A NExpTime-Complete Description Logic Strictly Contained in C 2

We examine the complexity and expressivity of the combination of the Description Logic ALCQI with a terminological formalism based on cardinality restrictions on concepts. This combination can naturally be embedded into C 2 , the two variable fragment of predicate logic with counting quanti(cid:12)ers. We prove that ALCQI has the same complexity as C 2 but does not reach its expressive power.


Introduction
Description Logic (DL) systems can be used in knowledge based systems to represent and reason about taxonomical knowledge of problem domain in a semantically well-de ned manner WS92].These systems usually consist at least of the following three components: (1) a DL that allows the de nition of complex concepts (unary predicates) and roles (binary relations) to be built from atomic ones by the application of a given set of constructors; for example the following concept describes those fathers having at least two daughters: Parent u Male u ( 2 hasChild Female) (2) a terminological component (TBox) that allows for the organisation of dened concepts and roles.The TBox formalisms studied in the DL context range from weak ones allowing only for the introduction of abbreviations for complex concepts, over TBoxes capable of expressing various forms of axioms, to cardinality restrictions that can express restrictions on the number of elements a concept may have.Consider the following three expressions: ( 2 Person u ( 0 hasChild 1 Parent)) The rst introduces BusyParent as an abbreviation for a more complex concept, the second is an axiom stating that Male and Female are exactly those persons having two parents, the third is a cardinality restriction expressing that in the domain of discourse there are at most two earliest ancestors.
(3) a reasoning service that performs task like subsumption or consistency test for the knowledge stored in the TBox and ABox.There exist sound and complete algorithms for reasoning in a large number of DLs and di erent TBox and ABox formalisms that meet the known worst-case complexity these problems (see DLNN97] for an overview).Generally, reasoning for DLs can be performed in four di erent ways: by structural comparison of syntactical normal forms of concepts BPS94].by tableaux algorithms that are hand-tailored to suit the necessities of the operators used to form the DL and the TBox formalism.Initially, these algorithm were designed to decide inference problems only for the DL without taking into account TBoxes, but it is possible to generalise these algorithms to deal with di erent TBox formalisms.Most DLs handled this way are at most PSpace complete but additional complexity may arise from the TBox.The complexity of the tableaux approach usually meets the known worst-case complexity of the problem SSS91, DLNN97].by perceiving the DL as a (fragment of a) modal logic such as PDL GL96]; many DLs handled in this manner are already ExpTime-complete, but axioms can be \internalised" Baa91] into the concepts and hence do not increase the complexity.by translation of the problem into a fragment or rst order other logic with a decidable decision problem Bor96, OSH96].From the fragments of predicate logic that are studied in the second context, only C 2 , the two variable fragment of rst order predicate logic augmented with counting quanti ers, is capable of dealing with counting expressions that are commonly used in DLs; similarly it is able to express cardinality restrictions.Another thing that comes \for free" when translating DLs into rst order logic is the ability to deal with inverse roles.
Combining all these construct into a single DL, one obtains the DL ALCQI| the well-known DL ALC SSS91] augmented by qualifying number restrictions (Q) and inverse roles (I).In this work we study both complexity and expressivity of ALCQI combined with TBoxes based on cardinality restrictions.
Regarding the complexity we show that ALCQI with cardinality restrictions already is NExpTime-hard and hence has the same complexity as C 2 PST97]1 .
To our knowledge this is the rst DL for which NExpTime-completeness has formally been proved.Since ALCQI with TBoxes consisting of axioms is still in ExpTime, this indicates that cardinality restrictions are algorithmically hard to handle.
Despite the fact that both ALCQI and C 2 have the same worst-case complexity we show that ALCQI lacks some of the expressive power of C 2 .Properties of binary predicates (e.g.re exivity) that are easily expressible in C 2 can not be expressed in ALCQI.We establish our result by giving an Ehrenfeucht-Fra ss e game that exactly captures the expressivity of ALCQI with cardinality restrictions.This is the rst time in the area of DL that a game-theoretic characterisation is used to prove an expressivity result involving TBox formalisms.
The game as it is presented here is not only applicable to ALCQI with cardinality restrictions; straightforward modi cations make it applicable to both ALCQ as well as to weaker TBox formalisms like terminological axioms.
In Bor96] a DL is presented that has the same expressivity as C 2 .This expressivity result is one of the main results of that paper and the DL combines a large number of constructs; the paper does not study the computational complexity of the presented logics.Our motivation is of a di erent nature: We study the complexity and expressivity of a DL consisting of a only minimal set of constructs that seem sensible when a reduction of that DL to C 2 is to be considered.

