ALESSANDRO ARTALE and VLADISLAV RYZHIKOV, KRDB Research Centre, Free University of Bozen-Bolzano, Italy
ROMAN KONTCHAKOV and MICHAEL ZAKHARYASCHEV, Department of Computer Science and Information Systems, Birkbeck, University of London, U.K.
B. PROOF OF THEOREM 4.1
THEOREM 4.1. ATU SDL-LiteNboolKBK = (T,A)is satisfiable iff theQT L 1
sentence K† is satisfiable.
PROOF. (⇐) LetMbe a first-order temporal model with acountabledomainDandM,0|=K†. Without loss of generality we may assume that theaM, fora∈ob
A, are all distinct. We are going to construct aTU SDL-LiteNboolinterpretationIsatisfyingKand based on some domain∆Ithat will
be inductively defined as the union
∆I =[
m≥0∆m, where ∆0=
aM|a∈obA ⊆D and ∆m⊆∆m+1, form≥0.
The interpretations of object names inI are given by their interpretations inM:aI =aM ∈∆
0.
Each set∆m+1, form≥0, is constructed by adding to∆msome new elements that are freshcopies
of certain elements fromD\∆0. If such a new elementuis a copy ofu0 ∈D\∆0then we write
cp(u) =u0, while foru∈∆0we letcp(u) =u.
The interpretationAI(n)of each concept nameAinIis defined by taking
AI(n)=
u∈∆I|M, n|=A∗[cp(u)] . (72) The interpretationSI(n)of each role nameSinIis constructed inductively as the union
SI(n)=[
m≥0S n,m
, whereSn,m⊆∆m×∆m, for allm≥0.
We require the following two definitions to guide our construction. TherequiredR-rank%R,nd of
d∈Dat momentnis
%R,nd = max {0} ∪ {q∈QT |M, n|=EqR[d]}.
By (5),%R,nd is a function and if%R,nd =qthenM, n|=Eq0R[d]for everyq0 ∈QT withq0 ≤q, and(M, n)|=¬Eq0R[d]for everyq0∈QT withq0 > q. We also define theactualR-rankτu,mR,nof
u∈∆Iat momentnand stepmby taking
τu,mR,n= max {0} ∪ {q∈QT |(u, u1), . . . ,(u, uq)∈Rn,mfor distinctu1, . . . , uq∈∆I}
,
whereRn,misSn,mifR=Sand{(u0, u)|(u, u0)∈Sn,m}ifR=S−, for a role nameS. For the basis of induction, for each role nameS, we set
Sn,0=
(aI, bI)∈∆0×∆0|S(a, b)∈ ASn , forn∈Z (73)
(note thatS(a, b)∈ AS
n for alln∈ZifkS(a, b)∈ A, for a rigid role nameS). It follows from
the definition ofA†that, for allR∈role
Kandu∈∆0,
τu,0R,n ≤ %R,ncp(u). (74)
c
YYYY ACM 1529-3785/YYYY/01-ARTA $15.00 DOI:http://dx.doi.org/10.1145/0000000.0000000
Suppose that∆mand theSn,mhave been defined for somem≥0. If, for all rolesRandu∈∆m,
we hadτR,n u,m = %
R,n
cp(u)then the interpretation of roles would have been constructed. However, in
general this is not the case because there may be some ‘defects’ in the sense that the actual rank of some elements is smaller than the required rank. Consider the following two sets of defects in
Sn,m: Λn,mR = u∈∆m\∆m−1|τu,mR,n< % R,n cp(u) , forR∈ {S, S −}
(for convenience, we assume∆−1 =∅). The purpose of, say,Λn,mS is to identify those ‘defective’
elementsu∈∆m\∆m−1from which precisely% S,n
cp(u)distinctS-arrows should start (according to M), but some arrows are still missing (onlyτS,n
u,marrows exist). To ‘repair’ these defects, we extend
∆mto∆m+1andSn,mtoSn,m+1according to the following rules:
(Λn,mS ) Letu ∈ ΛSn,m. Denoted = cp(u)andq = %S,ncp(u)−τu,mS,n. ThenM, n |= Eq0S[d] for someq0 ≥ q >0. By (5), we haveM, n |= E1S[d]and, by (7), there isd0 ∈ D such that M, n|=E1S−[d0]. In this case we takeqfreshcopiesu01, . . . , u0qofd0(socp(u0i) =d0), add
them to∆m+1and add the pairs(u, u01), . . . ,(u, u0q)toSn,m+1. If Sis rigid we add these
pairs to allSk,m+1, fork∈ Z. (Λn,mS− ) Letu∈ Λ n,m S− . Denoted=cp(u)andq=% S−,n cp(u)−τ S−,n
u,m . ThenM, n|=Eq0S−[d]for
q0 ≥q > 0. By (5),M, n|=E1S−[d]and, by (7), there isd0 ∈D withM, n |=E1S[d0].
In this case we takeqfresh copiesu01, . . . , u0q ofd0, add them to ∆m+1 and add the pairs
(u01, u), . . . ,(u0q, u)toSn,m+1. IfSis rigid we add these pairs to allSk,m+1, fork∈Z.
Now we observe the following property of the construction: for allm0≥0andu∈∆m0\∆m0−1,
τu,mR,n= 0, ifm < m0,
q, ifm=m0, for someq≤%R,ncp(u),
%R,ncp(u), ifm > m0.
