Rational closure in \mathcal{SHIQ}
Abstract
We define a notion of rational closure for the logic \mathcal{SHIQ}, which does not enjoys the finite model property, building on the notion of rational closure introduced by Lehmann and Magidor in [23]. We provide a semantic characterization of rational closure in \mathcal{SHIQ} in terms of a preferential semantics, based on a finite rank characterization of minimal models. We show that the rational closure of a TBox can be computed in ExpTime using entailment in \mathcal{SHIQ}.
1 Introduction
Recently, a large amount of work has been done in order to extend the basic formalism of Description Logics (for short, DLs) with nonmonotonic reasoning features [26, 1, 10, 11, 13, 17, 20, 4, 2, 6, 25, 22]; the purpose of these extensions is that of allowing reasoning about prototypical properties of individuals or classes of individuals. In these extensions one can represent, for instance, knowledge expressing the fact that the hematocrit level is usually under 50%, with the exceptions of newborns and of males residing at high altitudes, that have usually much higher levels (even over 65%). Furthermore, one can infer that an individual enjoys all the typical properties of the classes it belongs to. As an example, in the absence of information that Carlos and the son of Fernando are either newborns or adult males living at a high altitude, one would assume that the hematocrit levels of Carlos and Fernando’s son are under 50%. This kind of inferences apply to individual explicitly named in the knowledge base as well as to individuals implicitly introduced by relations among individuals (the son of Fernando).
In spite of the number of works in this direction, finding a solution to the problem of extending DLs for reasoning about prototypical properties seems far from being solved. The most well known semantics for nonmonotonic reasoning have been used to the purpose, from default logic [1], to circumscription [2], to Lifschitz’s nonmonotonic logic MKNF [10, 25], to preferential reasoning [13, 4, 17], to rational closure [6, 9].
In this work, we focus on rational closure and, specifically, on the rational closure for \mathcal{SHIQ}. The interest of rational closure in DLs is that it provides a significant and reasonable nonmonotonic inference mechanism, still remaining computationally inexpensive. As shown for \mathcal{ALC} in [6], its complexity can be expected not to exceed the one of the underlying monotonic DL. This is a striking difference with most of the other approaches to nonmonotonic reasoning in DLs mentioned above, with some exception such as [25, 22]. More specifically, we define a rational closure for the logic \mathcal{SHIQ}, building on the notion of rational closure in [23] for propositional logic. This is a difference with respect to the rational closure construction introduced in [6] for \mathcal{ALC}, which is more similar to the one by Freund [12] for propositional logic (for propositional logic, the two definitions of rational closure are shown to be equivalent [12]). We provide a semantic characterization of rational closure in \mathcal{SHIQ} in terms of a preferential semantics, by generalizing to \mathcal{SHIQ} the results for rational closure in \mathcal{ALC} presented in [18]. This generalization is not trivial, since \mathcal{SHIQ} lacks a crucial property of \mathcal{ALC}, the finite model property [19]. Our construction exploits an extension of \mathcal{SHIQ} with a typicality operator {\bf T}, that selects the most typical instances of a concept C, {\bf T}(C). We define a minimal model semantics and a notion of minimal entailment for the resulting logic, \mathcal{SHIQ}^{\textsf{R}}{\bf T}, and we show that the inclusions belonging to the rational closure of a TBox are those minimally entailed by the TBox, when restricting to canonical models. This result exploits a characterization of minimal models, showing that we can restrict to models with finite ranks. We also show that the rational closure construction of a TBox can be done exploiting entailment in \mathcal{SHIQ}, without requiring to reason in \mathcal{SHIQ}^{\textsf{R}}{\bf T}, and that the problem of deciding whether an inclusion belongs to the rational closure of a TBox is in ExpTime.
Concerning ABox reasoning, because of the interaction between individuals (due to roles) it is not possible to separately assign a unique minimal rank to each individual and alternative minimal ranks must be considered. We end up with a kind of skeptical inference with respect to the ABox, whose complexity in ExpTime as well.
2 A nonmonotonic extension of \mathcal{SHIQ}
Following the approach in [14, 17], we introduce an extension of \mathcal{SHIQ} [19] with a typicality operator {\bf T} in order to express typical inclusions, obtaining the logic \mathcal{SHIQ}^{\textsf{R}}{\bf T}. The intuitive idea is to allow concepts of the form {\bf T}(C), whose intuitive meaning is that {\bf T}(C) selects the typical instances of a concept C. We can therefore distinguish between the properties that hold for all instances of C (C\sqsubseteq D), and those that only hold for the typical such instances ({\bf T}(C)\sqsubseteq D). Since we are dealing here with rational closure, we attribute to {\bf T} properties of rational consequence relation [23]. We consider an alphabet of concept names \mathcal{C}, role names \mathcal{R}, transitive roles \mathcal{R}^{+}\subseteq\mathcal{R}, and individual constants \mathcal{O}. Given A\in\mathcal{C}, S\in\mathcal{R}, and n\in\mathbb{N} we define:
C_{R}:=A\mid\top\mid\bot\mid\lnot C_{R}\mid C_{R}\sqcap C_{R}\mid C_{R}\sqcup C% _{R}\mid\forall S.C_{R}\mid\exists S.C_{R}\mid(\geq nS.C_{R})\mid(\leq nS.C_{R}) C_{L}:=C_{R}\mid{\bf T}(C_{R}) S:=R\mid R^{}
As usual, we assume that transitive roles cannot be used in number restrictions [19]. A KB is a pair (TBox, ABox). TBox contains a finite set of concept inclusions C_{L}\sqsubseteq C_{R} and role inclusions R\sqsubseteq S. ABox contains assertions of the form C_{L}(a) and S(a,b), where a,b\in\mathcal{O}.
The semantics of \mathcal{SHIQ}^{\textsf{R}}{\bf T} is formulated in terms of rational models: ordinary models of \mathcal{SHIQ} are equipped with a preference relation < on the domain, whose intuitive meaning is to compare the “typicality” of domain elements, that is to say, x<y means that x is more typical than y. Typical instances of a concept C (the instances of {\bf T}(C)) are the instances x of C that are minimal with respect to the preference relation < (so that there is no other instance of C preferred to x)^{1}^{1}1As for the logic \mathcal{ALC}^{\textsf{R}}{\bf T} in [15], an alternative semantic characterization of {\bf T} can be given by means of a set of postulates that are essentially a reformulation of the properties of rational consequence relation [23]..
In the following definition we introduce the notion of
Definition 1 (Semantics of \mathcal{SHIQ}^{\textsf{R}}{\bf T})
A \mathcal{SHIQ}^{\textsf{R}}{\bf T} model ^{2}^{2}2In this paper, we follow the terminology in [23] for preferential and ranked models, and we use the term “model” to denote an an interpretation. \mathcal{M} is any structure \langle\Delta,<,I\rangle where:

\Delta is the domain;

< is an irreflexive, transitive, wellfounded, and modular relation over \Delta;

I is the extension function that maps each concept C to C^{I}\subseteq\Delta, and each role R to R^{I}\subseteq\Delta^{I}\times\Delta^{I}. For concepts of \mathcal{SHIQ}, C^{I} is defined as usual. For the {\bf T} operator, we have ({\bf T}(C))^{I}=Min_{<}(C^{I}), where Min_{<}(S)=\{u:u\in S and \nexists z\in S s.t. z<u\}.
We say that an irreflexive and transitive relation < is:

modular if, for all x,y,z\in\Delta, if x<y then x<z or z<y [23];

wellfounded if, for all S\subseteq\Delta, for all x\in S, either x\in Min_{<}(S) or \exists y\in Min_{<}(S) such that y<x.
It can be proved that an irreflexive and transitive relation < on \Delta is wellfounded if and only if there are no infinite descending chains \ldots x_{i+1}<^{*}x_{i}<^{*}\ldots<^{*}x_{0} of elements of \Delta (see Appendix B).
In [23] it is shown that, for a strict partial order < over a set W, the modularity requirement is equivalent to postulating the existence of a rank function k:W\rightarrow\Omega, such that \Omega is a totally ordered set. In the presence of the wellfoundedness condition above, the totally ordered set \Omega happens to be a wellorder, and we can introduce a rank function k_{\mathcal{M}}:\Delta\longmapsto\mathit{Ord} assigning an ordinal to each domain element in W, and let x<y if and only if k_{\mathcal{M}}(x)<k_{\mathcal{M}}(y). We call k_{\mathcal{M}}(x) the rank of element x in \mathcal{M}. Observe that, when the rank k_{\mathcal{M}}(x) is finite, it can be understood as the length of a chain x_{0}<\dots<x from x to a minimal x_{0} (i.e. an x_{0} s.t. for no {x^{\prime}}, {x^{\prime}}<x_{0}).
Notice that the meaning of {\bf T} can be split into two parts: for any x of the domain \Delta, x\in({\bf T}(C))^{I} just in case (i) x\in C^{I}, and (ii) there is no y\in C^{I} such that y<x. In order to isolate the second part of the meaning of {\bf T}, we introduce a new modality \square. The basic idea is simply to interpret the preference relation < as an accessibility relation. The wellfoundedness of < ensures that typical elements of C^{I} exist whenever C^{I}\neq\emptyset, by avoiding infinitely descending chains of elements. The interpretation of \square in \mathcal{M} is as follows:
Definition 2
Given a model \mathcal{M}, we extend the definition of I with the following clause:
(\square C)^{I}=\{x\in\Delta\mid for every y\in\Delta, if y<x then y\in C^{I}\}
It is easy to observe that x is a typical instance of C if and only if it is an instance of C and \square\lnot C, that is to say:
Proposition 1
Given a model \mathcal{M}, given a concept C and an element x\in\Delta, we have that
x\in({\bf T}(C))^{I}\ \mbox{iff}\ x\in(C\sqcap\square\neg C)^{I} 
Since we only use \square to capture the meaning of {\bf T}, in the following we will always use the modality \square followed by a negated concept, as in \square\neg C.
In the next definition of a model satisfying a knowledge base, we extend the function I to individual constants; we assign to each individual constant a\in\mathcal{O} a domain element a^{I}\in\Delta.
Definition 3 (Model satisfying a knowledge base)
Given a \mathcal{SHIQ}^{\textsf{R}}{\bf T} model \mathcal{M}=\langle\Delta,<,I\rangle, we say that:

a model \mathcal{M} satisfies an inclusion C\sqsubseteq D if C^{I}\subseteq D^{I}; similarly for role inclusions;

\mathcal{M} satisfies an assertion C(a) if a^{I}\in C^{I};