The Logic ALCQI
In this section we de ne syntax and semantics of the DL ALCQI.De nition 2.1 A signature is a pair = (N C ; N R ) where N C is a nite set of concepts names and N R is a nite set of role names.Concepts in ALCQI are built inductively from these using the following rules: All A 2 N C are concepts.If C, C 1 , and C 2 are concepts then also :C; C 1 u C 2 ; and ( n S C) with n 2 N, and S = R or S = R 1 for some R 2 N R are concepts.We de ne C 1 t C 2 as an abbreviation for :(:C 1 u :C 2 ) and ( n S C) as an abbreviation for :( (n + 1) S C).We also use (= n S C) as an abbreviation for ( n S C) u ( n S C).
A cardinality restriction of ALCQI is an expression of the form ( n C) or ( n C) where C is a concept and n 2 N; a TBox T of ALCQI is a nite set of cardinality restrictions.
The semantics of a concept is de ned relative to an interpretation I, which is a tuple consisting of a domain I and a valuation ( I ) which maps each concept name A to a subset A I of I and each role name R to a subset R I of I I .This valuation is extend to arbitrary concept de nitions inductively using the following rules, where ]M denotes the cardinality of a set M:  Please note that the NExpTime-completeness result from PST97] is only valid if we assume unary coding of numbers to in the input; this implies that a large number like 1000 may not be stored in logarithmic space in some kary representation but consumes 1000 units of storage.This is the standard assumption made by most results concerning the complexity of DLs.We will come back to this issue later in this paper.

ALCQI is NExpTime-complete
To show that NExpTime is also the lower bound for the complexity of TBox consistency we use a bounded version of the domino problem.Domino problems Wan63, Ber66] have successfully been employed to establish undecidability and complexity results for various description and modal logics Spa93, BS99].

Domino Systems
De nition 3.1 For an n 2 N let Z n denote the set f0; : : : ; n 1g and n denote the addition modulo n.A domino system is a triple D = (D; H; V ), where D is a nite set (of tiles) and H; V D D are relations expressing horizontal and vertical compatibility constraints between the tiles.For s; t 2 N let U(s; t) be the torus Z s Z t and w = w 0 ; : : : ; w n 1 be an n-tuple of tiles (with n s).We say that D tiles U(s; t) with initial condition w i there exists a mapping : U(s; t) !D such that for all (x; y) 2 U(s; t): If (x; y) = d and (x s 1; y) = d 0 then (d; d 0 ) 2 H; if (x; y) = d and (x; y t 1) = d 0 then (d; d 0 ) 2 V ; (i; 0) = w i for 0 i < n, Bounded domino systems as capable of expressing the computational behaviour of restricted, so called simple, Turing Machines (TM).This restriction is non-essential in the following sense: Every language accepted in time T(n) and space S(n) by some one-tape TM is accepted within the same time and space bounds by a simple TM, as long as S(n); T(n) 2n.
Theorem 3.2 ( BGG97], Theorem 6.1.2)Let M be a simple TM with input alphabet .Then there exists a domino system D = (D; H; V ) and a linear time reduction which takes any input x 2 to a word w 2 D with jxj = jwj such that If M accepts x in time t 0 with space s 0 , then D tiles U(s; t) with initial condition w for all s s 0 + 2; t t 0 + 2; if M does not accept x, then D does not tile U(s; t) with initial condition w for any s; t 2.
A simple corollary of this theorem will be used to prove that TBox consistency for ALCQI is NExpTime-hard: Corollary 3.3 Let M be a (w.l.o.g.simple) non-deterministic TM with time-(and hence space-) bound 2 n d (d constant) deciding an arbitrary NExpTimecomplete language L(M) over the alphabet .Let D and trans be the according domino system and reduction from Theorem 3.2.The following is a NExpTime-hard problem: Given an initial condition w = w 0 ; : : : ; w n 1 of length n.Does D tile U(2 n d +1 ; 2 n d +1 ) with initial condition w?
Proof.The function trans is a linear reduction from L(M) to the stated problem: For a v 2 with jvj = n it holds that v 2 L(M) i M accepts v in time and space 2 jvj d i D tiles U(2 n d +1 ; 2 n d +1 ) with initial condition trans(v)

De ning a Torus of Exponential Size
Just as de ning in nite grids is the key problem in proving undecidability by reduction from unbounded domino problems, de ning a torus of exponential size is the key to obtaining a NExpTime-completeness proof by reduction from bounded domino problems.
