$ and $y\notin \rho (C,>d_2)^\mathcal {J}$. But then, this implies that $x\notin \rho (\forall r.C,\geqslant d)^\mathcal {J}$.

Nominals: Consider the case where $C=\{d_1/o_1,\ldots ,d_m/o_m\}$ such that $o_1,\dots ,o_m\in \mathsf{N_I}$ and $d_1,\ldots d_m\in {\mathscr {C}}_{>0}$. Then, $C^\mathcal {I} (x)\geqslant {d}$ iff $x=o_i^\mathcal {I}$ for some $i\in \{1,\dots ,m\}$ with $d_i\geqslant {d}$, which in turn is equivalent to

$$\begin{aligned} x \in \{o_i^\mathcal {I} \mid d_i\geqslant d,\ i\in \{1,\dots ,m\}\} = \rho (C,\geqslant {d})^\mathcal {J}. \end{aligned}$$

Unqualified Number Restrictions: The fact that $({\ge }m\,r)^\mathcal {I} (x)\geqslant d$ is equivalent to the existence of m different elements $y_1,\dots ,y_m\in \Delta ^\mathcal {I}$ such that $r^\mathcal {I} (x,y_i)\geqslant d$ holds for all $i\in \{1,\dots ,m\}$. This is in turn equivalent to the existence of such $y_i$ with $(x,y_i)\in \rho (r,\geqslant d)^\mathcal {J}$ for all i, and hence to $x\in ({\ge }m\,\rho (r,\geqslant d))^\mathcal {J}$.

The proof for unqualified at-most restrictions can be obtained by a combination of previous arguments for ${\ge }m\,r$ and $\lnot C$.

Local Reflexivity: We have $(\exists r.\text {Self})^\mathcal {I} (x)\geqslant {d}$ iff $r^\mathcal {I} (x,x)\geqslant {d}$, which is equivalent to $(x,x)\in \rho (r,\geqslant {d})^\mathcal {J}$, and to $x\in (\exists \rho (r,{\geqslant {d}}).\text {Self})^\mathcal {J}$. $\square$

To finish the proof of the first direction of Theorem 12, it remains to show the following lemma.

Lemma 20

If $\mathcal {J}$ is a classical model of $\mathcal {O} _\mathsf {c}$, then $\mathcal {I}$ is a fuzzy model of $\mathcal {O}$.

Proof

We need to show that $\mathcal {I}$ satisfies all axioms in $\mathcal {O}$:

Concept Assertions: Suppose that $\mathcal {O}$ contains the concept assertion $C(a)\geqslant d$, and thus $\mathcal {O} _\mathsf {c}$ contains the $\rho (C,\geqslant {d})(a)$. Since $\mathcal {J}$ is a model of $\mathcal {O} _\mathsf {c}$, we have $a^\mathcal {J} \in \rho (C,\geqslant d)^\mathcal {J}$, and by Proposition 19 that $C^\mathcal {I} (a^\mathcal {I})\geqslant d$. Similarly, for an assertion $C(a)\leqslant d$ in $\mathcal {O}$, we have $a^\mathcal {J} \notin \rho (C,>d)^\mathcal {J}$, and thus $C^\mathcal {I} (a^\mathcal {I})\leqslant d$.

Other ABox Axioms: The proof for role assertions can be obtained by adapting the proof for concept assertions. Axioms of the form $a\ne b$, $a=b$ are trivially satisfied since $\Delta ^\mathcal {I} =\Delta ^\mathcal {J}$ and for every individual $a\in \mathsf{N_I}$ we have $a^\mathcal {I} =a^\mathcal {J}$.

Concept Inclusions: Suppose that our ontology contains the concept inclusion $\langle C\sqsubseteq D \geqslant d\rangle$ and assume that there is a $x\in \Delta ^\mathcal {I}$ such that $C^\mathcal {I} (x)\Rightarrow D^\mathcal {I} (x)<d$. Thus, there exists a pair $(d_1,d_2)\in {\Rightarrow ^-}(d)$ such that $C^\mathcal {I} (x)\geqslant d_1$ and $D^\mathcal {I} (x)\leqslant d_2$. Since $d_1>0$, Proposition 19 yields $x\in \rho (C,\geqslant d_1)^\mathcal {J}$ and $x\notin \rho (D,>d_2)^\mathcal {J}$, which contradicts the fact that $\mathcal {J}$ satisfies $\rho (C,\geqslant d_1)\sqsubseteq \rho (D,>d_2)$.