\mathcal{M} satisfies an assertion R(a,b) if (a^{I},b^{I})\in R^{I}.
Given a KB=(TBox,ABox), we say that: \mathcal{M} satisfies TBox if \mathcal{M} satisfies all inclusions in TBox; \mathcal{M} satisfies ABox if \mathcal{M} satisfies all assertions in ABox; \mathcal{M} satisfies KB (or, is a model of KB) if it satisfies both its TBox and its ABox.
As a difference with the approach in [17], we do no longer assume the unique name assumption (UNA), namely we do not assume that each a\in\mathcal{O} is assigned to a distinct element a^{I}\in\Delta. In \mathcal{ALC}+{\bf T}_{min} [17], in which we compare models that might have a different interpretation of concepts and that are not canonical, UNA avoids that models in which two named individuals are mapped into the same domain element are preferred to those in which they are mapped into distinct ones. UNA is not needed here as we compare models with the same domain and the same interpretation of concepts, while assuming that models are canonical (see Definition 9) and contain all the possible domain elements “compatible” with the KB.
The logic \mathcal{SHIQ}^{\textsf{R}}{\bf T}, as well as the underlying \mathcal{SHIQ}, does not enjoy the finite model property [19].
Given a KB, we say that an inclusion C_{L}\sqsubseteq C_{R} is entailed by KB, written KB \models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}C_{L}% \sqsubseteq C_{R}, if {C_{L}}^{I}\subseteq{C_{R}}^{I} holds in all models \mathcal{M}=\langle\Delta,<,I\rangle satisfying KB; similarly for role inclusions. We also say that an assertion C_{L}(a), with a\in\mathcal{O}, is entailed by KB, written KB \models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}C_{L}(a), if a^{I}\in{C_{L}}^{I} holds in all models \mathcal{M}=\langle\Delta,<,I\rangle satisfying KB.
Let us now introduce the notions of rank of a \mathcal{SHIQ} concept.
Definition 4 (Rank of a concept k_{\mathcal{M}}(C_{R}))
Given a model \mathcal{M}=\langle\Delta,<,I\rangle, we define the rank k_{\mathcal{M}}(C_{R}) of a concept C_{R} in the model \mathcal{M} as k_{\mathcal{M}}(C_{R})=min\{k_{\mathcal{M}}(x)\mid x\in{C_{R}}^{I}\}. If {C_{R}}^{I}=\emptyset, then C_{R} has no rank and we write k_{\mathcal{M}}(C_{R})=\infty.
Proposition 2
For any \mathcal{M}=\langle\Delta,<,I\rangle, we have that \mathcal{M} satisfies {\bf T}(C)\sqsubseteq D if and only if k_{\mathcal{M}}(C\sqcap D)<k_{\mathcal{M}}(C\sqcap\lnot D).
It is immediate to verify that the typicality operator {\bf T} itself is nonmonotonic: {\bf T}(C)\sqsubseteq D does not imply {\bf T}(C\sqcap E)\sqsubseteq D. This nonmonotonicity of {\bf T} allows to express the properties that hold for the typical instances of a class (not only the properties that hold for all the members of the class). However, the logic \mathcal{SHIQ}^{\textsf{R}}{\bf T} is monotonic: what is inferred from KB can still be inferred from any KB’ with KB \subseteq KB’. This is a clear limitation in DLs. As a consequence of the monotonicity of \mathcal{SHIQ}^{\textsf{R}}{\bf T}, one cannot deal with irrelevance. For instance, if typical VIPs have more than two marriages, we would like to conclude that also typical tall VIPs have more than two marriages, since being tall is irrelevant with respect to being married. However, KB=\{\mathit{VIP}\sqsubseteq\mathit{Person}, {\bf T}(\mathit{Person})\sqsubseteq\ \leq 1\ \mathit{HasMarried}.\mathit{Person}, {\bf T}(\mathit{VIP})\sqsubseteq\ \geq 2\ \mathit{HasMarried}.\mathit{Person}\} does not entail KB \models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{\bf T}(% \mathit{VIP}\sqcap\mathit{Tall})\sqsubseteq\ \geq 2\ \mathit{HasMarried}.% \mathit{Person}, even if the property of being tall is irrelevant with respect to the number of marriages. Observe that we do not want to draw this conclusion in a monotonic way from \mathcal{SHIQ}^{\textsf{R}}{\bf T}, since otherwise we would not be able to retract it when knowing, for instance, that typical tall VIPs have just one marriage (see also Example 1). Rather, we would like to obtain this conclusion in a nonmonotonic way. In order to obtain this nonmonotonic behavior, we strengthen the semantics of \mathcal{SHIQ}^{\textsf{R}}{\bf T} by defining a minimal models mechanism which is similar, in spirit, to circumscription. Given a KB, the idea is to: 1. define a preference relation among \mathcal{SHIQ}^{\textsf{R}}{\bf T} models, giving preference to the model in which domain elements have a lower rank; 2. restrict entailment to minimal \mathcal{SHIQ}^{\textsf{R}}{\bf T} models (w.r.t. the above preference relation) of KB.
Definition 5 (Minimal models)
Given \mathcal{M}=\langle\Delta,<,I\rangle and \mathcal{M}^{\prime}=\langle\Delta^{\prime},<^{\prime},I^{\prime}\rangle we say that \mathcal{M} is preferred to \mathcal{M}^{\prime} (\mathcal{M}<_{\mathit{FIMS}}\mathcal{M}^{\prime}) if (i) \Delta=\Delta^{\prime}, (ii) C^{I}=C^{I^{\prime}} for all concepts C, and (iii) for all x\in\Delta, k_{\mathcal{M}}(x)\leq k_{\mathcal{M}^{\prime}}(x) whereas there exists y\in\Delta such that k_{\mathcal{M}}(y)<k_{\mathcal{M}^{\prime}}(y). Given a KB, we say that \mathcal{M} is a minimal model of KB with respect to <_{\mathit{FIMS}} if it is a model satisfying KB and there is no \mathcal{M}^{\prime} model satisfying KB such that \mathcal{M}^{\prime}<_{\mathit{FIMS}}\mathcal{M}.
The minimal model semantics introduced above is similar to the one introduced in [17] for \mathcal{ALC}. However, it is worth noticing that the notion of minimality here is based on the minimization of the ranks of the worlds, rather then on the minimization of formulas of a specific kind. Differently from [17], here we only compare models in which the interpretation of concepts is the same. In this respect, the minimal model semantics above is similar to the minimal model semantics FIMS, introduced in [16] to provide a semantic characterization to rational closure in propositional logic. In FIMS, the interpretation of propositions in the models to be compared is fixed. In contrast, in the alternative semantic characterization VIMS, models are compared in which the interpretation of propositions may vary. Although fixing the interpretation of propositions (or concepts) can appear to be rather restrictive, for the propositional case, it has been proved in [16] that the two semantic characterizations (VIMS and FIMS) are equivalent under suitable assumptions and, in particular, under the assumption that in FIMS canonical models are considered. Similarly to FIMS, here we compare models by fixing the interpretation of concepts, and we also restrict our consideration to canonical models, as we will do in section 5^{3}^{3}3 Note that our language does not provide a direct way for minimizing roles. On the other hand, fixing roles does not appear to be very promising. Indeed, for circumscribed KBs, it has been proved in [2] that allowing role names to be fixed makes reasoning highly undecidabe. For the time being we have not studied the issue of allowing fixed roles in our minimal model semantics for \mathcal{SHIQ}^{\textsf{R}}{\bf T}. .
Let us define:
K_{F}=\{C\sqsubseteq D\in TBox:{\bf T}\ \mbox{does not occur in $C$}\}\cup
\mbox{\ \ \ \ \ \ \ \ \ \ \ \ }\{R\sqsubseteq S\in TBox\}\cup ABox
K_{D}=\{{\bf T}(C)\sqsubseteq D\in TBox\},
so that KB =K_{F}\cup K_{D}.
Proposition 3 (Existence of minimal models)
Let KB be a finite knowledge base, if KB is satisfiable then it has a minimal model.
Proof
Let \mathcal{M}=\langle\Delta,<,I\rangle be a model of KB, where we assume that k_{\mathcal{M}}:\Delta\longrightarrow Ord determines < and Ord is the set of ordinals. Define the relation
\mathcal{M}\approx\mathcal{M}^{\prime} if \mathcal{M}^{\prime}=\langle\Delta^{\prime},<^{\prime},I^{\prime}\rangle and \Delta=\Delta^{\prime} and I=I^{\prime}
where <^{\prime} is also determined by a rank k_{\mathcal{M}^{\prime}} on ordinals. Define further \mathit{Mod}_{\mathit{KB}}(\mathcal{M})=\{\mathcal{M}^{\prime}\mid\mathcal{M}^% {\prime}\models\mathit{KB}\ \mbox{and}\ \mathcal{M}^{\prime}\approx\mathcal{M}\}. Let us define finally \mathcal{M}_{min}=\langle\Delta,<^{min},I^{min}\rangle, where I^{min}=I and <^{min} is defined by the ranking, for any x\in\Delta:
k_{min}(x)=min\{k_{\mathcal{M}^{\prime}}(x)\mid\mathcal{M}^{\prime}\in\mathit{% Mod}_{\mathit{KB}}(\mathcal{M})\}
Observe that k_{min}(x) is welldefined for any concept C and
k_{\mathcal{M}_{min}}(C)=min\{k_{min}(x)\mid x\in C^{I^{min}}\}
is also welldefined (a set of ordinals has always a least element). We now show that \mathcal{M}_{min}\models KB. Since I is the same as in \mathcal{M}, it follows immediately that \mathcal{M}\models K_{F}.
We prove that \mathcal{M}\models K_{D}. Let {\bf T}(C)\sqsubseteq E\in F_{D}. Suppose by absurdity that \mathcal{M}_{min}\not\models{\bf T}(C)\sqsubseteq E, this means that k_{min}(C\sqcap\neg E)\leq k_{min}(C\sqcap E). Let \mathcal{M}_{1}\in\mathit{Mod}_{KB}(\mathcal{M}), such that k_{min}(C\sqcap\neg E)=k_{\mathcal{M}_{1}}(C\sqcap\neg E). \mathcal{M}_{1} exists. Similarly, let \mathcal{M}_{2}\in\mathit{Mod}_{KB}(\mathcal{M}), such that k_{min}(C\sqcap E)=k_{\mathcal{M}_{2}}(C\sqcap E). We then have k_{\mathcal{M}_{1}}(C\sqcap\neg E)=k_{min}(C\sqcap\neg E)\leq k_{min}(C\sqcap E)= k_{\mathcal{M}_{2}}(C\sqcap E)\leq k_{\mathcal{M}_{1}}(C\sqcap E), as k_{\mathcal{M}_{2}}(C\sqcap E) is minimal. Thus we get that k_{\mathcal{M}_{1}}(C\sqcap\neg E)\leq k_{\mathcal{M}_{1}}(C\sqcap E) against the fact that \mathcal{M}_{1} is a model of KB. \hfill\square
The following theorem says that reasoning in \mathcal{SHIQ}^{\textsf{R}}{\bf T} has the same complexity as reasoning in \mathcal{SHIQ}, i.e. it is in ExpTime. Its proof is given by providing an encoding of satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} into satisfiability \mathcal{SHIQ}, which is known to be an ExpTimecomplete problem.
Theorem 2.1
Satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} is an ExpTimecomplete problem.
The proof can be found in Appendix A.
3 Rational Closure for \mathcal{SHIQ}
In this section, we extend to \mathcal{SHIQ} the notion of rational closure proposed by Lehmann and Magidor [23] for the propositional case. Given the typicality operator, the typicality inclusions {\bf T}(C)\sqsubseteq D (all the typical C’s are D’s) play the role of conditional assertions C\mathrel{{\scriptstyle\mid\!\sim}}D in [23]. Here we define the rational closure of the TBox. In Section 6 we will discuss an extension of rational closure that also takes into account the ABox.
Definition 6 (Exceptionality of concepts and inclusions)
Let T_{B} be a TBox and C a concept. C is said to be exceptional for T_{B} if and only if T_{B}\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{\bf T% }(\top)\sqsubseteq\neg C. A Tinclusion {\bf T}(C)\sqsubseteq D is exceptional for T_{B} if C is exceptional for T_{B}. The set of Tinclusions of T_{B} which are exceptional in T_{B} will be denoted as \mathcal{E}(T_{B}).
Given a DL KB=(TBox,ABox), it is possible to define a sequence of non increasing subsets of TBox E_{0}\supseteq E_{1},E_{1}\supseteq E_{2},\dots by letting E_{0}=\mbox{TBox} and, for i>0, E_{i}=\mathcal{E}(E_{i1})\cup\{C\sqsubseteq D\in\mbox{TBox} s.t. {\bf T} does not occurr in C\}. Observe that, being KB finite, there is an n\geq 0 such that, for all m>n,E_{m}=E_{n} or E_{m}=\emptyset. Observe also that the definition of the E_{i}’s is the same as the definition of the C_{i}’s in Lehmann and Magidor’s rational closure [21], except for that here, at each step, we also add all the “strict” inclusions C\sqsubseteq D (where {\bf T} does not occur in C).
Definition 7 (Rank of a concept)
A concept C has rank i (denoted by \mathit{rank}(C)=i) for KB=(TBox,ABox), iff i is the least natural number for which C is not exceptional for E_{i}. If C is exceptional for all E_{i} then \mathit{rank}(C)=\infty, and we say that C has no rank.
The notion of rank of a formula allows to define the rational closure of the TBox of a KB. Let \models_{\mathcal{SHIQ}} be the entailment in \mathcal{SHIQ}. In the following definition, by KB \models_{\mathcal{SHIQ}}F we mean K_{F}\models_{\mathcal{SHIQ}}F, where K_{F} does not include the defeasible inclusions in KB.
Definition 8 (Rational closure of TBox)
Let KB=(TBox,ABox) be a DL knowledge base. We define, \overline{\mathit{TBox}}, the rational closure of TBox, as \mbox{$\overline{\mathit{TBox}}$}=\{{\bf T}(C)\sqsubseteq D\mid\mbox{either}\ % \mathit{rank}(C)<\mathit{rank}(C\sqcap\lnot D) \mbox{or}\ \mathit{rank}(C)=\infty\}\ \cup\ \{C\sqsubseteq D\mid\ \mbox{KB}\ % \models_{\mathcal{SHIQ}}C\sqsubseteq D\}, where C and D are arbitrary \mathcal{SHIQ} concepts.
Observe that, apart form the addition of strict inclusions, the above definition of rational closure is the same as the one by Lehmann and Magidor in [23]. The rational closure of TBox is a nonmonotonic strengthening of \mathcal{SHIQ}^{\textsf{R}}{\bf T}. For instance, it allows to deal with irrelevance, as the following example shows.
Example 1
Let TBox = \{{\bf T}(\mathit{Actor})\sqsubseteq\mathit{Charming}\}. It can be verified that {\bf T}(\mathit{Actor}\sqcap\mathit{Comic})\sqsubseteq\mathit{Charming}\in% \overline{\mathit{TBox}}. This is a nonmonotonic inference that does no longer follow if we discover that indeed comic actors are not charming (and in this respect are untypical actors): indeed given TBox’= TBox \cup\ \{{\bf T}(\mathit{Actor}\sqcap\mathit{Comic})\sqsubseteq\neg\mathit{% Charming}\}, we have that {\bf T}(\mathit{Actor}\sqcap\mathit{Comic})\sqsubseteq\mathit{Charming}\not\in% \overline{\mathit{TBox^{\prime}}}.
Furthermore, as for the propositional case, rational closure is closed under rational monotonicity [21]: from {\bf T}(\mathit{Actor})\sqsubseteq\mathit{Charming}\in\overline{\mathit{TBox}} and {\bf T}(\mathit{Actor})\sqsubseteq\mathit{Bold}\not\in\overline{\mathit{TBox}} it follows that {\bf T}(\mathit{Actor}\sqcap\lnot\mathit{Bold})\sqsubseteq\mathit{Charming}\in% \overline{\mathit{TBox}}.
Although the rational closure \overline{\mathit{TBox}} is an infinite set, its definition is based on the construction of a finite sequence E_{0},E_{1},\ldots,E_{n} of subsets of TBox, and the problem of verifying that an inclusion {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}} is in ExpTime. To prove this result we need to introduce some propositions.
First of all, let us remember that rational entailment is equivalent to preferential entailment for a knowledge base only containing positive nonmonotonic implications A\mathrel{{\scriptstyle\mid\!\sim}}B (see [23]). The same holds in preferential description logics with typicality. Let \mathcal{SHIQ}^{P}{\bf T} be the logic that we obtain when we remove the requirement of modularity in the definition of \mathcal{SHIQ}^{\textsf{R}}{\bf T}. In this logic the typicality operator has a preferential semantics [21], based on the preferential models of P rather then on the ranked models [23]. An extension of \mathcal{ALC} with typicality based on preferential logic P has been studied in [14]. As a TBox of a KB in \mathcal{SHIQ}^{\textsf{R}}{\bf T} is a set of strict inclusions and defeasible inclusions (i.e., positive nonmonotonic implications), it can be proved that:
Proposition 4
Given a KB with empty ABox, and an inclusion E\sqsubseteq D we have
KB\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}E\sqsubseteq D\ \mbox{iff}\ KB% \models_{\mathcal{SHIQ}^{P}{\bf T}}E\sqsubseteq D 
Proof
(sketch) The (if) direction is trivial, thus we consider the (only if) one. Suppose that KB\not\models_{\mathcal{SHIQ}^{P}{\bf T}}E\sqsubseteq D, let \mathcal{M}=\langle\Delta,<,I\rangle a preferential model of KB, where < is transitive, irreflexive, and wellfounded, which falsifies E\sqsubseteq D. Then for some element x\in E and x\not\in D. Define first a model \mathcal{M}_{1}=\langle\mathcal{W},<_{1},I\rangle, where the relation <_{1} is defined as follows:
<_{1}=<\ \cup\ \{(u,v)\mid(u=x\lor u<x)\land v\not=x\land v\not<x\} 
It can be proved that:

<_{1} is transitive and irreflexive

<_{1} is wellfounded

if u<v then u<_{1}v

if u<_{1}x then u<x.
We can show that \mathcal{M}_{1} is a model of KB. This is obvious for inclusions that do not involve {\bf T}, as the interpretation I is the same. Given an inclusion {\bf T}(G)\sqsubseteq F\in KB, if it holds in \mathcal{M} then it holds also in \mathcal{M}_{1} as Min^{\mathcal{M}_{1}}_{<_{1}}(G)\subseteq Min^{\mathcal{M}}_{<}(G). Moreover \mathcal{M}_{1} falsifies E\sqsubseteq D by x, in particular (the only interesting case) when E={\bf T}(C). To this regard, we know that x\not\in D^{\mathcal{M}_{1}}, suppose by absurd that x\not\in({\bf T}(C))^{\mathcal{M}_{1}}, since x\in({\bf T}(C))^{\mathcal{M}}, we have that x\in C^{\mathcal{M}}=C^{\mathcal{M}_{1}}, thus there must be a y<_{1}x with y\in C^{\mathcal{M}_{1}}=C^{\mathcal{M}}. But then by 4 y<x and we get a contradiction. Thus x\in({\bf T}(C))^{\mathcal{M}_{1}} and x\not\in D^{\mathcal{M}_{1}}, that is x falsifies E\sqsubseteq D in \mathcal{M}_{1}.
Observe that <_{1} in model \mathcal{M}_{1} satisfies:
(*)\ \forall z\not=x\ (z<_{1}x\lor x<_{1}z) 
As a next step we define a modular model \mathcal{M}_{2}=\langle\mathcal{W},<_{2},I\rangle, where the relation <_{2} is defined as follows. Considering \mathcal{M}_{1} where <_{1} is wellfounded, we can define by recursion the following function k from \mathcal{M} to ordinals:

k(u)=0 if u is minimal in \mathcal{M}_{1}

k(u)=max\{k(y)\mid y<_{1}u\}+1 if the set \{y\mid y<_{1}u\} is finite

k(u)=sup\{k(y)\mid y<_{1}u\} if the set \{y\mid y<_{1}u\} is infinite.
Observe that if u<_{1}v then k(u)<k(v). We now define:
u<_{2}v\ \mbox{iff}\ k(u)<k(v) 
Notice that <_{2} is clearly transitive, modular, and wellfounded; moreover u<_{1}v implies u<_{2}v. We can prove as before that \mathcal{M}_{2} is a model of KB and that it falsifies E\sqsubseteq D by x. For the latter, we consider again the only interesting case when E={\bf T}(C). Suppose by absurd that x\not\in({\bf T}(C))^{\mathcal{M}_{2}}, since x\in({\bf T}(C))^{\mathcal{M}_{1}}, we have that x\in C^{\mathcal{M}_{2}}=C^{\mathcal{M}_{1}}, thus there must be a y<_{2}x with y\in C^{\mathcal{M}_{2}}=C^{\mathcal{M}_{1}}. But y<_{2}x means that k(y)<k(x). We can conclude that it must be also y<_{1}x, otherwise by (*) we would have x<_{1}y which entails k(x)<k(y), a contradiction. We have shown that y<_{1}x, thus x\not\in({\bf T}(C))^{\mathcal{M}_{1}} a contradiction. Therefore x\in({\bf T}(C))^{\mathcal{M}_{2}} and x\not\in D^{\mathcal{M}_{2}}, that is x falsifies E\sqsubseteq D in \mathcal{M}_{2}. We have shown that KB\not\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}E\sqsubseteq D. \hfill\square
The proof above also extends to a KB with a nonempty ABox, but it must not contain positive typicality assertions on individuals.
Proposition 5
Let KB=(TBox,\emptyset) be a knowledge base with empty ABox. KB\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C_{L}\sqsubseteq C_{R} iff KB^{\prime}\models_{\mathcal{SHIQ}}C^{\prime}_{L}\sqsubseteq C^{\prime}_{R}, where KB^{\prime}, C^{\prime}_{L} and C^{\prime}_{R} are polynomial encodings in \mathcal{SHIQ} of KB, C_{L} and C_{R}, respectively.
Proof
By Proposition 4, we have that
KB\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C_{L}\sqsubseteq C_{R} iff KB\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R}
where C_{L}\sqsubseteq C_{R} is any (strict or defeasible) inclusion in \mathcal{SHIQ}^{\textsf{R}}{\bf T}.
To prove the thesis it suffices to show that for all inclusions C_{L}\sqsubseteq C_{R} in \mathcal{SHIQ}^{\textsf{R}}{\bf T}:
KB\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R} iff KB^{\prime}\models_{\mathcal{SHIQ}}{C^{\prime}_{L}}\sqsubseteq{C^{\prime}_{R}}
for some polynomial encoding KB^{\prime}, C^{\prime}_{L} and C^{\prime}_{R} in \mathcal{SHIQ}.
The idea, on which the encoding is based, exploits the definition of the typicality operator {\bf T} introduced in [14], in terms of a GödelLöb modality \Box as follows: {\bf T}(C) is defined as C\sqcap\Box\neg C where the accessibility relation of the modality \Box is the preference relation < in preferential models.
We define the encoding KB’=(TBox’, ABox’) of KB in \mathcal{SHIQ} as follows. First, ABox’=\emptyset.
For each A\sqsubseteq B\in TBox, not containing {\bf T}, we introduce A\sqsubseteq B in TBox’.
For each {\bf T}(A) occurring in the TBox, we introduce a new atomic concept \Box_{\neg A} and, for each inclusion {\bf T}(A)\sqsubseteq B\in TBox, we add to TBox’ the inclusion
A\sqcap\Box_{\neg A}\sqsubseteq B 
Furthermore, to capture the properties of the \Box modality, a new role R is introduced to represent the relation < in preferential models, and the following inclusions are introduced in TBox’:
\Box_{\neg A}\sqsubseteq\forall R.(\neg A\sqcap\Box_{\neg A})
\neg\Box_{\neg A}\sqsubseteq\exists R.(A\sqcap\Box_{\neg A})
The first inclusion accounts for the transitivity of <. The second inclusion accounts for the smoothness (see [23, 14]): the fact that if an element is not a typical A element then there must be a typical A element preferred to it.
For the encoding of the inclusion C_{L}\sqsubseteq C_{R}: if C_{L}\sqsubseteq C_{R} is a strict inclusion in \mathcal{SHIQ}^{\textsf{R}}{\bf T}, then C^{\prime}_{L}=C_{L} and C^{\prime}_{R}=C_{R}; if C_{L}\sqsubseteq C_{R} is a defeasible inclusion in \mathcal{SHIQ}^{\textsf{R}}{\bf T}, i.e. C_{L}={\bf T}(A), then, we define C^{\prime}_{L}=A\sqcap\Box_{\neg A} and C^{\prime}_{R}=C_{R}.
It is clear that the size of KB’ is polynomial in the size of the KB (and the same holds for C^{\prime}_{L} and C^{\prime}_{R}, assuming the size of C_{L} and C_{R} polynomial in the size of the KB). Given the above encoding, we can prove that:
KB\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R} iff KB^{\prime}\models_{\mathcal{SHIQ}}{C^{\prime}_{L}}\sqsubseteq{C^{\prime}_{R}}
(If) By contraposition, let us assume that KB\not\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R}. We want to prove that KB^{\prime}\not\models_{\mathcal{SHIQ}}{C^{\prime}_{L}}\sqsubseteq{C^{\prime}_% {R}}. From the hypothesis, there is a preferential model \mathcal{M}=(\Delta,<,I) satisfying KB such that for some element x\in\Delta, x\in(C_{L})^{I} and x\in(\neg C_{R})^{I}. We build a \mathcal{SHIQ} model \mathcal{M}^{\prime}=(\Delta^{\prime},I^{\prime}) satisfying KB’ as follows:
\Delta^{\prime}=\Delta;
C^{I^{\prime}}=C^{I}, for all concepts C in the language of \mathcal{SHIQ};
R^{I}=R^{I^{\prime}}, for all roles R;
(x,y)\in R^{I^{\prime}} if and only if y<x in the model \mathcal{M}.
By construction it follows that {\bf T}(A)^{I}=(A\sqcap\Box_{\neg A})^{I^{\prime}}. Also, it can be easily verified that \mathcal{M} satisfies all the inclusions in KB’ and that x\in(C^{\prime}_{L})^{I^{\prime}} and x\in(\neg C^{\prime}_{R})^{I^{\prime}}. Hence KB^{\prime}\not\models_{\mathcal{SHIQ}}{C^{\prime}_{L}}\sqsubseteq{C^{\prime}_% {R}}.
(Only\>if) By contraposition, let us assume that KB^{\prime}\not\models_{\mathcal{SHIQ}}{C^{\prime}_{L}}\sqsubseteq{C^{\prime}_% {R}}. We want to prove that KB\not\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R}. From the hypothesis, we know there is a model \mathcal{M}^{\prime}=(\Delta^{\prime},I^{\prime}) satisfying KB’, such that x\in(C^{\prime}_{L})^{I^{\prime}} and x\in(\neg C^{\prime}_{R})^{I^{\prime}}. We build a model \mathcal{M}=(\Delta,<,I) satisfying KB such that some element of \mathcal{M} does not satisfy the inclusion C_{L}\sqsubseteq C_{R}. We let:
\Delta=\Delta^{\prime};
C^{I}=C^{I^{\prime}}, for all concepts C in the language of \mathcal{SHIQ};
R^{I}=R^{I^{\prime}}, for all roles R;
y<x if and only if (x,y)\in(R^{I^{\prime}})^{*} (the transitive closure of R^{I^{\prime}}).
By construction, it is easy to show that {\bf T}(A)^{I}=(A\sqcap\Box_{\neg A})^{I^{\prime}} and we can easily verify that \mathcal{M} satisfies all the inclusions in KB and that x\in(C_{L})^{I} and x\in(\neg C_{R})^{I}.
The relation < is transitive, as it is defined as the transitive closure of R, but < is not guaranteed to be wellfounded. However, we can modify the relation < in \mathcal{M} to make it wellfounded, by shortening the descending chains.
For any y\in\Delta, we let \Box_{y}=\{\Box C\;\mid\;y\in(\Box C)^{I}\}. Observe that for the elements x_{i} in a descending chain \ldots,x_{i1},x_{i},x_{i+1},\ldots, the set \Box_{x_{i}} is monotonically increasing (i.e., \Box_{x_{i}}\subseteq\Box_{x_{i+1}}).
We define a new model \mathcal{M}^{\prime\prime}=(\Delta,<^{\prime\prime},I) by changing the preference relation < in \mathcal{M} to <^{\prime\prime} as follows:
y<^{\prime\prime}x iff (y<x and \Box_{x}\subset\Box_{y}) or
(y<x and \Box_{x}=\Box_{y} and \forall w\in\Delta such that x<w, \Box_{w}\subset\Box_{x})
In essence, for a pair of elements (x,y) such that y<x but x and y are instances of exactly the same boxed concepts (\Box_{x}=\Box_{y}) and x is not the first element in the descending chain which is instance of all the boxed concepts in \Box_{x}, we do not include the pair (x,y) in <^{\prime\prime} (so that x and y will not be comparable in the preorder <^{\prime\prime}). The relation <^{\prime\prime} is transitive and wellfounded. \mathcal{M}^{\prime\prime} can be shown to be a model of KB, and x to be an instance of C_{L} but not of C_{R}. Hence, KB\not\models_{\mathcal{SHIQ}^{P}{\bf T}}C_{L}\sqsubseteq C_{R}. \hfill\square
Theorem 3.1 (Complexity of rational closure over TBox)
Given a TBox, the problem of deciding whether {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}} is in ExpTime.
Proof
Checking if {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}} can be done by computing the finite sequence E_{0},E_{1},\ldots,E_{n} of non increasing subsets of TBox inclusions in the construction of the rational closure. Note that the number n of the E_{i} is O(KB), where KB is the size of the knowledge base KB. Computing each E_{i}={\cal E}(E_{i1}), requires to check, for all concepts A occurring on the left hand side of an inclusion in the TBox, whether E_{i1}\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{% \bf T}(\top)\sqsubseteq\neg A. Regarding E_{i1} as a knowledge base with empty ABox, by Proposition 5 it is enough to check that {E^{\prime}_{i1}}\models_{\mathcal{SHIQ}}\top\sqcup\Box_{\neg\top}\sqsubseteq\neg A, which requires an exponential time in the size of {E^{\prime}_{i1}} (and hence in the size of KB). If not already checked, the exceptionality of C and of C\sqcap\neg D have to be checked for each E_{i}, to determine the ranks of C and of C\ \sqcap\ \neg D (which also can be computed in \mathcal{SHIQ} and requires an exponential time in the size of KB). Hence, verifying if {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}} is in ExpTime. \Box
The above proof provides an ExpTime complexity upper bound for computing the rational closure over a TBox in \mathcal{SHIQ} and shows that the rational closure of a TBox can be computed simply using the entailment in \mathcal{SHIQ}.
4 Infinite Minimal Models with finite ranks
In the following we provide a characterization of minimal models of a KB in terms of their rank: intuitively minimal models are exactly those ones where each domain element has rank 0 if it satisfies all defeasible inclusions, and otherwise has the smallest rank greater than the rank of any concept C occurring in a defeasible inclusion {\bf T}(C)\sqsubseteq D of the KB falsified by the element. Exploiting this intuitive characterization of minimal models, we are able to show that, for a finite KB, minimal models have always a finite ranking function, no matter whether they have a finite domain or not. This result allows us to provide a semantic characterization of rational closure of the previous section to logics, like \mathcal{SHIQ}, that do not have the finite model property.
Given a model \mathcal{M}=\langle\Delta,<,I\rangle, let us define the set S^{\mathcal{M}}_{x} of defeasible inclusions falsified by a domain element x\in\Delta, as S^{\mathcal{M}}_{x}=\{{\bf T}(C)\sqsubseteq D\in K_{D}\mid x\in(C\sqcap\neg D)% ^{I}\}\}.
Proposition 6
Let \mathcal{M}=\langle\Delta,<,I\rangle be a model of KB and x\in\Delta, then: (a) if k_{\mathcal{M}}(x)=0 then S^{\mathcal{M}}_{x}=\emptyset; (b) if S^{\mathcal{M}}_{x}\not=\emptyset then k_{\mathcal{M}}(x)>k_{\mathcal{M}}(C) for every C such that, for some D, {\bf T}(C)\sqsubseteq D\in S^{\mathcal{M}}_{x}.
Proof
Observe that (a) follows from (b). Let us prove (b). Suppose for a contradiction that (b) is false, so that S^{\mathcal{M}}_{x}\not=\emptyset and for some C such that, for some D, {\bf T}(C)\sqsubseteq D\in S^{\mathcal{M}}_{x}, we have k_{\mathcal{M}}(x)\leq k_{\mathcal{M}}(C). We have also that x\in(C\sqcap\neg D)^{I}. But \mathcal{M}\models KB, in particular \mathcal{M}\models{\bf T}(C)\sqsubseteq D, thus it must be x\not\in({\bf T}(C))^{I}, but x\in C^{I}, so that we get that k_{\mathcal{M}}(x)>k_{\mathcal{M}}(C) a contradiction. \hfill\Box
Proposition 7
Let KB =K_{F}\cup K_{D} and \mathcal{M}=\langle\Delta,<,I\rangle be a model of K_{F}; suppose that for any x\in\Delta it holds:

(a) if k_{\mathcal{M}}(x)=0 then S^{\mathcal{M}}_{x}=\emptyset

(b) if S^{\mathcal{M}}_{x}\not=\emptyset then k_{\mathcal{M}}(x)>k_{\mathcal{M}}(C) for every C such that, for some D, {\bf T}(C)\sqsubseteq D\in S^{\mathcal{M}}_{x}.
then \mathcal{M}\models KB.
Proof
Let {\bf T}(C)\sqsubseteq D\in K_{D}, suppose that for some x\in C, it holds x\in({\bf T}(C))^{I}D^{I}, then {\bf T}(C)\sqsubseteq D\in S^{\mathcal{M}}_{x}. By hypothesis, we have k_{\mathcal{M}}(x)>k_{\mathcal{M}}(C), against the fact that x\in{\bf T}(C). \hfill\Box
Proposition 8
Let KB =K_{F}\cup K_{D} and \mathcal{M}=\langle\Delta,<,I\rangle a minimal model of KB, for every x\in\Delta, it holds:

(a) if S^{\mathcal{M}}_{x}=\emptyset then k_{\mathcal{M}}(x)=0

(b) if S^{\mathcal{M}}_{x}\not=\emptyset then k_{\mathcal{M}}(x)=1+max\{k_{\mathcal{M}}(C)s.t.{\bf T}(C)\sqsubseteq D\in S^{% \mathcal{M}}_{x}\}.
Proof
Let \mathcal{M}=\langle\Delta,<,I\rangle be a minimal model of KB. Define another model \mathcal{M}^{\prime}=\langle\Delta,<^{\prime},I\rangle, where <^{\prime} is determined by a ranking function k_{\mathcal{M}^{\prime}} as follows:

k_{\mathcal{M}^{\prime}}(x)=0 if S^{\mathcal{M}}_{x}=\emptyset,

k_{\mathcal{M}^{\prime}}(x)=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D% \in S^{\mathcal{M}}_{x}\} if S^{\mathcal{M}}_{x}\not=\emptyset.
It is easy to see that (i) for every x k_{\mathcal{M}^{\prime}}(x)\leq k_{\mathcal{M}}(x). Indeed, if S^{\mathcal{M}}_{x}=\emptyset then it is obvious; if S^{\mathcal{M}}_{x}\not=\emptyset, then k_{\mathcal{M}^{\prime}}(x)=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D% \in S^{\mathcal{M}}_{x}\}\leq k_{\mathcal{M}}(x) by Proposition 6. It equally follows that (ii) for every concept C, k_{\mathcal{M}^{\prime}}(C)\leq k_{\mathcal{M}}(C). To see this: let z\in C^{I} such that k_{\mathcal{M}}(z)=k_{\mathcal{M}}(C), either k_{\mathcal{M}^{\prime}}(C)=k_{\mathcal{M}^{\prime}}(z)\leq k_{\mathcal{M}}(z) and we are done, or there exists y\in C^{I}, such that k_{\mathcal{M}^{\prime}}(C)=k_{\mathcal{M}^{\prime}}(y)<k_{\mathcal{M}^{\prime% }}(z)\leq k_{\mathcal{M}}(z).
Observe that S^{\mathcal{M}}_{x}=S^{\mathcal{M}^{\prime}}_{x}, since the evaluation function I is the same in the two models. By definition of \mathcal{M}^{\prime}, we have \mathcal{M}^{\prime}\models K_{F}; moreover by (i) and (ii) it follows that:
(iii) if k_{\mathcal{M}^{\prime}}(x)=0 then S^{\mathcal{M}^{\prime}}_{x}=\emptyset.
(iv) if S^{\mathcal{M}^{\prime}}_{x}\not=\emptyset: k_{\mathcal{M}^{\prime}}(x)=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D% \in S^{\mathcal{M}}_{x}\}\geq 1+max\{k_{\mathcal{M}^{\prime}}(C)\mid{\bf T}(C)% \sqsubseteq D\in S^{\mathcal{M}^{\prime}}_{x}\}, that is k_{\mathcal{M}^{\prime}}(x)>k_{\mathcal{M}^{\prime}}(C) for every C such that for some D, {\bf T}(C)\sqsubseteq D\in S^{\mathcal{M}^{\prime}}_{x}.
By Proposition 7 we obtain that \mathcal{M}^{\prime}\models KB; but by (i) k_{\mathcal{M}^{\prime}}(x)\leq k_{\mathcal{M}}(x) and by hypothesis \mathcal{M} is minimal. Thus it must be that for every x\in\Delta, k_{\mathcal{M}^{\prime}}(x)=k_{\mathcal{M}}(x) (whence k_{\mathcal{M}^{\prime}}(C)=k_{\mathcal{M}}(C)) which entails that \mathcal{M} satisfies (a) and (b) in the statement of the theorem. \hfill\Box
Also the opposite direction holds:
Proposition 9
Let KB =K_{F}\cup K_{D}, let \mathcal{M}=\langle\Delta,<,I\rangle be a model of K_{F}, suppose that for every x\in\Delta, it holds:

(a) S^{\mathcal{M}}_{x}=\emptyset iff k_{\mathcal{M}}(x)=0

(b) if S^{\mathcal{M}}_{x}\not=\emptyset then k_{\mathcal{M}}(x)=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D\in S^{% \mathcal{M}}_{x}\}.
then \mathcal{M} is a minimal model of KB.
Proof
In light of previous Propositions 6 and 7, it is sufficient to show that \mathcal{M} is minimal. To this aim, let \mathcal{M}^{\prime}=\langle\Delta,<^{\prime},I\rangle, with associated ranking function k_{\mathcal{M}^{\prime}}, be another model of KB, we show that for every x\in\Delta, it holds k_{\mathcal{M}}(x)\leq k_{\mathcal{M}^{\prime}}(x). We proceed by induction on k_{\mathcal{M}^{\prime}}(x). If S^{\mathcal{M}}_{x}=S^{\mathcal{M}^{\prime}}_{x}=\emptyset, we have that k_{\mathcal{M}}(x)=0\leq k_{\mathcal{M}^{\prime}}(x) (no need of induction). If S^{\mathcal{M}}_{x}=S^{\mathcal{M}^{\prime}}_{x}\not=\emptyset, then since \mathcal{M}^{\prime}\models KB, by Proposition 6: k_{\mathcal{M}^{\prime}}(x)\geq 1+max\{k_{\mathcal{M}^{\prime}}(C)\mid{\bf T}(% C)\sqsubseteq D\in S^{\mathcal{M}^{\prime}}_{x}\}. Let S^{\mathcal{M}^{\prime}}_{x}=S^{\mathcal{M}}_{x}=\{{\bf T}(C_{1})\sqsubseteq D% _{1},\ldots,{\bf T}(C_{u})\sqsubseteq D_{u}\}. For i=1,\ldots,u let k_{\mathcal{M}^{\prime}}(C_{i})=k_{\mathcal{M}^{\prime}}(y_{i}) for some y_{i}\in\Delta. Observe that k_{\mathcal{M}^{\prime}}(y_{i})<k_{\mathcal{M}^{\prime}}(x), thus by induction hypothesis k_{\mathcal{M}}(y_{i})\leq k_{\mathcal{M}^{\prime}}(y_{i}), for i=1,\ldots,u. But then k_{\mathcal{M}}(C_{i})\leq k_{\mathcal{M}}(y_{i}), so that we finally get:
\displaystyle k_{\mathcal{M}^{\prime}}(x)  \displaystyle\geq  \displaystyle 1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D\in S^{% \mathcal{M}^{\prime}}_{x}\}  
\displaystyle=  \displaystyle 1+max\{k_{\mathcal{M}^{\prime}}(C_{1}),\ldots,k_{\mathcal{M}^{% \prime}}(C_{u})\}  
\displaystyle=  \displaystyle 1+max\{k_{\mathcal{M}^{\prime}}(y_{1}),\ldots,k_{\mathcal{M}^{% \prime}}(y_{u})\}  
\displaystyle\geq  \displaystyle 1+max\{k_{\mathcal{M}}(y_{1}),\ldots,k_{\mathcal{M}}(y_{u})\}  
\displaystyle\geq  \displaystyle 1+max\{k_{\mathcal{M}}(C_{1}),\ldots,k_{\mathcal{M}}(C_{u})\}  
\displaystyle=  \displaystyle 1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D\in S^{% \mathcal{M}}_{x}\}  
\displaystyle=  \displaystyle k_{\mathcal{M}^{\prime}}(x) 
\hfill\Box
Putting Propositions 8 and 9 together, we obtain the following theorem which provides a characterization of minimal models.
Theorem 4.1
Let KB =K_{F}\cup K_{D}, and let \mathcal{M}=\langle\Delta,<,I\rangle be a model of K_{F}. The following are equivalent:

\mathcal{M} is a minimal model of KB

For every x\in\Delta it holds: (a) S^{\mathcal{M}}_{x}=\emptyset iff k_{\mathcal{M}}(x)=0 (b) if S^{\mathcal{M}}_{x}\not=\emptyset then k_{\mathcal{M}}(x)=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D\in S^{% \mathcal{M}}_{x}\}.
The following proposition shows that in any minimal model the rank of each domain element is finite.
Proposition 10
Let KB =K_{F}\cup K_{D} and \mathcal{M}=\langle\Delta,<,I\rangle a minimal model of KB, for every x\in\Delta, k_{\mathcal{M}}(x) is a finite ordinal (k_{\mathcal{M}}(x)<\omega).
Proof
Let k_{\mathcal{M}}(x)=\alpha, we proceed by induction on \alpha. If S^{\mathcal{M}}_{x}=\emptyset, then by Proposition 8 \alpha=0 and we are done (no need of induction). Otherwise if S^{\mathcal{M}}_{x}\not=\emptyset, by Proposition 8, we have that k_{\mathcal{M}}(x)=\alpha=1+max\{k_{\mathcal{M}}(C)\mid{\bf T}(C)\sqsubseteq D% \in S^{\mathcal{M}}_{x}\}. Let S^{\mathcal{M}}_{x}=\{{\bf T}(C_{1})\sqsubseteq D_{1},\ldots,{\bf T}(C_{u})% \sqsubseteq D_{u}\}. For i=1,\ldots,u let k_{\mathcal{M}}(C_{i})=\beta_{i}=k_{\mathcal{M}}(y_{i}) for some y_{i}\in\Delta. So that we have k_{\mathcal{M}}(x)=\alpha=1+max\{\beta_{1},\ldots,\beta_{u}\}. Since k_{\mathcal{M}}(y_{i})=\beta_{i}<\alpha, by induction hypothesis we have that \beta_{i}<\omega, thus also \alpha<\omega. \hfill\Box
The previous proposition is essential for establishing a correspondence between the minimal model semantics of a KB and its rational closure. From now on, we can assume that the ranking function assigns to each domain element in \Delta a natural number, i.e. that k_{\mathcal{M}}:\Delta\longrightarrow\mathbb{N}.
5 A Minimal Model Semantics for Rational Closure in \mathcal{SHIQ}
In previous sections we have extended to \mathcal{SHIQ} the syntactic notion of rational closure introduced in [23] for propositional logic. To provide a semantic characterization of this notion, we define a special class of minimal models, exploiting the fact that, by Proposition 10, in all minimal \mathcal{SHIQ}^{\textsf{R}}{\bf T} models the rank of each domain element is always finite. First of all, we can observe that the minimal model semantics in Definition 5 as it is cannot capture the rational closure of a TBox.
Consider the following KB=(TBox,\emptyset), where TBox contains:
\mathit{VIP}\sqsubseteq\mathit{Person},
{\bf T}(\mathit{Person})\sqsubseteq\ \leq 1\ \mathit{HasMarried}.\mathit{Person},
{\bf T}(\mathit{VIP}) \sqsubseteq\ \geq 2\ \mathit{HasMarried}. \mathit{Person}.
We observe that {\bf T}(\mathit{VIP}\sqcap\mathit{Tall})\sqsubseteq\ \geq 2\ \mathit{% HasMarried}.\mathit{Person} does not hold in all minimal \mathcal{SHIQ}^{\textsf{R}}{\bf T} models of KB w.r.t. Definition 5. Indeed there can be a model \mathcal{M}=\langle\Delta,<,I\rangle in which \Delta=\{x,y,z\}, \mathit{VIP}^{I}=\{x,y\}, \mathit{Person}^{I}=\{x,y,z\}, (\leq 1\ \mathit{HasMarried}.\mathit{Person})^{I}=\{x,z\}, (\geq 2\ \mathit{HasMarried}.\mathit{Person})^{I}=\{y\}, \mathit{Tall}^{I}=\{x\}, and z<y<x. \mathcal{M} is a model of KB, and it is minimal. Also, x is a typical tallVIP in \mathcal{M} (since there is no other tall VIP preferred to him) and has no more than one spouse, therefore {\bf T}(\mathit{VIP}\sqcap\mathit{Tall})\sqsubseteq\ \geq 2\ \mathit{% HasMarried}.\mathit{Person} does not hold in \mathcal{M}. On the contrary, it can be verified that {\bf T}(\mathit{VIP}\sqcap\mathit{Tall})\sqsubseteq\ \geq 2\ \mathit{% HasMarried}.\mathit{Person}\in\overline{\mathit{TBox}}.
Things change if we consider the minimal models semantics applied to models that contain a domain element for each combination of concepts consistent with KB. We call these models canonical models. Therefore, in order to semantically characterize the rational closure of a \mathcal{SHIQ}^{\textsf{R}}{\bf T} KB, we restrict our attention to minimal canonical models. First, we define \mathcal{S} as the set of all the concepts (and subconcepts) not containing {\bf T}, which occur in KB or in the query F, together with their complements.
In order to define canonical models, we consider all the sets of concepts \{C_{1},C_{2},\dots, C_{n}\}\subseteq\mathcal{S} that are consistent with KB, i.e., s.t. KB \not\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C_{1}\sqcap C_{2}\sqcap\dots% \sqcap C_{n}\sqsubseteq\bot.
Definition 9 (Canonical model with respect to \mathcal{S})
Given KB=(TBox,ABox) and a query F, a model \mathcal{M}=\langle\Delta,<,I\rangle satisfying KB is canonical with respect to \mathcal{S} if it contains at least a domain element x\in\Delta s.t. x\in(C_{1}\sqcap C_{2}\sqcap\dots\sqcap C_{n})^{I}, for each set of concepts \{C_{1},C_{2},\dots,C_{n}\}\subseteq\mathcal{S} that is consistent with KB.
Next we define the notion of minimal canonical model.
Definition 10 (Minimal canonical models (w.r.t. \mathcal{S}))
Proposition 11 (Existence of minimal canonical models)
Let KB be a finite knowledge base, if KB is satisfiable then it has a minimal canonical model.
Proof
Let \mathcal{M}=\langle\Delta,<,I\rangle be a minimal model of KB (which exists by Proposition 3), and let \{C_{1},C_{2},\dots,C_{n}\}\subseteq\mathcal{S} any subset of \mathcal{S} consistent with KB.
We show that we can expand \mathcal{M} in order to obtain a model of KB that contains an instance of C_{1}\sqcap C_{2}\sqcap\dots\sqcap C_{n}. By repeating the same construction for all maximal subsets \{C_{1},C_{2},\dots,C_{n}\} of \mathcal{S}, we eventually obtain a canonical model of KB.
For each \{C_{1},C_{2},\dots,C_{n}\} consistent with KB, it holds that KB \not\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C_{1}\sqcap C_{2}\sqcap\dots% \sqcap C_{n}\sqsubseteq\bot, i.e. there is a model \mathcal{M}^{\prime}=\langle\Delta^{\prime},<^{\prime},I^{\prime}\rangle of KB that contains an instance of \{C_{1},C_{2},\dots,C_{n}\}.
Let \mathcal{M}^{{}^{\prime}*} be the union of \mathcal{M} and \mathcal{M}^{\prime}, i.e. \mathcal{M}^{{}^{\prime}*}=\langle\Delta^{{}^{\prime}*},<^{{}^{\prime}*},I^{{}% ^{\prime}*}\rangle, where \Delta^{{}^{\prime}*}=\Delta\cup\Delta^{*}. As far as individuals named in the ABox, I^{{}^{\prime}*}=I, whereas for the concepts and roles, I^{{}^{\prime}*}=I on \Delta and I^{{}^{\prime}*}=I^{\prime} on \Delta^{\prime}. Also, k_{\mathcal{M}^{{}^{\prime}*}}=k_{\mathcal{M}} for the elements in \Delta, and k_{\mathcal{M}^{{}^{\prime}*}}=k_{\mathcal{M}^{\prime}} for the elements in \Delta^{\prime}. <^{{}^{\prime}*} is straightforwardly defined from k_{\mathcal{M}^{{}^{\prime}*}} as described just before Definition 4.
The model \mathcal{M}^{{}^{\prime}*} is still a model of KB. For the set K_{F} in the previous definition this is obviously true. For K_{D}, for each {\bf T}(C)\sqsubseteq D in K_{D}, if x\in Min_{<^{\prime}*}(C) in \mathcal{M}^{\prime}*, also x\in Min_{<}(C) in \mathcal{M} or x\in Min_{<^{\prime}}(C) in \mathcal{M}^{\prime}. In both cases x is an instance of D (since both \mathcal{M} and \mathcal{M}^{\prime} satisfy K_{D}), therefore x\in D^{I^{{}^{\prime}*}}, and \mathcal{M}^{{}^{\prime}*} satisfies K_{D}.
By repeating the same construction for all maximal subsets \{C_{1},C_{2},\dots,C_{n}\} of \mathcal{S}, we obtain a canonical model of KB, call it \mathcal{M}^{*}. We do not know whether the model is minimal. However by applying the construction used in the proof of Proposition 3, we obtain {\mathcal{M}^{*}}_{min} that is a minimal model of KB with the same domain and interpretation function than \mathcal{M}^{*}. {\mathcal{M}^{*}}_{min} is therefore a canonical model of KB, and furthermore it is minimal. Therefore KB has a minimal canonical model. \hfill\square
To prove the correspondence between minimal canonical models and the rational closure of a TBox, we need to introduce some propositions. The next one concerns all \mathcal{SHIQ}^{\textsf{R}}{\bf T} models. Given a \mathcal{SHIQ}^{\textsf{R}}{\bf T} model \mathcal{M}=\langle\Delta,<,I\rangle, we define a sequence \mathcal{M}{}_{0}, \mathcal{M}{}_{1},\mathcal{M}{}_{2},\ldots of models as follows: We let \mathcal{M}{}_{0}=\mathcal{M} and, for all i, we let \mathcal{M}{}_{i}=\langle\Delta,<_{i},I\rangle be the \mathcal{SHIQ}^{\textsf{R}}{\bf T} model obtained from \mathcal{M} by assigning a rank 0 to all the domain elements x with k_{\mathcal{M}}(x)<i, i.e., k_{{\mathcal{M}}_{i}}(x)=k_{\mathcal{M}}(x)i if k_{\mathcal{M}}(x)>i, and k_{{\mathcal{M}}_{i}}(x)=0 otherwise. We can prove the following:
Proposition 12
Let KB=\langle TBox,ABox\rangle and let \mathcal{M}=\langle\Delta,<,I\rangle be any \mathcal{SHIQ}^{\textsf{R}}{\bf T} model of TBox. For any concept C, if rank(C) \geq i, then 1) k_{\mathcal{M}}(C)\geq i, and 2) if {\bf T}(C)\sqsubseteq D is entailed by E_{i}, then \mathcal{M}{}_{i} satisfies {\bf T}(C)\sqsubseteq D.
Proof
By induction on i. For i=0, 1) holds (since it always holds that k_{\mathcal{M}}(C)\geq 0). 2) holds trivially as \mathcal{M}{}_{0}=\mathcal{M}.
For i>0, 1) holds: if rank(C)\geq i, then, by Definition 7, for all j<i, we have that E_{j}\models{\bf T}(\top)\sqsubseteq\neg C. By inductive hypothesis on 2), for all j<i, \mathcal{M}{}_{j}\models{\bf T}(\top)\sqsubseteq\neg C. Hence, for all x with k_{\mathcal{M}}(x)<i, x\not\in C^{I}, and k_{\mathcal{M}}(C)\geq i.
To prove 2), we reason as follows. Since E_{i}\subseteq E_{0}, \mathcal{M} \models E_{i}. Furthermore by definition of rank, for all {\bf T}(C)\sqsubseteq D\in E_{i}, rank(C) \geq i, hence by 1) just proved k_{\mathcal{M}}(C)\geq i. Hence, in \mathcal{M}, Min_{<}(C^{I})\geq i, and also \mathcal{M}{}_{i}\models{\bf T}(C)\sqsubseteq D. Therefore \mathcal{M}{}_{i}\models E_{i}. \hfill\Box
Let us now focus our attention on minimal canonical models by proving the correspondence between rank of a formula (as in Definition 7) and rank of a formula in a model (as in Definition 4). The following proposition is proved by induction on the rank i:
Proposition 13
Given KB and \mathcal{S}, for all C\in\mathcal{S}, if \mathit{rank}(C)=i, then: 1. there is a \{C_{1}\dots C_{n}\}\subseteq\mathcal{S} maximal and consistent with KB such that C\in\{C_{1}\dots C_{n}\} and \mathit{rank}(C_{1}\sqcap\dots\sqcap C_{n})=i; 2. for any \mathcal{M} minimal canonical model of KB, k_{\mathcal{M}}(C)=i.
Proof
By induction on i. Let us first consider the base case in which i=0. We have that KB \not\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{\bf T% }(\top)\sqsubseteq\neg C. Then there is a minimal model \mathcal{M}{}_{1} of KB with a domain element x such that k_{{\mathcal{M}}_{1}}(x)=0 and x satisfies C. For 1): consider the maximal consistent set of concepts in \mathcal{S} of which x is an instance in \mathcal{M}{}_{1}. This is a maximal consistent \{C_{1}\dots C_{n}\}\subseteq\mathcal{S} containing C. Furthermore, \mathit{rank}(C_{1}\sqcap\dots\sqcap C_{n})=0 since clearly KB \not\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{\bf T% }(\top)\sqsubseteq\neg(C_{1}\sqcap\dots\sqcap C_{n}). For 2): by definition of canonical model, in any canonical model \mathcal{M} of KB, \{C_{1}\dots C_{n}\} is satisfiable by an element x. Furthermore, in any minimal canonical \mathcal{M}, k_{\mathcal{M}}(x)=0, since otherwise we could build \mathcal{M}^{\prime} identical to \mathcal{M} except from the fact that k_{\mathcal{M}^{\prime}}(x)=0. It can be easily proven that \mathcal{M}^{\prime} would still be a model of KB (indeed \{C_{1}\dots C_{n}\} was already satisfiable in \mathcal{M}{}_{1} by an element with rank 0) and \mathcal{M}^{\prime}<_{\mathit{FIMS}}\mathcal{M}, against the minimality of \mathcal{M}. Therefore, in any minimal canonical model \mathcal{M} of KB, it holds k_{\mathcal{M}}(C)=0.
For the inductive step, consider the case in which i>0. We have that E_{i}\not\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{% \bf T}(\top)\sqsubseteq\neg C, then there must be a model \mathcal{M}_{1}=\langle\Delta_{1},<_{1},I_{1}\rangle of E_{i}, and a domain element x such that k_{{\mathcal{M}}_{1}}(x)=0 and x satisfies C. Consider the maximal consistent set of concepts \{C_{1},\dots Cn\}\subseteq\mathcal{S} of which x is an instance in \mathcal{M}{}_{1}. C\in\{C_{1},\dots Cn\}. Furthermore, \mathit{rank}(C_{1}\sqcap\dots\sqcap C_{n})=i. Indeed E_{i1}\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{% \bf T}(\top)\sqsubseteq\neg(C_{1}\sqcap\dots\sqcap C_{n}) (since E_{i1}\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{% \bf T}(\top)\sqsubseteq\neg C and C\in\{C_{1},\dots Cn\}), whereas clearly by the existence of x, E_{i}\not\models_{\scriptsize\mathcal{SHIQ}^{\scriptsize{\textsf{R}}}{\bf T}}{% \bf T}(\top)\sqsubseteq\neg(C_{1}\sqcap\dots\sqcap C_{n}). In order to prove 1) we are left to prove that the set \{C_{1},\dots Cn\} (that we will call \Gamma in the following) is consistent with KB.