To be able to apply Corollary 3.3 to TBox consistency for ALCQI we have to be able to characterise the torus Z 2 n Z 2 n with a TBox of polynomial size.
To characterize this torus we will use the 2 n atomic concepts X 0 ; : : : ; X n 1 and Y 0 ; : : : ; Y n 1 ; these will be used to code the horizontal and vertical position of an element in the torus as follows: For a interpretation I and an element a 2 I we de ne pos(a) by pos(a) := (xpos(a); ypos(a We use a well-known characterisation of binary addition BGG97] to relate the positions of the elements in the torus: Lemma 3.4 Let x; x 0 be a natural numbers with binary representations Then it holds that: where the empty conjunction and disjunction are interpreted as true and false respectively.
We de ne the TBox T n to consist of the following cardinality restrictions: where we use the following abbreviations: (8 C) is an abbreviation for the cardinality restriction ( 0 :C).> stands for an arbitrary concept that is satis ed in all interpretations like A t :A.
The concept C 0 is satis ed by all elements a of the domain for which pos(a) = (0; 0) holds.C 1 is a similar concept, which is satis ed if pos(a) = (2 n 1; 2 n 1): The concept D 1 (D 2 ) enforces that along the role succ 1 (succ 2 ) the value of xpos (ypos) increases by one while the value of ypos (xpos) stays the same.They exactly resemble the formula from Lemma 3.4: The concept D 2 is similar to D 1 where the roles succ 1 and succ 2 as well as the variables X i and Y i have been swapped.
The following lemma is a simple consequence of the de nition of pos and Lemma 3. The TBox T n de nes a torus of exponential size in the following sense: Lemma 3.6 Let T n be the TBox as introduced above.Let I = ( I ; I ) be an interpretation such that I j = T n .This implies where U(2 n ; 2 n ) is the torus Z 2 n Z 2 n and S 1 ; S 2 are the rst (resp.second) successor relation on the torus.Proof.This lemma is established by showing that the function pos is an isomorphism from I to U(2 n ; 2 n ).
Firstly we will prove that pos is injective.We do this by induction on the \Manhattan distance" of the pos-value of the elements to then pos-value of the upper right corner.We will show by induction that for each (x; y) 2 U(2 n ; 2 n ) there is an element a 2 I such that pos(a) = (x; y).d 0 (x; y) = 0 implies x = y = 0. Since I j = ( 1 C 0 ) there is an element a 2 I such that f(a) = (0; 0).Now consider (x; y) 2 U(2 n ; 2 n ) with d 0 (x; y) > 0. Without loss of generality we assume x > 0 (if x = 0 then y > 0 must hold).Hence (x 1; y) 2 U(2 n ; 2 n ) and d 0 (x 1; y) < d 0 (x; y).From the induction hypothesis it follows that there is an element a 2 I such that pos(a) = (x 1; y).Then there must be an element a 1 such that (a; a 1 ) 2 succ I 1 and from Lemma 3.5 it follows that pos(a 1 ) = (x; y).Hence f is also surjective.
Finally we have to show that (a; b) 2 succ I i i (pos(a); pos(b)) 2 S i .This is an immediate consequence of Lemma 3.5.
It is interesting to note that we need inverse roles only to guarantee that pos is injective.The same can be achieved by adding the cardinality restriction ( (2 n 2 n ) >) to T n , from which the injectivity of pos follows from its surjectivity and simple cardinality considerations.Of course the size of this cardinality restriction would only be polynomial in n if we allow binary coding of numbers.
Also note that we have made explicit use of the special expressive power of cardinality restrictions by stating that in any model of T n the extension of C 1 must have at most one element.This can not be expressed with a TBox consisting of axioms.Obviously the following holds: Lemma 3.7 The size jT n j (i.e., the number of symbols necessary to write down T n ) is quadratic in n.

Reducing Domino Problems to TBox Consistency
Once Lemma 3.6 has been proved it is easy to reduce the bounded domino problem to TBox consistency.We use the standard reduction that, has been applied in the DL context e.g. in BS99].
Lemma 3.8 Let D = (D; V; H) be a domino system.Let w = w 0 ; : : : ; w n 1 2 D .There is a TBox T(n; D; w) such that: T(n; D; w) is consistent i D tiles U(2 n ; 2 n ) with initial condition w. jT(n; D; w)j = O(n 2 ) and T(n; D; w) can be computed in time polynomial in n.
Proof.We de ne T(n; D; where pos = (x; y) is a concept which is satis ed by an element a i pos(a) = (x; y).