Role Inclusions: Suppose that our ontology contains the role inclusion $\langle r_1r_2\sqsubseteq r\geqslant d\rangle$ and it holds that

$$\begin{aligned} \big (r_1^\mathcal {I} (x,y)\otimes r_2^\mathcal {I} (y,z)\big ) \Rightarrow r^\mathcal {I} (x,z) < d, \end{aligned}$$

or equivalently,

$$\begin{aligned} r_1^\mathcal {I} (x,y)\Rightarrow \big (r_2^\mathcal {I} (y,z)\Rightarrow r^\mathcal {I} (x,z)\big ) < d, \end{aligned}$$

for some $x,y,z\in \Delta ^\mathcal {I}$. Then, there exist $(d_1,d')\in {\Rightarrow ^-}(d)$ such that $r_1^\mathcal {I} (x,y)\geqslant d_1$ and $r_2^\mathcal {I} (y,z)\Rightarrow r^\mathcal {I} (x,z)\leqslant d'$. The latter implies the existence of $(d_2,d_3)\in {\Rightarrow ^-}(d'_{\mathrm{next}})$ with $r_2^\mathcal {I} (y,z)\geqslant d_2$ and $r^\mathcal {I} (x,z)\leqslant d_3$. Proposition 19 yields that $(x,y)\in \rho (r_1,\geqslant d_1)^\mathcal {J}$, $(y,z)\in \rho (r_2,\geqslant d_2)^\mathcal {J}$, and $(x,z)\notin \rho (r_3,>d_3)^\mathcal {J}$, which contradicts the fact that $\mathcal {J}$ satisfies $\rho (r_1,\geqslant d_1)\rho (r_2,\geqslant d_2)\sqsubseteq \rho (r,>d_3)$.

Disjoint Role Axioms: Suppose that our ontology contains the axiom $\text {dis}(r_1,r_2)$. We show that for all $x,y\in \Delta ^\mathcal {I}$, either $r_1^\mathcal {I} (x,y)=0$ or $r_2^\mathcal {I} (x,y)=0$. Since $\mathcal {J}$ satisfies $\mathcal {O} _\mathsf {c}$, $\rho (r_1,>0)^\mathcal {J} \cap \rho (r_2,>0)^\mathcal {J} =\emptyset$. By Proposition 19, there can be no pair $x,y\in \Delta ^\mathcal {I}$ such that $r_1^\mathcal {I} (x,y)>0$ and $r_2^\mathcal {I} (x,y)>0$, as we wanted to show.

The proofs for the other role axioms are similar.$\square$

1.2 Completeness

Conversely, we consider a fuzzy model $\mathcal {I}$ of $\mathcal {O}$, and define the classical interpretation $\mathcal {J}$ as follows (for all $x,y\in \Delta ^{\mathcal {I}}$, $a\in \mathsf{N_I}$, $A\in \mathsf{N_C}$, $r\in \mathsf{N_R}$, and $d\in {\mathscr {C}}_{>0}$):

$$\begin{aligned}&\Delta ^{\mathcal {J}} := \Delta ^{\mathcal {I}} \\&a^\mathcal {J} \nonumber := a^\mathcal {I} \\&A_{\geqslant {d}}^{\mathcal {J}}:=\{x\mid A^{\mathcal {I}}\left( x\right) \geqslant d \}\\&r_{\geqslant {d}}^{\mathcal {J}} := \{(x,y)\mid r^{\mathcal {I}}\left( x,y\right) \geqslant d\} \end{aligned}$$

We again prove a connection similar to the one of Proposition 19.

Proposition 21

Let C be a concept, r a role, $x,y\in \Delta ^\mathcal {I}$, and $d\in {\mathscr {C}} _>0$. Then, we have

$$\begin{aligned}&x\in \rho \left( C,\geqslant {d}\right) ^\mathcal {J} \text { iff } {C}^{\mathcal {I}}\left( x\right) \geqslant {d} \ \text { and}\\&(x,y)\in \rho \left( r,\geqslant {d}\right) ^{\mathcal {J}}\text { iff } {r}^{\mathcal {I}}\left( x,y\right) \geqslant {d}. \end{aligned}$$

Proof

The proof is nearly the same as for Proposition 19, the only difference being the induction base cases. But it is easy to show the claim for concept and role names, given the definition of $\mathcal {J}$.$\square$

Lemma 22

If $\mathcal {I}$ is a fuzzy model of $\mathcal {O}$, then $\mathcal {J}$ is a classical model of $\mathcal {O} _\mathsf {c}$.