To prove this, take any minimal canonical model \mathcal{M}=\langle\Delta,<,I\rangle of KB. By inductive hypothesis we know that for all concepts C^{\prime} such that \mathit{rank}(C^{\prime})<i, there is a maximal consistent set of concepts \{C^{\prime}_{1},\dots C^{\prime}_{n}\} with C^{\prime}\in\{C^{\prime}_{1},\dots C^{\prime}_{n}\} and \mathit{rank}(C^{\prime}_{1}\sqcap\dots\sqcap C^{\prime}_{n})=j<i. Furthermore, we know that k_{\mathcal{M}}(C^{\prime})=j<i. For a contradiction, if \mathcal{M} did not contain any element satisfying \Gamma we could expand it by adding to \mathcal{M} a portion of the model \mathcal{M}_{1} including x\in\Delta_{1}. More precisely, we add to \mathcal{M} a new set of domain elements \Delta_{x}\subseteq\Delta_{1}, containing the domain element x of \mathcal{M}_{1} and all the domain elements of \Delta_{1} which are reachable from x in \mathcal{M}_{1} through a sequence of relations R_{i}^{I_{1}}s or (R^{}_{i})^{I_{1}}s. Let \mathcal{M}^{\prime} be the resulting model. We define I^{\prime} on the elements of \Delta as in \mathcal{M}, while we define I^{\prime} on the element of \Delta_{x} as in I_{1}. Finally, we let, for all w\in\Delta, k_{\mathcal{M}^{\prime}}(w)=k_{\mathcal{M}}(w) and, for all y\in\Delta_{x}, k_{\mathcal{M}^{\prime}}(y)=i+k_{\mathcal{M}_{1}}(y). In particular, k_{\mathcal{M}^{\prime}}(x)=i. The resulting model \mathcal{M}^{\prime} would still be a model of KB. Indeed, the ABox would still be satisfied by the resulting model (being the \mathcal{M} part unchanged). For the TBox: all domain elements already in \mathcal{M} still satisfy all the inclusions. For all y\in\Delta_{x} (including x): for all inclusions in E_{i}, y satisfies them (since it did it in \mathcal{M}_{1}); for all typicality inclusions {\bf T}(D)\sqsubseteq G\in KB E_{i}, \mathit{rank}(D)<i, hence by inductive hypothesis k_{\mathcal{M}}(D)<i, hence k_{\mathcal{M}^{\prime}}(D)<i, and y is not a typical instance of D and trivially satisfies the inclusion. It is easy to see that \mathcal{M}^{\prime} also satisfies role inclusions R\sqsubseteq S and that, for each transitive roles R, R^{I^{\prime}} is transitive.
We have then built a model of KB satisfying \Gamma. Therefore \Gamma is consistent with KB, and therefore by definition of canonical model, \Gamma must be satisfiable in \mathcal{M}. Up to now we have proven that \Gamma is maximal and consistent with KB, it contains C and has rank i, therefore point 1) holds.
In order to prove point 2) we need to prove that any minimal canonical model \mathcal{M} of KB not only satisfies \Gamma but it satisfies it with rank i, i.e. k_{\mathcal{M}}(C_{1}\sqcap\dots\sqcap C_{n})=i, which entails k_{\mathcal{M}}(C)=i (since C\in\{C_{1},\dots C_{n}\}). By Proposition 12 we know that k_{\mathcal{M}}(C_{1}\sqcap\dots\sqcap C_{n})\geq i. We need to show that also k_{\mathcal{M}}(C_{1}\sqcap\dots\sqcap C_{n})\leq i. We reason as above: for a contradiction suppose k_{\mathcal{M}}(C_{1}\sqcap\dots\sqcap C_{n})>i, i.e., for all the minimal domain elements y instances of C_{1}\sqcap\dots\sqcap C_{n}, k_{\mathcal{M}}(y)>i. We show that this contradicts the minimality of \mathcal{M}. Indeed consider \mathcal{M}^{\prime} obtained from \mathcal{M} by letting k_{\mathcal{M}^{\prime}}(y)=i, for some minimal domain element y instance of C_{1}\sqcap\dots\sqcap C_{n}, and leaving all the rest unchanged. \mathcal{M}^{\prime} would still be a model of KB: the only thing that changes with respect to \mathcal{M} is that y might have become in \mathcal{M}^{\prime} a minimal instance of a concept of which it was only a nontypical instance in \mathcal{M}. This might compromise the satisfaction in \mathcal{M} of a typical inclusion as {\bf T}(E)\sqsubseteq G. However: if \mathit{rank}(E)<i, we know by inductive hypothesis that k_{\mathcal{M}}(E)<i hence also k_{\mathcal{M}^{\prime}}(E)<i and y is not a minimal instance of E in \mathcal{M}^{\prime}. If \mathit{rank}(E)\geq i, then {\bf T}(E)\sqsubseteq G\in E_{i}. As y\in C_{1}\sqcap\dots\sqcap C_{n} (where \{C_{1},\dots C_{n}\} is maximal consistent with KB), we have that: y\in F^{I} iff x\in F^{I_{1}}, for all concepts F. If y\in E^{I}, then E\in\{C_{1},\dots C_{n}\}. Hence, in \mathcal{M}_{1}, x\in E^{I_{1}}. But \mathcal{M}_{1} is a model of E_{i}, and satisfies all the inclusions in E_{i}. Therefore x\in G^{I_{1}} and, thus, y\in G^{I}.
It follows that \mathcal{M}^{\prime} would be a model of KB, and \mathcal{M}^{\prime}<_{\mathit{FIMS}}\mathcal{M}, against the minimality of \mathcal{M}. We are therefore forced to conclude that k_{\mathcal{M}}(C_{1}\sqcap\dots\sqcap C_{n})=i, and hence also k_{\mathcal{M}}(C)=i, and 2) holds.
\hfill\Box
The following theorem follows from the propositions above:
Theorem 5.1
Let KB=(TBox,ABox) be a knowledge base and C\sqsubseteq D a query. We have that C\sqsubseteq D\in \overline{\mathit{TBox}} if and only if C\sqsubseteq D holds in all minimal canonical models of KB with respect to \mathcal{S}.
Proof
(Only if part) Assume that C\sqsubseteq D holds in all minimal canonical models of KB with respect to \mathcal{S}, and let \mathcal{M}=\langle\Delta,<,I\rangle be a minimal canonical model of KB satisfying C\sqsubseteq D. Observe that C and D (and their complements) belong to \mathcal{S}. We consider two cases: (1) the left end side of the inclusion C does not contain the typicality operator, and (2) the left end side of the inclusion is {\bf T}(C).
In case (1), if the minimal canonical model \mathcal{M} of KB satisfies C\sqsubseteq D. Then, C^{I}\subseteq D^{I}. For a contradiction, let us assume that C\sqsubseteq D\not\in \overline{\mathit{TBox}}. Then, by definition of \overline{\mathit{TBox}}, it must be: KB \not\models_{\mathcal{SHIQ}}C\sqsubseteq D. Hence, KB \not\models_{\mathcal{SHIQ}}C\sqcap\neg D\sqsubseteq\bot, and the set of concepts \{C,\neg D\} is consistent with KB. As \mathcal{M} is a canonical model of KB, there must be a element x\in\Delta such that x\in(C\sqcap\neg D)^{I}. This contradicts the fact that C^{I}\subseteq D^{I}.
In case (2), assume \mathcal{M} satisfies {\bf T}(C)\sqsubseteq D. Then, {\bf T}(C)^{I}\subseteq D^{I}, i.e., for each x\in Min_{<}(C^{I}), x\in D^{I}. If Min_{<}(C^{I})=\emptyset, then there is no x\in C^{I} (by the smoothness condition), hence C has no rank in \mathcal{M} and, by Proposition 13, C has no rank (\mathit{rank}(C)=\infty). In this case, by Definition 8, {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}}. Otherwise, let us assume that k_{\mathcal{M}}(C)=i. As k_{\mathcal{M}}(C\sqcap D)<k_{\mathcal{M}}(C\sqcap\neg D), then k_{\mathcal{M}}(C\sqcap\neg D)>i. By Proposition 13, rank(C)=i and rank(C\sqcap\neg D)>i. Hence, by Definition 8, {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}}.
(If part) If C\sqsubseteq D\in\overline{\mathit{TBox}}, then, by definition of \overline{\mathit{TBox}}, KB \models_{\mathcal{SHIQ}}C\sqsubseteq D. Therefore, each minimal canonical model \mathcal{M} of KB satisfies C\sqsubseteq D.
If {\bf T}(C)\sqsubseteq D\in\overline{\mathit{TBox}}, then by Definition 8, either (a) rank(C)<rank(C\sqcap\neg D), or (b) C has no rank. Let \mathcal{M} be any minimal canonical model of KB. In the case (a), by Proposition 13, k_{\mathcal{M}}(C)<k_{\mathcal{M}}(C\sqcap\neg D), which entails k_{\mathcal{M}}(C\sqcap D)<k_{\mathcal{M}}(C\sqcap\neg D). Hence \mathcal{M} satisfies {\bf T}(C)\sqsubseteq D. In case (b), by Proposition 13, C has no rank in \mathcal{M}, hence \mathcal{M} satisfies {\bf T}(C)\sqsubseteq D. \hfill\Box
6 Rational Closure over the ABox
The definition of rational closure in Section 3 takes only into account the TBox. We address the issue of ABox reasoning first by the semantical side: as for any domain element, we would like to attribute to each individual constant named in the ABox the lowest possible rank. Therefore we further refine Definition 10 of minimal canonical models with respect to TBox by taking into account the interpretation of individual constants of the ABox.
Definition 11 (Minimal canonical model w.r.t. ABox)
Given KB=(TBox,ABox), let \mathcal{M}=\langle\Delta,<,I\rangle and \mathcal{M}^{\prime}=\langle\Delta^{\prime},<^{\prime},I^{\prime}\rangle be two canonical models of KB which are minimal w.r.t. Definition 10. We say that \mathcal{M} is preferred to \mathcal{M}^{\prime} w.r.t. ABox (\mathcal{M}<_{\mathit{ABox}}\mathcal{M}^{\prime}) if, for all individual constants a occurring in ABox, k_{\mathcal{M}}(a^{I})\leq k_{\mathcal{M}^{\prime}}(a^{I^{\prime}}) and there is at least one individual constant b occurring in ABox such that k_{\mathcal{M}}(b^{I})<k_{\mathcal{M}^{\prime}}(b^{I^{\prime}}).
As a consequence of Proposition 11 we can prove that:
Theorem 6.1
For any KB=(TBox,ABox) there exists a minimal canonical model of KB with respect to ABox.
In order to see the strength of the above semantics, consider our example about marriages and VIPs.
Example 2
Suppose we have a KB=(TBox,ABox) where: TBox=\{{\bf T}(\mathit{Person})\sqsubseteq\ \leq 1\ \mathit{HasMarried}.\mathit{% Person}, {\bf T}(\mathit{VIP})\sqsubseteq\ \geq 2\ \mathit{HasMarried}.\mathit{Person}, \mathit{VIP}\sqsubseteq\mathit{Person}\}, and ABox = \{\mathit{VIP}(\mathit{demi}),\mathit{Person}(\mathit{marco})\}. Knowing that Marco is a person and Demi is a VIP, we would like to be able to assume, in the absence of other information, that Marco is a typical person, whereas Demi is a typical VIP, and therefore Marco has at most one spouse, whereas Demi has at least two. Consider any minimal canonical model \mathcal{M} of KB. Being canonical, \mathcal{M} will contain, among other elements, the following:
x\in(\mathit{Person})^{I}, x\in(\leq 1\ \mathit{HasMarried}.\mathit{Person})^{I}, x\in(\neg\mathit{VIP})^{I}, k_{\mathcal{M}}(x)=0;
y\in(\mathit{Person})^{I}, y\in(\geq 2\ \mathit{HasMarried}.\mathit{Person})^{I}, y\in(\neg\mathit{VIP})^{I}, k_{\mathcal{M}}(y)=1;
z\in(\mathit{VIP})^{I}, z\in(\mathit{Person})^{I}, z\in(\geq 2\ \mathit{HasMarried}.\mathit{Person})^{I} , k_{\mathcal{M}}(z)=1;
w\in(\mathit{VIP})^{I}, w\in(\mathit{Person})^{I}, w\in(\leq 1\ \mathit{HasMarried}.\mathit{Person})^{I} , k_{\mathcal{M}}(w)=2.
so that x is a typical person and z is a typical VIP. Notice that in the definition of minimal canonical model there is no constraint on the interpretation of constants \mathit{marco} and \mathit{demi}. As far as Definition 10 is concerned, for instance, \mathit{marco} can be mapped onto x ((\mathit{marco})^{I}=x) or onto y ((\mathit{marco})^{I}=y): the minimality of \mathcal{M} w.r.t. Definition 10 is not affected by this choice. However in the first case it would hold that Marco is a typical person, in the second Marco is not a typical person. According to Definition 11, we prefer the first case, and there is a unique minimal canonical model w.r.t. ABox in which (\mathit{marco})^{I}=x and (\mathit{demi})^{I}=z.
We next provide an algorithmic construction for the rational closure of ABox. The idea is that of considering all the possible minimal consistent assignments of ranks to the individuals explicitly named in the ABox. Each assignment adds some properties to named individuals which can be used to infer new conclusions. We adopt a skeptical view by considering only those conclusions which hold for all assignments. The equivalence with the semantics shows that the minimal entailment captures a skeptical approach when reasoning about the ABox. More formally, in order to calculate the rational closure of ABox, written \overline{\mathit{ABox}}, for all individual constants of the ABox we find out which is the lowest possible rank they can have in minimal canonical models with respect to Definition 10: the idea is that an individual constant a_{i} can have a given rank k_{j}(a_{i}) just in case it is compatible with all the inclusions {\bf T}(A)\sqsubseteq D of the TBox whose antecedent A’s rank is \geq k_{j}(a_{i}) (the inclusions whose antecedent A’s rank is <k_{j}(a_{i}) do not matter since, in the canonical model, there will be an instance of A with rank <k_{j}(a_{i}) and therefore a_{i} will not be a typical instance of A). The algorithm below computes all minimal rank assignments k_{j}s to all individual constants: \mu^{j}_{i} contains all the concepts that a_{i} would need to satisfy in case it had the rank attributed by k_{j} (k_{j}(a_{i})). The algorithm verifies whether \mu^{j} is compatible with (\overline{\mathit{TBox}}, ABox) and whether it is minimal. Notice that, in this phase, all constants are considered simultaneously (indeed, the possible ranks of different individual constants depend on each other). For this reason \mu^{j} takes into account the ranks attributed to all individual constants, being the union of all \mu^{j}_{i} for all a_{i}, and the consistency of this union with (\overline{\mathit{TBox}}, ABox) is verified.
Definition 12 (\overline{\mathit{ABox}}: rational closure of ABox)
Let a_{1},\dots,a_{m} be the individuals explicitly named in the ABox. Let k_{1},k_{2},\dots,k_{h} be all the possible rank assignments (ranging from 1 to n) to the individuals occurring in ABox.
– Given a rank assignment k_{j} we define:

for each a_{i}: \mu^{j}_{i}=\{(\neg C\sqcup D)(a_{i}) s.t. C,D\in\mathcal{S}, {\bf T}(C)\sqsubseteq D in \overline{\mathit{TBox}}, and k_{j}(a_{i})\leq rank(C)\}\cup\{(\neg C\sqcup D)(a_{i}) s.t. C\sqsubseteq D in TBox \};

let \mu^{j}=\mu^{j}_{1}\cup\dots\cup\mu^{j}_{m} for all \mu^{j}_{1}\dots\mu^{j}_{m} just calculated for all a_{1},\dots,a_{m} in ABox
– k_{j} is minimal and consistent with (\overline{\mathit{TBox}}, ABox), i.e.: (i) TBox \cup ABox \cup\mu^{j} is consistent in \mathcal{SHIQ}^{\textsf{R}}{\bf T}; (ii) there is no k_{i} consistent with (\overline{\mathit{TBox}}, ABox) s.t. for all a_{i}, k_{i}(a_{i})\leq k_{j}(a_{i}) and for some b, k_{i}(b)<k_{j}(b).
– The rational closure of ABox ( \overline{\mathit{ABox}}) is the set of all assertions derivable in \mathcal{SHIQ}^{\textsf{R}}{\bf T} from TBox \cup ABox \cup\mu^{j} for all minimal consistent rank assignments k_{j}, i.e:
\overline{\mathit{ABox}}=\bigcap_{k_{j}\mbox{{\tiny minimal consistent}}}\{C(a% ):\; TBox \cup ABox \cup\ \mu^{j}\ \models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C(a)\}
The example below is the syntactic counterpart of the semantic Example 2 above.
Example 3
Consider the KB in Example 2. Computing the ranking of concepts we get that rank(\mathit{Person})=0, rank(\mathit{VIP})=1, rank(\mathit{Person}\ \sqcap\ \geq 2\ \mathit{HasMarried}.\mathit{Person})=1, rank(\mathit{VIP}\ \sqcap\ \leq 1\ \mathit{HasMarried}.\mathit{Person})=2. It is easy to see that a rank assignment k_{0} with k_{0}(\mathit{demi})=0 is inconsistent with KB as \mu^{0} would contain (\neg\mathit{VIP}\sqcup\mathit{Person})(\mathit{demi}), (\neg\mathit{Person}\ \sqcup\ \leq 1\ \mathit{HasMarried}.\mathit{Person})(% \mathit{demi}), (\neg\mathit{VIP}\sqcup\ \geq\ 2\ \mathit{HasMarried}. \mathit{Person})(\mathit{demi}) and \mathit{VIP}(\mathit{demi}). Thus we are left with only two ranks k_{1} and k_{2} with respectively k_{1}(\mathit{demi})=1,k_{1}(\mathit{marco})=0 and k_{2}(\mathit{demi})=k_{2}(\mathit{marco})=1.
The set \mu^{1} contains, among the others, (\neg\mathit{VIP}\ \sqcup\ \geq 2\ \mathit{HasMarried}.\mathit{Person})(% \mathit{demi}) , (\neg\mathit{Person}\ \sqcup\ \leq 1\ \mathit{HasMarried}.\mathit{Person})(% \mathit{marco}). It is tedious but easy to check that KB \cup\mu^{1} is consistent and that k_{1} is the only minimal consistent assignment (being k_{1} preferred to k_{2}), thus both (\geq 2\ \mathit{HasMarried}.\mathit{Person})(\mathit{demi}) and (\leq 1\ \mathit{HasMarried}.\mathit{Person}) (\mathit{marco}) belong to \overline{\mathit{ABox}}.
We are now ready to show the soundness and completeness of the algorithm with respect to the semantic definition of rational closure of ABox.
Theorem 6.2 (Soundness of \overline{\mathit{ABox}})
Given KB=(TBox, ABox), for each individual constant a in ABox, we have that if C(a)\in \overline{\mathit{ABox}} then C(a) holds in all minimal canonical models with respect to ABox of KB.
Proof (Sketch)
Let C(a)\in \overline{\mathit{ABox}}, and suppose for a contradiction that there is a minimal canonical model \mathcal{M} with respect to ABox of KB s.t. C(a) does not hold in \mathcal{M}. Consider now the rank assignment k_{j} corresponding to \mathcal{M} (such that k_{j}(a_{i})=k_{\mathcal{M}}(a_{i})). By hypothesis \mathcal{M} \models TBox \cup ABox. Furthermore it can be easily shown that \mathcal{M} \models\mu^{j}.
Since by hypothesis \mathcal{M}\not\models C(a), it follows that TBox \cup ABox \cup\ \mu^{j}\not\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C(a), and by definition of \overline{\mathit{ABox}}, C(a)\not\in\overline{\mathit{ABox}}, against the hypothesis. \hfill\Box
Theorem 6.3 (Completeness of \overline{\mathit{ABox}})
Given KB=(TBox, ABox), for all individual constant a in ABox, we have that if C(a) holds in all minimal canonical models with respect to ABox of KB, then C(a)\in \overline{\mathit{ABox}}.
Proof (Sketch)
We show the contrapositive. Suppose C(a)\not\in \overline{\mathit{ABox}}, i.e. there is a minimal k_{j} consistent with (\overline{\mathit{TBox}}, ABox) s.t. TBox \cup ABox \cup\mu^{j}\not\models_{\mathcal{SHIQ}^{\textsf{R}}{\bf T}}C(a). This means that there is an \mathcal{M}^{\prime}=\langle\Delta^{\prime},<,I^{\prime}\rangle such that for all a_{i}\in ABox, k_{\mathcal{M}^{\prime}}(a_{i})=k_{j}(a_{i}), \mathcal{M}^{\prime}\models TBox \cup ABox \cup\mu^{j} and \mathcal{M}^{\prime}\not\models C(a). From \mathcal{M}^{\prime} we build a minimal canonical model with respect to ABox \mathcal{M}=\langle\Delta,<,I\rangle of KB, such that C(a_{i}) does not hold in \mathcal{M}.
Since we do not know whether \mathcal{M}^{\prime}=\langle\Delta^{\prime},<^{\prime},I^{\prime}\rangle is minimal or canonical, we cannot use it directly; rather, we only use it as a support to the construction of \mathcal{M}. As the TBox is satisfiable, by Theorem 46, we know that there exists a minimal canonical model \mathcal{M}^{\prime\prime}=\langle\Delta^{\prime\prime},<^{\prime\prime},I^{\prime\prime}\rangle of the TBox. We extend such a model with domain elements from \Delta^{\prime} including those elements interpreting the individuals a_{1},\ldots,a_{m} explicitly named in the ABox. Let \Delta=\Delta_{1}\cup\Delta^{\prime\prime} where \Delta_{1}=\{(a_{i})^{I^{\prime}}: a_{i} in ABox \}\cup \{x\in\Delta^{\prime}: x is reachable from some (a_{i})^{I^{\prime}} in \mathcal{M}^{\prime} by a sequence of R^{I^{\prime}} or (R^{})^{I^{\prime}}\}.
We define the rank k_{\mathcal{M}} of each domain element in \Delta as follows. For the elements y\in\Delta^{\prime\prime}, k_{\mathcal{M}}(y)=k_{\mathcal{M}^{\prime\prime}}(y). For the elements x\in\Delta_{1}, if x=(a_{i})^{I^{\prime}}, then k_{\mathcal{M}}(x)=k_{\mathcal{M}^{\prime}}(x); if x\neq(a_{i})^{I^{\prime}}, then k_{\mathcal{M}}(x)=k_{\mathcal{M}^{\prime\prime}}(X), for some X\in\Delta^{\prime\prime} such that for all concepts C^{\prime}\in{\cal S}, we have x\in(C^{\prime})^{I^{\prime}} if and only if X\in(C^{\prime})^{I^{\prime\prime}}. We then define I as follows. First, for all a_{i} in ABox we let a_{i}^{I}=(a_{i})^{I^{\prime}}. We define the interpretation of each concept as in \Delta^{\prime} on the elements of \Delta_{1} and as in \Delta^{\prime\prime} on the elements of \Delta^{\prime\prime}. Last, we define the interpretation of each role R as in \mathcal{M}^{\prime} on the pairs of elements of \Delta_{1} and as in \mathcal{M}^{\prime\prime} on the pairs of elements of \Delta^{\prime\prime}. I is extended to quantified concepts in the usual way.
It can be proven that \mathcal{M} satisfies ABox (by definition of I and since \mathcal{M}^{\prime} satisfies it). Furthermore it can be proven that \mathcal{M} satisfies TBox (the full proof is omitted due to space limitations). C(a) does not hold in \mathcal{M}, since it does not hold in \mathcal{M}^{\prime}. Last, \mathcal{M} is canonical by construction. It is minimal with respect to Definition 10: for all X\in\Delta_{2} k_{\mathcal{M}}(X) is the lowest possible rank it can have in any model (by Proposition 13); for all a_{i}\in\Delta_{1}, this follows by minimality of k_{j}. From minimality of k_{j} it also follows that \mathcal{M} is a minimal canonical model with respect to ABox. Since in \mathcal{M} C(a) does not hold, the theorem follows by contraposition. \hfill\Box
Theorem 6.4 (Complexity of rational closure over the ABox)
Given a knowledge base KB=(TBox,ABox) in \mathcal{SHIQ}^{\textsf{R}}{\bf T}, an individual constant a and a concept C, the problem of deciding whether C(a)\in\overline{\mathit{ABox}} is ExpTimecomplete.
We omit the proof, which is similar to the one for rational closure over ABox in \mathcal{ALC} (Theorem 5 [18]).
7 Extending the correspondence to more expressive logics
A natural question is whether the correspondence between the rational closure and the minimal canonical model semantics of the previous section can be extended to stronger DLs. We give a negative answer for the logic {\cal SHOIQ}. This depends on the fact that, due to the interaction of nominals with number restriction, a consistent {\cal SHOIQ} knowledge base may have no canonical models (whence no minimal canonical ones). Let us consider for instance the following example:
Example 4
Consider the KB, where TBox=\{\{o\}\sqsubseteq\leq 1R^{}.\top,\;\neg\{o\}\sqsubseteq\ \geq 1R.\{o\}\}, and ABox=\{\neg A(o),\neg B(o)\}.
KB is consistent and, for instance, the model \mathcal{M}_{1}=\langle\Delta,<,I\ ra, where \Delta=\{x,y\}, < is the empty relation, A^{I}=B^{I}=(\neg\{o\})^{I}=\{x\}, and (\{o\})^{I}=\{y\}, is a model of KB. In particular, x\in(A\sqcap B)^{I}.
Also, there is a model \mathcal{M}_{2} of KB similar to \mathcal{M} (with \Delta_{2}=\{x_{2},y\}) in which x_{2}\in(A\sqcap\neg B)^{I}, another one \mathcal{M}_{3} (with \Delta_{3}=\{x_{2},y\}) in which x_{3}\in(\neg A\sqcap B)^{I}, and so on. Hence, \{A,B\}, \{A,\neg B\}, \{\neg A,B\}, \{\neg A,\neg B\} are all sets of concepts \mathcal{S} that are consistent with KB. Nevertheless, there is no canonical model for KB containing x_{1}, x_{2} and x_{3} all together. as the inclusions in the TBox prevent models from containing more than two domain elements.
The above example shows that the notion of canonical model as defined in this paper is too strong to capture the notion of rational closure for logics which are as expressive as {\cal SHOIQ}. Beacause of this negative result, we can regard the correspondence result for {\cal SHIQ} only as a first step in the definition of a semantic characterization of rational closure for expressive description logics. A suitable refinement of the semantics is needed, and we leave its definition for future work.
8 Related Works
There are a number of works which are closely related to our proposal.
In [14, 17] nonmonotonic extensions of DLs based on the T operator have been proposed. In these extensions, focused on the basic DL \mathcal{ALC}, the semantics of T is based on preferential logic P. Moreover and more importantly, the notion of minimal model adopted here is completely independent from the language and is determined only by the relational structure of models.
[6] develop a notion of rational closure for DLs. They propose a construction to compute the rational closure of an \mathcal{ALC} knowledge base, which is not directly based on Lehmann and Magidor definition of rational closure, but is similar to the construction of rational closure proposed by Freund [12] at a propositional level. In a subsequent work, [8] introduces an approach based on the combination of rational closure and Defeasible Inheritance Networks (INs). In [7], a work on the semantic characterization of a variant of the notion of rational closure introduced in [6] has been presented, based on a generalization to \mathcal{ALC} of our semantics in [16].
An approach related to ours can be found in [3]. The basic idea of their semantics is similar to ours, but it is restricted to the propositional case. ÊFurthermore, their construction relies on a specific representation of models and it provides a recipe to build a model of the rational closure, rather than a characterization of its properties. Our semantics, defined in terms of Êstandard Kripke models, can be more easily generalized to richer languages, as we have done here for \mathcal{SHIQ}.
In [5] the semantics of the logic of defeasible subsumptions is strengthened by a preferential semantics. Intuitively, given a TBox, the authors first introduce a preference ordering \ll on the class of all subsumption relations \textstyle\sqsubset \widetilde{\vrule height 0.0pt depth 0.0pt width 7.499886pt} including TBox, then they define the rational closure of TBox as the most preferred relation \textstyle\sqsubset \widetilde{\vrule height 0.0pt depth 0.0pt width 7.499886pt} with respect to \ll, i.e. such that there is no other relation \mathchoice{{\vtop{\hbox{$\displaystyle\sqsubset$}\hbox{$\widetilde{\vrule hei% ght 0.0pt depth 0.0pt width 7.499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$% \textstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0pt depth 0.0pt width 7% .499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$\scriptstyle\sqsubset$}\hbox{$% \widetilde{\vrule height 0.0pt depth 0.0pt width 5.24992pt}$}\vskip4.3pt}}}{{% \vtop{\hbox{$\scriptscriptstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0% pt depth 0.0pt width 3.749943pt}$}\vskip4.3pt}}}^{\prime} such that TBox \subseteq\mathchoice{{\vtop{\hbox{$\displaystyle\sqsubset$}\hbox{$\widetilde{% \vrule height 0.0pt depth 0.0pt width 7.499886pt}$}\vskip4.3pt}}}{{\vtop{% \hbox{$\textstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0pt depth 0.0pt % width 7.499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$\scriptstyle\sqsubset$}\hbox{% $\widetilde{\vrule height 0.0pt depth 0.0pt width 5.24992pt}$}\vskip4.3pt}}}{% {\vtop{\hbox{$\scriptscriptstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0% pt depth 0.0pt width 3.749943pt}$}\vskip4.3pt}}}^{\prime} and \mathchoice{{\vtop{\hbox{$\displaystyle\sqsubset$}\hbox{$\widetilde{\vrule hei% ght 0.0pt depth 0.0pt width 7.499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$% \textstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0pt depth 0.0pt width 7% .499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$\scriptstyle\sqsubset$}\hbox{$% \widetilde{\vrule height 0.0pt depth 0.0pt width 5.24992pt}$}\vskip4.3pt}}}{{% \vtop{\hbox{$\scriptscriptstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0% pt depth 0.0pt width 3.749943pt}$}\vskip4.3pt}}}^{\prime}\ll\mathchoice{{% \vtop{\hbox{$\displaystyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0pt dep% th 0.0pt width 7.499886pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$\textstyle\sqsubset$% }\hbox{$\widetilde{\vrule height 0.0pt depth 0.0pt width 7.499886pt}$}\vskip4% .3pt}}}{{\vtop{\hbox{$\scriptstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0% .0pt depth 0.0pt width 5.24992pt}$}\vskip4.3pt}}}{{\vtop{\hbox{$% \scriptscriptstyle\sqsubset$}\hbox{$\widetilde{\vrule height 0.0pt depth 0.0pt% width 3.749943pt}$}\vskip4.3pt}}}. Furthermore, the authors describe an ExpTime algorithm in order to compute the rational closure of a given TBox in \mathcal{ALC}. [5] does not address the problem of dealing with the ABox. In [24] a plugin for the Protégé ontology editor implementing the mentioned algorithm for computing the rational closure for a TBox for OWL ontologies is described.
Recent works discuss the combination of open and closed world reasoning in DLs. In particular, formalisms have been defined for combining DLs with logic programming rules (see, for instance, [11] and [25]). A grounded circumscription approach for DLs with local closed world capabilities has been defined in [22].
9 Conclusions
In this work we have proposed an extension of the rational closure defined by Lehmann and Magidor to the Description Logic \mathcal{SHIQ}, taking into account both TBox and ABox reasoning. Defeasible inclusions are expressed by means of a typicality operator {\bf T} which selects the typical instances of a concept. One of the contributions is that of extending the semantic characterization of rational closure proposed in [16] for propositional logic, to \mathcal{SHIQ}, which does not enjoy the finite model property. In particular, we have shown that in all minimal models of a finite KB in \mathcal{SHIQ} the rank of domain elements is always finite, although the domain might be infinite, and we have exploited this result to establish the correspondence between the minimal model semantics and the rational closure construction for \mathcal{SHIQ}. The (defeasible) inclusions belonging to the rational closure of a \mathcal{SHIQ} KB correspond to those that are minimally entailed by the KB, when restricting to canonical models. We have provided some complexity results, namely that, for \mathcal{SHIQ}, the problem of deciding whether an inclusion belongs to the rational closure of the TBox is in ExpTime as well as the problem of deciding whether C(a) belongs to the rational closure of the ABox. Finally, we have shown that the rational closure of a TBox can be computed simply using entailment in \mathcal{SHIQ}.
The rational closure construction in itself can be applied to any description logic. We would like to extend its semantic characterization to stronger logics, such as {\cal SHOIQ}, for which the notion of canonical model as defined in this paper is too strong, as we have seen in section 7.
It is well known that rational closure has some weaknesses that accompany its wellknown qualities. Among the weaknesses is the fact that one cannot separately reason property by property, so that, if a subclass of C is exceptional for a given aspect, it is exceptional “tout court” and does not inherit any of the typical properties of C. Among the strengths there is its computational lightness, which is crucial in Description Logics. Both the qualities and the weaknesses seems to be inherited by its extension to Description Logics. To address the mentioned weakness of rational closure, we may think of attacking the problem from a semantic point of view by considering a finer semantics where models are equipped with several preference relations; in such a semantics it might be possible to relativize the notion of typicality, whence to reason about typical properties independently from each other.
Acknowledgement. We thank the anonymous referees for their helpful comments. This work has been partially supported by the Compagnia di San Paolo and by the project “CONDESC: deduzione automatica per logiche CONDizionali e DESCrittive”.
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Appendix A APPENDIX: Encoding \mathcal{SHIQ}^{\textsf{R}}{\bf T} in \mathcal{SHIQ}
In this section, we provide an encoding of \mathcal{SHIQ}^{\textsf{R}}{\bf T} in \mathcal{SHIQ} and show that reasoning in \mathcal{SHIQ}^{\textsf{R}}{\bf T} has the same complexity as reasoning in \mathcal{SHIQ}. To this purpose, we first need to show that among \mathcal{SHIQ}^{\textsf{R}}{\bf T} models, we can restrict our consideration to models where the rank of each element is finite and less than the number of (sub)concepts occurring in the KB, which is polynomial in the size of the KB.
Proposition 14
Given a knowledge base KB= (TBox, ABox) in \mathcal{SHIQ}^{\textsf{R}}{\bf T}, there is an h_{KB}\in\mathbb{N} such that, for each model M of the KB in \mathcal{SHIQ}^{\textsf{R}}{\bf T} satisfying a concept C, there exists a model M^{\prime} of the KB such that the rank of each element in M^{\prime} is finite and less then h_{KB}, satisfying the concept C. Also, h_{KB} is polynomial in the size of the KB.
Proof
(Sketch) Given a \mathcal{SHIQ}^{\textsf{R}}{\bf T} model \mathcal{M}=\langle\Delta,<,I\rangle, observe that:

it is not the case that an element x\in\Delta is an instance of concept \Box\neg C and another domain element y\in\Delta, with y<x is an instance of concept \neg\Box\neg C;