From the de nition of T(n; D; w) and Theorem 3.6 it follows that each model of T(n; D; w) immediately induces a tiling of U(2 n ; 2 n ) and vice versa.Also, for a xed domino system D the size of T(n; D; w) is obviously polynomial in n.
The nexttheorem is an immediate consequence of Lemma 3.8 and Corollary 3.3: Theorem 3.9 Consistency of ALCQI-TBoxes is NExpTime-hard.
Recalling the note below Lemma 3.8, we see that the same argument also applies to ALCQ if we allow binary coding of numbers.Corollary 3.10 Consistency for ALCQ-TBoxes is NExpTime-hard, if binary coding is used to represent numbers in cardinality restrictions.
Note that for unary coding we needed both inverse roles and cardinality restrictions for the reduction.This is consistent with the fact that satis ability for ALCQI concepts with respect to TBoxes consisting of terminological axioms is still in ExpTime, which can be shown by a reduction to Converse-PDL GM99].This shows that cardinality restrictions on concepts are an additional source of complexity.

Expressiveness of ALCQI
Since reasoning for ALCQI has the same (worst-case) complexity as for C 2 , naturally the question arises how the two logics are related concerning their expressivity.As it turns out, ALCQI is strictly less expressive then C 2 .

A De nition of Expressiveness
There are di erent approaches to de ne the expressivity of DLs Baa96, Bor96, AdR98], but only the one presented in Baa96] is capable of handling TBoxes.We will use a de nition that is equivalent to the one given in Baa96] restricted to a special case.It bases the notion of expressivity on the classes of interpretations de nable by a sentence (or TBox).
De nition 4.1 Let = (N C ; N R ) be a nite signature.A class C of -interpretations is called characterisable by a logic L, if there is a sentence ' C over such that C = fI j I j = ' C g.The class C is called projectively characterisable, if there is a sentence ' 0 C over a signature 0 such that C = fIj j I j = ' 0 C g, where Ij denotes the -reduct of I.
A logic L 1 is called as expressive as another logic L 2 (L 1 L 2 ) if for any nite signature any L 2 -characterisable class C can be projectively characterised in L 1 .
Since C 2 is usually restricted to a relational signature with relation symbols of arity at most two, this de nition is appropriate to relate the expressiveness of ALCQI and C 2 .First of all it is worth noting that clearly ALCQI is strictly more expressive then ALCQ, because ALCQ has the nite model property BBH96], while the following is an ALCQI TBox that has no nite models: The rst cardinality restriction requires an outgoing R-edge for every element of a model and thus each R-path in the model in in nite.The second and third restriction require the existence of an R-path in the model that contains no cycle, which implies the existence of in nitely many elements in the model.Since ALCQ has the nite model property, the class C inf := fI j I j = T inf g, which contains only models with in nitely many elements, can not be projectively characterised by an ALCQ-TBox.
The translation from ALCQI-TBoxes to C 2 sentences given in Fig. 1 not only preserves satis ability, but the translation also has exactly the same models as the initial TBox.This implies that ALCQI C 2 .

A Game for ALCQI
Usually, the separation of two logics with respect to their expressivity is a hard task and not as easily accomplished as we have just done with ALCQ and ALCQI.Even for logics of very restricted expressivity, proofs of separation results may become involved and complex Baa96] and usually require a detailed analysis of the classes of models a logic is able to characterise.A valuable tool for this analysis are Ehrenfeucht-Fra ss e games.In this section we present an Two interpretations I and J are called n-m-equivalent (I n m J ) i for all TBoxes T in L n m it holds that I j = T i J j = T. Similarly, for x 2 I and y 2 J we say that I; x and J ; y are n-m-equivalent (I; x n m J ; y) i for all C 2 C n m it holds that x 2 C I i y 2 C J .Lemma 4.3 There are only nitely many pairwise inequivalent de nitions in C n m .
Proof.Since we chose N C , the set of atomic concepts, to be nite, the claim obviously holds for C n 0 , since this set only contains concept de nitions which are Boolean combinations of atomic concepts.Since we only have nitely many roles and an upper bound on the numbers which can appear in number restrictions we only get nitely many pairwise inequivalent concept de nitions for all m > 0.
Corollary 4.4 For all structures I; J and each x 2 I there is a concept de nition C x such that y 2 C J x i I; x n m J ; y.We will now de ne an Ehrenfeucht-Fraisse game for ALCQI to capture the expressivity of concepts in the classes C n m : The game is played by two players.Player I is called the spoiler while Player II is called the duplicator.The spoiler's aim is to prove two structures not to be n-m-equivalent, while Player II tries to prove the contrary.The game consists of a number of rounds in which the players move pebbles on the elements of the two structures.