Proof

We need to show that $\mathcal {J}$ satisfies all axioms in $\mathcal {O} _\mathsf {c}$:

Concept Assertions: Suppose that $\mathcal {O} _\mathsf {c}$ contains the concept assertion $\rho (C,\geqslant {d})(a)$. By the construction of $\mathcal {O} _\mathsf {c}$, $C(a)\geqslant d$ appears in $\mathcal {O}$. Since $\mathcal {I}$ is a model of $\mathcal {O}$, we have $C^\mathcal {I} (a)\geqslant d$, and by Proposition 21 we get $a^\mathcal {J} \in \rho (C,\geqslant d)^\mathcal {J}$, as we wanted to show.

If $\mathcal {O} _\mathsf {c}$ contains an assertion $\lnot \rho (C,> {d})(a)$, then $C(a)\leqslant d$ appears in $\mathcal {O}$ and consequently $C^\mathcal {I} (a)\leqslant d$. By Proposition 21, we have that $a\not \in \rho (C,>d)^\mathcal {J}$, as we wanted to show.

Other ABox Axioms: The proof for role assertions can be obtained by adapting the proof for concept assertions. Axioms of the form $a\ne b$, $a=b$ are trivially satisfied since $\Delta ^\mathcal {I} =\Delta ^\mathcal {J}$ and for every individual $a\in \mathsf{N_I}$ we have $a^\mathcal {I} =a^\mathcal {J}$.

Concept Inclusions: Suppose that our ontology contains a concept inclusion $\rho (C,\geqslant d_1)\sqsubseteq \rho (D,>d_2)$ that is not satisfied. Thus, there exists some $x\in \Delta ^\mathcal {J}$ such that $x\in \rho (C,\geqslant d_1)^\mathcal {J}$ and $x\not \in \rho (D,> d_2)^\mathcal {J}$ and, by Proposition 21, $C^\mathcal {I} (x)\geqslant d_1$ and $D^\mathcal {I} (x)\leqslant d_2$. By the construction of $\mathcal {O} _\mathsf {c}$, we have $\langle C\sqsubseteq D \geqslant d\rangle$ in $\mathcal {O}$ and $(d_1\Rightarrow d_2)< d$ for some $d\in {\mathscr {C}}_{>0}$. By the properties of $\Rightarrow$, we get $C^\mathcal {I} (x)\Rightarrow D^\mathcal {I} (x)\leqslant d_1\Rightarrow d_2<d$, which contradicts our assumption that $\mathcal {I}$ is a model of $\mathcal {O}$.

All concept inclusions of the form $A_{\geqslant d_{\mathrm{next}}}\sqsubseteq A_{\geqslant d}$ are trivially satisfied by the construction of $\mathcal {J}$.

Role Inclusions: Assume that a role inclusion

$$\begin{aligned} \rho (r_1,\geqslant d_1)\rho (r_2,\geqslant d_2)\sqsubseteq \rho (r,>d_3) \in \mathcal {O} _\mathsf {c} \end{aligned}$$

is violated, i.e., there are three elements $x,y,z\in \Delta ^\mathcal {J}$ such that we have $(x,y)\in \rho (r_1,\geqslant d_1)^\mathcal {J}$, $(y,z)\in \rho (r_2,\geqslant d_2)^\mathcal {J}$, and $(x,z)\not \in \rho (r,> d_3)^\mathcal {J}$. Proposition 21 implies that:

$$\begin{aligned} r_1^\mathcal {I} (x,y)&\geqslant d_1,\quad&r_2^\mathcal {I} (y,z)&\geqslant d_2,\quad&r(x,z)&\leqslant d_3. \end{aligned}$$

(8)

By construction of $\mathcal {O} _\mathsf {c}$, we have that $\langle r_1r_2\sqsubseteq r\geqslant d\rangle \in \mathcal {O}$, $(d_1\Rightarrow d')<d$, and $(d_2\Rightarrow d_3)<d'_{\mathrm{next}}$ for some $d,d'\in {\mathscr {C}}_{>0}$. We obtain

$$\begin{aligned} \big (d_1\Rightarrow (d_2\Rightarrow d_3)\big ) \leqslant (d_1\Rightarrow d') < d, \end{aligned}$$

and hence $\big ((d_1\otimes d_2)\Rightarrow d_3\big ) < d$.