given two domain elements x and y such that x and y have different ranks (for instance, k_{\mathcal{M}}(x)=i, k_{\mathcal{M}}(y)=j and i<j), if they are instances of exactly the same concepts of the form \Box\neg C (i.e., x\in(\Box\neg C)^{I} iff y\in(\Box\neg C)^{I}) for all concepts C occurring in the KB, then y can be assigned the same rank as x without changing the set of concepts of which y is an instance. Note that {\bf T} cannot occur in the scope of a \Box modality.
By changing the rank of (possibly infinite many) domain elements according item (2), we can transform any \mathcal{SHIQ}^{\textsf{R}}{\bf T} model into another \mathcal{SHIQ}^{\textsf{R}}{\bf T} model \mathcal{M}^{\prime}=\langle\Delta,{<^{\prime}},I^{\prime}\rangle where each domain element has a finite rank.
For each domain element x\in\Delta, let
x^{\mathcal{M}}_{\Box}=\{\Box\neg C\mid x\in(\Box\neg C)^{I}\} 
We let I^{\prime}=I and we define <^{\prime} by the following ranking function, for any y\in\Delta:
k_{\mathcal{M}^{\prime}}(y)=min\{k_{\mathcal{M}}(x)\mid x\in\Delta\mbox{ and }% x^{\mathcal{M}}_{\Box}=y^{\mathcal{M}}_{\Box}\}
Observe that k_{min}(y) is welldefined for any element y\in\Delta (a set of ordinals has always a least element). We can show that \mathcal{M}^{\prime}\models KB. Since I^{\prime} is the same as I in \mathcal{M}, it follows immediately that \mathcal{M}^{\prime} satisfies strict concept inclusions, role inclusions and ABox assertions. Also, for each transitive role R, R^{I^{\prime}} is transitive (as R^{I} is transitive).
We prove that \mathcal{M}^{\prime}\models K_{D}. Let {\bf T}(C)\sqsubseteq E\in F_{D}. Suppose, by absurdum, that \mathcal{M}^{\prime}\not\models{\bf T}(C)\sqsubseteq E, this means that there is a z\in\Delta such that z\in(T(C))^{I^{\prime}} and z\not\in E^{I^{\prime}}. We show that in \mathcal{M}, z\in(T(C))^{I^{\prime}} and z\not\in E^{I^{\prime}}. As I^{\prime}=I, from z\not\in E^{I^{\prime}} it follows that z\not\in E^{I}. Let z\in(T(C))^{I^{\prime}}. Then, by definition of T(C) as C\sqcap\Box\neg C, it must be that z\in(C)^{I^{\prime}} and z\in(\Box\neg C)^{I^{\prime}}. Observe that, by construction, z_{\Box}^{\mathcal{M}^{\prime}}=z_{\Box}^{\mathcal{M}}, since z has been assigned in \mathcal{M}^{\prime} the same rank as an element x such that x^{\mathcal{M}}_{\Box}=z^{\mathcal{M}}_{\Box}. Therefore, z\in(\Box\neg C)^{I}. Also, since I^{\prime}=I, z\in(C)^{I}. Hence, z\in(C\sqcap\Box\neg C)^{I}, and z\in(T(C))^{I}. We can then conclude that \mathcal{M}\not\models{\bf T}(C)\sqsubseteq E, against the fact that \mathcal{M} is a model of KB.
Hence, \mathcal{M}^{\prime} is a model of KB. Similarly, it can be easily shown that if C is satisfiable in \mathcal{M}, i.e. there is an x\in\Delta such that x\in C^{I}, then x\in C^{I^{\prime}} and therefore, C is satisfiable in \mathcal{M}^{\prime}.
Observe that, in \mathcal{M}^{\prime}, any pair of domain element with different ranks cannot be instances of the same concepts \Box\neg C for all the C occurring in the KB (not containing the {\bf T} operator). This is true, in particular, for the pairs v and w of domain elements with adjacent ranks, i.e., such that k_{\mathcal{M}}(v)=i+1 and k_{\mathcal{M}}(w)=i, for some i. For such a pair, there must be at least a concept C such that v is an instance of \neg\Box\neg C while w is an instance of \Box\neg C (the converse, that w is an instance of \neg\Box\neg C while v is an instance of \Box\neg C, is not possible by the transitivity of \Box, as w<v).
As a consequence, for each domain element w with rank i, there is at least a concept C occurring in the KB such that: all the domain elements with rank i+1 are instances of \neg\Box\neg C, while w is an instance of \Box\neg C. Informally, the number of \Box formulas of which a domain element is an instance increases, when the rank decreases. For a given KB, an upper bound h_{KB} on the rank of all domain elements can thus be determined as the number of (sub)concepts occurring in the KB, which is polynomial in the size of the KB.
\hfill\Box
In the following, we can restrict our consideration to models of the KB with finite ranks whose value is less or equal to h_{KB}, the number of (sub)concepts occurring in the KB (which is polynomial in the size of the KB).
The following theorem says that reasoning in \mathcal{SHIQ}^{\textsf{R}}{\bf T} has the same complexity as reasoning in \mathcal{SHIQ}, i.e. it is in ExpTime. Its proof is given by providing an encoding of satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} into satisfiability \mathcal{SHIQ}, which is known to be an ExpTimecomplete problem.
Theorem 2.1. Satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} is an ExpTimecomplete problem.
Proof
(Sketch) The hardness comes from the fact that satisfiability in \mathcal{SHIQ} is ExpTimehard. We show that satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} can be solved in ExpTime by defining a polynomial reduction of satisfiability in \mathcal{SHIQ}^{\textsf{R}}{\bf T} to satisfiability in \mathcal{SHIQ}.
Let KB=(TBox,ABox) be a knowledge base, and C_{0} a concept in \mathcal{SHIQ}^{\textsf{R}}{\bf T}. We define an encoding (TBox’, ABox’) of KB and C^{\prime}_{0} of C_{0} in \mathcal{SHIQ} as follows.
First, we introduce new atomic concepts Zero and W in the language and a new role R, where R is intended to model the relation < of \mathcal{SHIQ}^{\textsf{R}}{\bf T} models. We let TBox’ contain the inclusions
\top\sqsubseteq\leq 1R.\top \top\sqsubseteq\leq 1R^{}.\top
so that R allows to represent linear sequences. We will consider the linear sequences of elements of the domain reachable trough R^{} from the Zero elements, i.e., those sequences w_{0},w_{1},w_{2},\ldots, with w_{0}\in Zero^{I} and (w_{i},w_{i+1})\in(R^{})^{I}. Given Proposition 14, we can restrict our consideration to finite linear sequences with length less or equal to h, the number of subconcepts of the KB (which is polynomial in the size of KB). We introduce h new atomic concepts S_{1},\ldots,S_{h} such that the instances of S_{i} are the domain elements reachable form a Zero element by a chain of length i of R^{}successors. We introduce in TBox’ the following inclusions:
Zero\sqsubseteq\forall R^{}.S_{1} S_{1}\sqsubseteq\exists R.Zero S_{i}\sqsubseteq\forall R^{}.S_{i+1} S_{i+1}\sqsubseteq\exists R.S_{i}
Zeroelements have no Rsuccessor and S_{h}elements have no Rpredecessors.
Zero\sqsubseteq\neg\exists R.\top. S_{h}\sqsubseteq\neg\exists R^{}.\top.
All the elements in a sequences w_{0},w_{1},w_{2},\ldots, as introduced above, are instances of concept W:
W\sqsubseteq Zero\sqcup S_{1}\sqcup\ldots\sqcup S_{n} Zero\sqcup S_{1}\sqcup\ldots\sqcup S_{n}\sqsubseteq W
From the sequences w_{0},w_{1},w_{2},\ldots starting from Zero elements, we can encode in \mathcal{SHIQ} the structure of ranked models of \mathcal{SHIQ}^{\textsf{R}}{\bf T}, by associating rank i to all the elements w_{i} in S_{i}.
We have to provide an encoding for the inclusions in TBox. For each A\sqsubseteq B\in TBox, not containing {\bf T}, we introduce A\sqsubseteq B in TBox’.
For each {\bf T}(A) occurring in the TBox, we introduce a new atomic concept \Box_{\neg A} and, for each inclusion {\bf T}(A)\sqsubseteq B\in TBox, we add to TBox’ the inclusion
A\sqcap\Box_{\neg A}\sqsubseteq B 
To capture the properties of the \Box modality, the following equivalences are introduced in TBox’:
\Box_{\neg A}\equiv\forall R.(\neg A\sqcap\Box_{\neg A})
\top\sqsubseteq\forall U.(\neg S_{i}\sqcup\Box_{\neg A})\sqcup\forall U.(\neg S% _{i}\sqcup\neg\Box_{\neg A})
for all i=0,\ldots,h and for all concept names A\in{\cal C}, where U is the universal role (which can be defined in \mathcal{SHIQ} [19]). The first inclusion, says that if a domain element of rank i is an instance of concept \Box_{\neg A}, the elements of rank i1 (in the same sequence) are instances of both the concepts {\neg A} and \Box_{\neg A}. (this is to account for the transitivity of the \Box modality). The second inclusion forces the S_{i}elements (i.e. all the domain elements with rank i) to be instances of the same boxed concepts \Box_{\neg A}, for all A\in{\cal C}.
For each named individual a\in N_{I}, we add to ABox’ the assertion W(a), to guarantee the interpretation of a to be a Welement.
For all the assertions C_{R}(a) in ABox, we add C_{R}(a) to ABox’. For all the assertions {\bf T}(C)(a) in ABox, we add (A\ \sqcap\ \Box_{\neg A})(a) to ABox’. For all the assertions R(a,b)\inABox, we add R(a,b) to ABox’.
Given a \mathcal{SHIQ}^{\textsf{R}}{\bf T} concept C_{0}, whose size is assumed to be polynomial in the size of the KB, we encode by introducing the following \mathcal{SHIQ} concept C^{\prime}_{0}
\exists U.(W\sqcap[C_{0}]) 
where U is the universal role and [C_{0}] is obtained from C_{0} by replacing each occurrence of {\bf T}(A) in C_{0} with A\sqcap\Box_{\neg A}. [C_{0}] is a \mathcal{SHIQ} concept and we require it to be satisfied in some Welement. We can then prove the following:

The size of KB’ and size of C^{\prime}_{0} are polynomial in the size of KB.

Concept C_{0} is satisfiable with respect to KB in \mathcal{SHIQ}^{\textsf{R}}{\bf T} if and only if C^{\prime}_{0} is satisfiable with respect to KB’ in \mathcal{SHIQ}.
The proof of this result can be done by showing that a \mathcal{SHIQ}^{\textsf{R}}{\bf T} model of KB satisfying C_{0} can be transformed into a \mathcal{SHIQ}^{\textsf{R}}{\bf T} model of KB’ satisfying C^{\prime}_{0}. And viceversa.
We can therefore conclude that the satisfiability problem in \mathcal{SHIQ}^{\textsf{R}}{\bf T} can be polynomially reduced to the satisfiability problem in \mathcal{SHIQ}, which is in ExpTime. \hfill\square
Appendix B APPENDIX: Wellfounded relations
A few definitions.
Definition 13
Let S be a nonempty set and <^{*} a transitive, irreflexive relation on S (a strict preorder). Let U\subseteq S, with U\not=\emptyset, we say that x\in S is a minimal element of U with respect to <^{*} if it holds:
x\in U and \forall y\in U we have y\not<^{*}x.
Given U\subseteq S, we denote by Min_{<^{*}}(U) the set of minimal elements of U with respect to <^{*}.
Definition 14
Let S and <^{*} as in previous definition. We say that <^{*} is wellfounded on S if for every nonempty U\subseteq S, we have Min_{<^{*}}(U)\not=\emptyset.
Proposition 15
Let S and <^{*} as above. The following are equivalent:

<^{*} is wellfounded on S;

there are no infinite descending chains: \ldots x_{i+1}<^{*}x_{i}<^{*}\ldots<^{*}x_{0} of elements of S.
Proof

(1)\Rightarrow(2). Suppose that <^{*} is wellfounded on S and by absurd that there is an infinite descending chain \ldots x_{i+1}<^{*}x_{i}<^{*}\ldots<^{*}x_{0} of elements of S. Let U be the set of elements of such a chain. Clearly for every x_{i}\in U there is a x_{j}\in U with x_{j}<^{*}x_{i}. But this means that Min_{<^{*}}(U)=\emptyset against the hypothesis that <^{*} is wellfounded on S.

(2)\Rightarrow(1). Suppose by absurd that for a nonempty U\subseteq S, we have that Min_{<^{*}}(U)=\emptyset. Thus:
\forall x\in U\ \ \exists y\in U\ y<^{*}x We can assume that there is a function f:U\longrightarrow U such that f(x)<^{*}x.
[If S is enumerable then f can be defined by means of an enumeration of S (e.g. take the smallest y<^{*}x in the enumeration); otherwise and more generally, by using the axiom of choice we can proceed as follows: given x\in U, let U_{\downarrow x}=\{y\in U\mid y<^{*}x\}, thus for every x\in U the set U_{\downarrow x} is nonempty. By the axiom of choice, there is a function g:
g:\{U_{\downarrow x}\mid x\in U\}\longrightarrow\bigcup_{x\in U}U_{\downarrow x% }\ (=U) such that for every x\in U, g(U_{\downarrow x})\in U_{\downarrow x}. We then define f(x)=g(U_{\downarrow x}).]
Since f(x)<x we also have f(f(x))<^{*}f(x)<^{*}x and so on. Using the notation f^{i}(x) for the i  iteration of f, we can immediatly define an infinite descending chain by fixing x_{0}\in U and by taking x_{i}=f^{i}(x_{0}) for all i>0.
\hfill\square
Theorem B.1
Let S be a nonempty set and <^{*} a binary relation on S. The following are equivalent:

<^{*} is (i) irreflexive, (ii) transitive, (iii) modular, (iv) wellfounded.

there exists a function k:S\longrightarrow Ord such that x<^{*}y iff k(x)<k(y) (where Ord is the set of ordinals).
Proof

(2)\Rightarrow(1). Suppose that there is a function k:S\longrightarrow Ord such that x<^{*}y iff k(x)<k(^{\prime}y). We can easily check that properties (i)–(iv) holds: irreflexivity and transitivity are immediate. For (iii) modularity: let x<^{*}y and z be any element in S. Suppose that x\not<^{*}z, thus k(x)\not<k(z); then it must be either k(x)=k(z) or k(z)<k(x), whence k(z)<k(y) in both cases, thus z<^{*}y.
For (iv) wellfoundedness, suppose by absurd that there is a nonempty U\subseteq S such that Min_{<^{*}}(U)=\emptyset, then for every x\in U there is y\in U such that y<^{*}x. Let us consider the image of U under k: U_{k}=\{k(x)\mid x\in U\}. The set of ordinals U_{k} has a least element, say \beta\in Ord (this by property of ordinals: every nonempty set of ordinals has a least element). Let z\in U such that k(z)=\beta. By hypothesis, there is y\in U such that y<^{*}u, but then k(y)\in U_{k} and k(y)<\beta, against the fact that \beta is the least ordinal in U_{k}.

(1)\Rightarrow(2) (Sketch). Suppose that <^{*} satisfies properties (i)–(iv). Let us consider the following sequence of sets indexed on Ordinals:
S_{\alpha}=S\bigcup_{\beta<\alpha}A_{\beta}
A_{\alpha}=Min_{<^{*}}(S_{\alpha})Thus S_{0}=S and A_{0}=Min_{<^{*}}(S). Observe that if S_{\alpha}\not=\emptyset then also A_{\alpha}\not=\emptyset (by wellfoundness); moreover the sequence is decreasing: S_{\alpha}\subset S_{\beta} for \beta<\alpha. But for cardinality reasons there must be a least ordinal \lambda such that S_{\lambda}=\emptyset, this means that S_{\lambda}=S\bigcup_{\beta<\lambda}A_{\beta}=\emptyset, so that we get
S=\bigcup_{\beta<\lambda}A_{\beta} It can be easily shown that:

for \alpha<\beta<\lambda, \forall x\in A_{\alpha},\forall y\in A_{\beta}\ x<^{*}y, and also A_{\alpha}\cap A_{\beta}=\emptyset

for each x\in S, there exists a unique A_{\alpha} with \alpha<\lambda such that x\in A_{\alpha}

x<^{*}y iff for some \alpha,\beta<\lambda x\in A_{\alpha} and y\in A_{\beta} and \alpha<\beta.
We can then define k(x)= the unique \alpha such that x\in A_{\alpha} and the result follows.

\hfill\square