De nition 4.5 Let be a nonempty set.Let x be an element of and X a subset of .For any binary relation R we write xRX to denote the fact that, for all x 0 2 X it holds that (x; x 0 ) 2 R.
For the set N R of role names let N R be the union of N R and fR 1 j R 2 N R g.
Let I and J be two interpretations.Two elements x 2 I and y 2 J are called locally equivalent (A; x l B; y), i for all A 2 N C : x 2 A I i y 2 A J .
A con guration captures the state of a game in progress.It is of the form G n m (I; x; J ; y), where n is an arbitrary natural number, m denotes the number of moves which still have to be played, and x and y are the elements of I resp.
J on which the pebbles are placed.
For the con guration G n m (I; x; J ; y) the rules are as follows: 1.If I; x 6 l J ; y, then Player II loses; if m = 0 and I; x l J ; y, then Player II wins.
2. If m > 0, then Player I selects one of the interpretations; assume this is I (the case J is handled dually).He then picks a role S 2 N R and a number l n.He picks a set X I such that xS I X and ]X = l.The duplicator has to answer with a set Y J with yS J Y and ]Y = l.If there is no such set, then she loses.
3. If Player II was able to pick such a set Y , then Player I picks an element y 0 2 Y .Player II has to answer with an element x 0 2 X. 4. The game progresses with G n m 1 (I; x 0 ; J ; y 0 ).
We say that Player II has a winning strategy for G n m (I; x; J ; y) i she can always reach a winning position no matter which moves Player I plays.We write I; x = n m J ; y to denote this fact.Theorem 4.6 For two structures I; J and two elements x 2 I ; y 2 J it holds that I; x = n+1 m J ; y i I; x n m J ; y.
Proof.We proof this theorem inductively over m.Concept de nitions of role depth 0 can only be (Boolean combinations of) atomic concepts.The claim follows immediately from the de nition of local equivalence and the winning condition.Now let m be greater then 0. We rstly show the implication from left to right.Let I; x = n+1 m J ; y hold.We show x 2 C I i y 2 C J for all concept de nitions C 2 C n m by induction over the structure of C. The only interesting cases are when C = ( l R C 0 ) or C = ( l R C 0 ) with l n, R 2 N R , and C 0 2 C n m 1 .Firstly we consider the case C = ( l R C).
Assume that x 2 ( l R C 0 ) I .Then there must be a set X I with ]X = l, xR I X, and X (C 0 ) I .If player I chooses this set, then player II can answer with an according set Y J which has the properties required by the rules of the game (]Y = l and yR J Y ).Player I can now pick an arbitrary element y 0 2 Y and player II can answer with an appropriate element x 0 2 X.The game progresses with G n+1 m 1 (I; x 0 ; J ; y 0 ) for which player II still has a winning strategy.From the induction hypothesis it follows that I; x 0 n m 1 J ; y 0 , and hence Y (C 0 ) J must hold.For symmetry reasons it holds that y 2 C J implies x 2 C I .
The case C = ( l R C) can be reduced to the above case.For this we note that ( l R C) :( (l + 1) R C).Since we have considered a game where the spoiler was allowed to pick sets of size at most n + 1 and the proof only required a set of size at most n we can use this reduction and immediately obtain a proof for the latter case.
To show the implication from right to left we show its contraposition.Let I; x 6 = n+1 m J ; y hold.Hence there must be a move of player I, that player II cannot answer appropriately.Let this be the set X I , reachable by role R 2 N R with ]X = l; l n + 1, and xR I X.Player II cannot answer with an appropriate set.If player II cannot even pick a reachable set of the needed size, we have that y 2 ( (l 1) R (C t :C)) J for any C 2 C, while obviously x 6 2 ( (l 1) R (C t :C)) I .Hence it follows that I; x 6 n m J ; y.For any set Y J player II can pick, which satis es ]Y = l and yR J Y there must be an y 0 2 Y which player I can pick such that, for all x 0 2 X player II has no winning strategy for the game G n+1 m 1 (I; x 0 ; J ; y 0 ).Using the induction hypothesis this implies that I; x 0 6 n m 1 J ; y 0 .We use this fact to construct a concept de nition D 2 C n m which distinguishes x and y in their respective structures.We simply use D = ( (l 1) R F fC x 0 j x 0 2 Xg).Claim 1: x 6 2 D I .Proof of claim 1: For all x 0 2 I we have x 0 2 C I x 0 .This yields X F fC x 0 j x 0 2 Xg.Since xR I X and ]X = l claim 1 follows.