Along with (8) and the monotonicity and antitonicity properties of the operators $\otimes$ and $\Rightarrow$, this implies that

$$\begin{aligned} \left( \left( r_1^\mathcal {I} (x,y)\otimes r_2^\mathcal {I} (y,z)\right) \Rightarrow r^\mathcal {I} (x,z)\right) <d \end{aligned}$$

which is absurd since $\mathcal {I}$ is a model of $\mathcal {O}$ and $\langle r_1r_2\sqsubseteq r\geqslant d\rangle \in \mathcal {O}$.

All role inclusions of the form $r_{\geqslant d_{\mathrm{next}}}\sqsubseteq r_{\geqslant d}$ are trivially satisfied by the construction of $\mathcal {J}$.

Disjoint Role Axioms: Suppose that $\mathcal {O} _\mathsf {c}$ contains the disjoint role axiom $\text {dis}(\rho (r_1,>0),\rho (r_2,>0))$. By construction of $\mathcal {O} _\mathsf {c}$, we also have that $\text {dis}(r_1,r_2)\in \mathcal {O}$, and therefore either $r_1(x,y)=0$ or $r_2(x,y)=0$ for all $x,y\in \Delta ^\mathcal {I}$. Proposition 21 now implies that the axiom

$$\begin{aligned} \text {dis}(\rho (r_1,>0),\rho (r_2,>0)) \end{aligned}$$

is satisfied.

The proofs for the other role axioms are similar.$\square$

Proof of Lemma 13

To determine the size of $\mathcal {O} _\mathsf {c}$ for a normalized ${\mathscr {C}}$-${\mathcal {SROIN}}$ ontology $\mathcal {O}$, we start by analyzing the size of the sets ${\otimes ^-}(d)$, ${\oplus ^-}(d)$ and ${\Rightarrow ^-}(d)$ (defined in the beginning of the proof of Theorem 12). It is clear that for every $d_1\in {\mathscr {C}}$ there can be at most one element $d_2\in {\mathscr {C}}$ such that $(d_1,d_2)$ is contained in any of these sets. This is due to the minimization conditions in their definitions. Thus, the size of these sets is at most linear in the size of ${\mathscr {C}}$. Consequently, the size of any expression of the form $\rho (C,\geqslant d)$, where C is a complex concept that contains only one concept constructor, is at most linear in the sizes of C and ${\mathscr {C}}$ (cf. Table 7).

Since $\mathcal {O}$ is normalized, ABox axioms contain no complex concepts, and hence the size of $\kappa (\alpha )$ for any such axiom $\alpha$ is the same as the size of $\alpha$.

Consider now a GCI $\alpha :=\langle C\sqsubseteq D\geqslant d\rangle$ and its reduced form, containing a GCI $\rho (C,\geqslant d_1)\sqsubseteq \rho (D,>d_2)$ for each pair $(d_1,d_2)\in {\Rightarrow ^-}(d)$. Since $\alpha$ contains at most one concept constructor, the size of each reduced axiom is linear in the sizes of $\alpha$ and ${\mathscr {C}}$. Moreover, there are at most linearly many such axioms (in the size of ${\mathscr {C}}$), bringing the total size of $\kappa (\alpha )$ to at most linear in the size of $\alpha$ and quadratic in the size of ${\mathscr {C}}$.

Likewise, for a role inclusion $\alpha :=\langle r_1r_2\sqsubseteq r\geqslant d\rangle$ the number of pairs $(d_1,d')\in {\Rightarrow ^-}(d)$ is linear in the size of ${\mathscr {C}}$, and for each of these pairs we additionally have to consider linearly many pairs of the form $(d_2,d_3)\in {\Rightarrow ^-}(d'_{\mathrm{next}})$. Thus, the same bounds are valid for role inclusions. The proof for the remaining role axioms is trivial.

In summary, the total size of $\mathcal {O} _\mathsf {c}$ is bounded linearly in the size of $\mathcal {O}$ and quadratically in the size of ${\mathscr {C}}$. $\square$