Proof of claim 2: As stated above, in any set Y J with yR J Y and ]Y = l there is an y 0 2 Y which is not n-(m 1)-equivalent to any x 0 2 X.
Hence there are at most l 1 elements in J which are reachable from y via the relation R J which are n-(m 1)-equivalent to an element of X.This yields y 2 D J .
Hence D is a concept which distinguishes the two elements in their structures.Since D 2 C n m it follows that I; x 6 n m J ; y.
The game as it has been presented so far is suitable only if we have already placed pebbles on the interpretations.To get a game that characterises n m as a relation between interpretations we have to introduce an additional rule that governs the placement of the rst pebbles.Since a TBox consists of cardinality restrictions which solely talk about concept membership, we introduce an unconstrained set move as the rst move of the game G n m (I; J ). De nition 4.7 For two interpretations I; J the game G n m (I; J ) is played as follows: 1. Player I picks one of the structures; assume he picks I (the case J is handled dually).He then picks a set X I with ]X = l where l n.Player II must pick a set Y J of equal size.If this is impossible then she loses.
2. Player I picks a element y 2 Y , Player II must answer with an x 2 X. 3. The game progresses with G n m (I; x; I; y).Again we say that Player II has a winning strategy for G n m (I; J ) i she can always reach a winning positions no matter which moves Player I chooses.We write I = n m J do denote this fact.
From Theorem 4.6 is can easily be derived that we have a similar statement for the game G n m (I; J ): Theorem 4.8 For two structures I; J it holds that I n m J i I = n+1 m J .Proof.Firstly we show the implication from right to left.Let I = n+1 m J hold.Let T 2 L n m a TBox with I j = T. Let C 2 T be a value restriction.From I j = T is follows that I j = C.We need to show that this implies J j = C.We distinguish two cases: C = ( l C) with l n and rd(C) m.If player I choses a set X I with X C I and ]X = l (such a set exists because I j = C) then player II can answer with an appropriate set Y J .From the rules of the game and the de nition of the winning condition we have that for each y 2 Y there must be an x 2 X such that player two has a winning strategy for the game G n+1 m (I; x; J ; y).This implies I; x n+1 m J ; y.It follows that Y C J and hence we have J j = C.For symmetry reasons we have also that J j = C implies I j = C. C = ( l C) with l n and rd(C) m.Again we can reduce this case to the above one by noting that I j = ( l C) if and only if I j = :( (l+1) C).
Since we chose C to be an arbitrary element of T we have I j = T if and only if J j = T.This concludes the proof of the rst implication.
The to show the implication from left to right we show the contraposition.
Let I 6 = n+1 m hold.If player II has no winning strategy for G n+1 m (I; J ) there must be a move of player I, she cannot answer.Without loss of generality let this be X I with ]X = l; l n + 1.If player II cannot even pick a set of size l we have that I 6 j = ( (l 1) (At:A)) while J j = ( (l 1) (At:A)) and hence I 6 n m J .If there is a set Y J with ]Y = l there must be a y 2 Y player I can pick such that for all x 2 X player II has no winning strategy for G n+1 m (I; x; J ; y).This implies I; x 6 n m J ; y for one y and all x 2 X.Again we use this fact to construct a cardinality restriction C 2 L n m which distinguishes the two structures.This is built in a similar fashion as in the previous prove.We choose C = ( (l 1) G fC x j x 2 Xg): Claim 1: I 6 j = C.
Proof of claim 1: Since X ( F fC x j x 2 Xg) I and ]X = l this follows immediately by construction.
Proof of claim 2: In any set Y J with ]Y = l there is at least one element y which is not n-m-equivalent to any x 2 X.Hence there are at most l 1 elements in Y which are n-m-equivalent to an element of X.This yields J j = C. Since C 2 L n m we have a TBox which distiguishes the two structures and hence we have I 6 n m J .This concludes the proof.

The Main Result
We will now use this game-theoretic characterisation of the expressivity of ALCQI to proof the main theorem of this section.Even though we have introduced the powerful tool of Ehrenfeucht-Fra ss e games, the proof is still rather complicated.This is mainly due to the fact that we use a very general de nition of expressiveness that allows for the introduction of arbitrary additional roleand concept-names into the signature.Theorem 4.9 ALCQI is not as expressive as C 2 .Proof.To prove this Theorem we have to show that there is a class C that is characterisable in C 2 but that cannot be projectively characterised in ALCQI: Claim 1: For an arbitrary R 2 N R the class C re := fI j R I is re exiveg is not projectively characterisable in ALCQI.Obviously, C re is characterisable in C 2 .
Proof of Claim 1: Assume Claim 1 does not hold and that C re is projectively characterised by the TBox T re 2 L n m over an arbitrary (but nite) signature = (N C ; N R ) with R 2 N R .We have derived a contradiction once we have shown that there are two -interpretations A; B such that A 2 C re , B 6 2 C re , but A n m B. In fact, A n m B implies B j = T re and hence B 2 C re , a contradiction.
In particular, C re contains all interpretations A such that R A = f(x; x) j x 2 A g, i.e. interpretations in which R is interpreted as equality.Since C n m contains only nitely many pairwise inequivalent concepts and C re contains interpretations of arbitrary size, there is also such an A such that there are two elements x 1 ; x 2 2 A such that x 1 6 = x 2 and A; x 1 n m A; x 2 .We de ne B from A as follows: B := A ; A B := A A for each A 2 N C ; S B := S A for each S 2 N R n fRg; R B := (R A n f(x 1 ; x 1 ); (x 2 ; x 2 )g) f(x 1 ; x 2 ); (x 2 ; x 1 )g: Since R B is no longer re exive it holds that B 6 2 C re as desired.It remains to be shown that A n m B holds.We prove this by showing that A = n+1 m B holds, which is equivalent to A n m B by Theorem 4.8.Any opening move of Player I can be answered by Player II in a way that leads to the con guration G n+1 m (A; x; B; x), where x depends on the choices of Player I. We have to show that for any con guration of this type, Player II has a winning strategy.Since certainly A; x = n+1 m A; x this follows from Claim 2: Claim 2: For all k m it holds that A; x = n+1 k A; y implies A; x = n+1 k B; y.
Proof of Claim 2: We show Claim 2 by induction over k.Denote Player II's strategy for the con guration G n+1 k (A; x; A; y) by S.
For the case k = 0, Claim 2 follows immediately from the construction of B: A; x = n+1 0 A; y implies A; x l A; y and A; y l B; y since B agrees with A on the interpretation of all atomic concepts.It follows that A; x l B; y which means that Player II wins the game G n+1 0 (A; x; B; y).For 0 < k m assume that Player I selects an arbitrary structure and a legal subset of the respective domain.Player II tries to answer that move according to S which provides her with a move on the interpretation A since S is a winning strategy for the game G n+1 k (A; x; A; y).There are two possibilities: The move provided by S is a valid move also for the game G n+1 k (A; x; B; y): Player II can answer the choice of Player I according to S without violating the rules, which yields a con guration G n+1 k 1 (A; x 0 ; B; y 0 ) such that for x 0 ; y 0 it holds that A; x 0 = n+1 k 1 A; y 0 (because Player II moved according to S).From the induction hypothesis it follows that A; x 0 = n+1 k 1 B; y 0 .The move provided by S is not a valid move for the game G n+1 k (A; x; B; y) This requires more detailed analysis: Assume Player I has chosen to move in A and has chosen an S 2 N R and a set X of size l n + 1 such that xS A X. Let Y be the set that Player II would choose according S.This implies that Y has also l elements and that yS A Y .That this choice is not valid in the game G n+1 k (A; x; B; y) implies that there is an element z 2 Y such that (y; z) 6 2 S B .This implies y 2 fx 1 ; x 2 g and S 2 fR; R 1 g, because these are the only elements and relations that are di erent in A and B. W.l.o.g.assume y = x 1 and S = R. Then also z = x 1 must hold, because this is the only element such that (x 1 ; z) 2 R A and (x 1 ; z) 6 2 R B .Thus, the choice Y 0 := (Y nfx 1 g) fx 2 g is a valid one for Player II in the game G n+1 m (A; x; B; y): x 1 R B Y 0 and jY 0 j = l because (x 1 ; x 2 ) 6 2 R A .There are two possibilities for Player I to choose an element y 0 2 Y 0 : 1. y 0 6 = x 2 : Player II chooses x 0 2 X according to S. This yields a con guration G n+1 k 1 (A; x 0 ; B; y 0 ) such that A; x 0 = n+1 k 1 A; y 0 .2. y 0 = x 2 : Player II answers with the x 0 2 X that is the answer to the move x 1 of Player I according to S. For the obtained con guration G n+1 k 1 (A; x 0 ; B; y 0 ) it also holds that A; x 0 = n+1 k 1 A; y 0 : By the choice of x 1 ; x 2 it holds that A; x 1 n m A; x 2 and since k 1 < m also A; x 1 n k 1 A; x 2 holds which implies A; x 1 = n+1 k 1 A; x 2 by Theorem 4.8.Since Player II chose x 0 according to S it holds that A; x 0 = n+1 k 1 A; x 1 and hence A; x 0 = n+1 k 1 A; x 2 since = n+1 k 1 is transitive.
In both cases we can apply the induction hypothesis; this yields A; x 0 = n+1 k 1 B; y 0 and hence Player II has a winning strategy for G n+1 k (A; x; B; y).The case that Player I chooses from B instead of A can be handled similarly.
By adding constructs to ALCQI that allow to form more complex role expressions one can obtain a DL that has the same expressive power as C 2 , such a DL is presented in Bor96].It has the ability to express a universal role that makes it possible to internalise both TBoxes based on terminological axioms and cardinality restrictions on concepts.

Conclusion
We have shown that with a rather limited set of constructors one can de ne a DL whose reasoning problems as as hard as that of C 2 without reaching the expressive power of the latter.This shows that cardinality restrictions, although interesting for knowledge representation, are inherently hard to handle algorithmically.At a rst glance, this makes ALCQI with cardinality restrictions on concepts obsolete for knowledge representation, because C 2 delivers more expressive power at the same computational price.Yet, is is likely that a dedicated algorithm for ALCQI will have better average complexity than the C 2 algorithm; such an algorithm has yet to be developed.An interesting question lies in the coding of numbers: If we allow binary coding of numbers, the translation approach together with the result from PST97] leads to a 2-NExpTime algorithm.It is an open question whether this additional exponential blow-up is necessary.A positive answer would settle the same question for C 2 while a proof of the negative answer might give hints how the result for C 2 might be improved.

For
an element a 2 I we de ne d(a) by: d(a) = (2 n 1 xpos(a)) + (2 n 1 ypos(a)) Note that pos(a) = pos(b) implies d(a) = d(b).There is at most one element a 2 I such that d(a) = 0, otherwise I 6 j = ( 1 C 1 ).Hence there is exactly one element a such that pos(a) = (2 n 1; 2 n 1).Now assume there are elements a; b 2 I such that pos(a) = pos(b) and d(a) = d(b) > 0. Hence either xpos(a) < 2 n 1 or ypos(a) < 2 n 1.Without loss of generality we may assume xpos(a) < 2 n 1 .From I j = T n it follows that a; b 2 (9succ 1 :>) I .Let a 1 ; b 1 be elements such that (a; a 1 ) 2 succ I 1 and (b; b 1 ) 2 succ I 1 .From Lemma 3.5 it follows that xpos(a 1 ) xpos(b 1 ) xpos(a) + 1 (mod 2 n ) ypos(a 1 ) = ypos(b 1 ) = ypos(a) Since d(a 1 ) = d(b 1 ) < d(a) we can use the induction hypothesis which yields that a 1 = b 1 because pos(a 1 ) = pos(b 1 ).This also implies a = b, because a 1 2 (= 1 succ 1 1 :>) I and f(a; a 1 ); (b; a 1 )g succ I use the distance from the lower left corner.For each element (x; y) 2 U(2 n ; 2 n ) we de ne: d 0 (x; y) = x + y w) := T n T D T w , where T n is de ned as above, T D captures the vertical and horizontal compatibility constraints of the domino systems D, and T w enforces the initial condition.We use an atomic concept C d for each tile d 2 D. T D consists of the following cardinality restrictions: of the cardinality restrictions (8 (pos = (0; 0) !C w 0 )); : : : ; (8 (pos = (n 1; 0) !C w n 1 )

Ehrenfeucht-
Fra ss e game that exactly captures the expressivity of ALCQI: De nition 4.2 For an ALCQI concept C, the role depth rd(C) counts the maximum number of nested cardinality restrictions.Formally we de ne rd as follows: rd(A) := 0 for A 2 N C rd(:C) := rd(C) rd(C 1 u C 2 ) := maxfrd(C 1 ); rd(C 2 )g rd( n R C) := 1 + rd(C) The set C n m is de ned to consist of exactly those ALCQI concepts that have a role depth of at most m, and in which the numbers appearing in number restrictions are bounded by n; the set L n m is de ned to consist of all ALCQI-TBoxes T that contain only cardinality restriction of the form (./ k C) with k n and C 2 C n m .
2 C I g, where C n m only contains one de nition from each equivalence class in C n m .