(75)
To prove this property, consider all possible cases. Ifm < m0thenu /∈ ∆m, i.e., it has not been
added to∆myet, and soτu,mR,n = 0. Ifm=m0 = 0thenτu,mR,n ≤% R,n
cp(u)by (74). Ifm=m0 >0
then u was added at stepm0 to repair a defect of some u0 ∈ ∆m0−1. This means that either
(u0, u) ∈ Sn,m0 andu0 ∈ Λn,m0−1
S , or(u, u0) ∈ S
n,m0 andu0 ∈ Λn,m0−1
S− , for a role nameS. Consider the first case. Since freshwitnesses uare picked up every time the rule (Λn,mS 0−1)is applied and those witnesses satisfyM, n |= E1S−[cp(u)], we obtainτu,mS,n0 = 0,τ
S−,n
u,m0 = 1and
%Scp−(u),n≥1. The second case is similar. Ifm=m0+1then all defects ofuare repaired at stepm0+1
by applying the rules(Λn,m0
S )and(Λ n,m0 S− ). Therefore,τ R,n u,m0 =% R,n cp(u). Ifm > m0+ 1then (75)
follows from the observation that no new arrows involvingucan be added after stepm0+ 1.
It follows that, for allR∈roleK,q∈QT,n∈Zandu∈∆I,
M, n|=EqR[cp(u)] iff u∈(≥q R)I(n). (76)
Indeed, if M, n |= EqR[cp(u)] then, by definition,%cpR,n(u) ≥ q. Let u ∈ ∆m0 \∆m0−1. Then, by (75),τu,mR,n = %
R,n
cp(u) ≥ q, for all m > m0. It follows from the definition ofτ R,n
u,m andRI(n)
that u ∈ (≥q R)I(n). Conversely, letu ∈ (≥q R)I(n) andu ∈ ∆
m0 \∆m0−1. Then, by (75), we haveq ≤ τu,mR,n = %
R,n
cp(u), for allm > m0. So, by the definition of% R,n
cp(u) and (5), we obtain M, n|=EqR[cp(u)].
Now we show by induction on the construction of conceptsCinKthat
M, n|=C∗[cp(u)] iff u∈CI(n), for alln∈Zandu∈∆I.
The basis of induction is trivial forC=⊥and follows from (72) ifC=Aiand (76) ifC=≥q R.
The induction step for the Booleans (C = ¬C1 andC = C1uC2) and the temporal operators
(C=C1UC2andC=C1SC2) follows from the induction hypothesis. Thus,I |=T.
It only remains to show that I |= A. IfnA(a) ∈ A then, by the definition ofA† and (72),
I |= nA(a). If n¬A(a) ∈ A then, analogously,I |= n¬A(a). If nS(a, b) ∈ Athen, by (73),(aI, bI)∈Sn,0, whence, by the definition ofSI(n),I |=nS(a, b). Ifn¬S(a, b)∈ A then, by (73),(aI, bI)∈/ Sn,0, and so, as no new arrows can be added between ABox individuals,
I |=n¬S(a, b).
(⇒)is straightforward.
C. PROOF OF THEOREM 4.5
THEOREM 4.5. The satisfiability problem for the core fragment of TU SDL-LiteNbool KBs is
PSPACE-complete.
PROOF. The proof is by reduction of the halting problem for Turing machines with a polyno- mial tape. We recall that, given a deterministic Turing machineM =hQ,Γ,#,Σ, δ, q0, qfiand a
polynomials(n), we construct a TBoxTM containing concept inclusions (8)–(13), which we list
here for the reader’s convenience:
Hiqv ⊥ UH(i+1)q0, Hiqv ⊥ USia0, ifδ(q, a) = (q0, a0, R)andi < s(n), (8) Hiqv ⊥ UH(i−1)q0, Hiqv ⊥ USia0, ifδ(q, a) = (q0, a0, L)andi >1, (9) Hiqv ⊥ UDi, (10) DiuDjv ⊥, ifi6=j, (11) SiavSiaUDi, (12) Hiqf v ⊥. (13)
For an input~a=a1. . . anof lengthn, we take the following ABoxA~a:
H1q0(d), Siai(d), for1≤i≤n, Si#(d), forn < i≤s(n).
We show that(TM,A~a)is unsatisfiable iffMaccepts~a. We represent configurations ofMas tuples
of the formc=hb1. . . bs(n), i, qi, whereb1. . . bs(n)is the contents of the firsts(n)cells of the tape
withbj ∈Γ, for allj, the head position isi,1≤i≤s(n), andq∈Qis the control state. LetIbe
an interpretation forKM,~a. We say thatIencodes configurationc=hb1. . . bs(n), i, qiat momentk
ifdI∈HiqI(k)anddI ∈SjbI(k)
j , for all1≤j≤s(n). We note here that, in principle, many different configurations can be encoded at momentkinI. Nevertheless, any prefix of a model of(TM,A~a)
contains the computation ofM on the given input~a:
LEMMA C.1. Letc0, . . . ,cmbe a sequence of configurations representing a partial computa- tion ofMon~a. Then every modelIof(TM,Aa~)encodesckat momentk, for0≤k≤m.
PROOF. The proof is by induction onk. For k = 0, the claim follows fromI |= A~a. For
the induction step, let I encode ck = hb1. . . bi. . . bs(n), i, qi at moment k, and let ck+1 be
hb1. . . b0i. . . bs(n), i0, q0i. Then we haveq∈Q\ {qf}. Consider firstδ(q, bi) = (q0, b0i, L), in which
casei > 1 and i0 = i−1. SincedI ∈ HiqI(k) we have, by (10), dI ∈ DiI(k+1) and, by (9),
dI ∈ H(iI(k+1)−1)q0 anddI ∈ S I(k+1) ib0
i . Consider cell
j,1 ≤ j ≤ s(n), such that j 6= i. By (11),
dI∈/DIj(k+1), and so, sincedI∈SjbI(k)
j , we obtain, by (12),d
I∈SI(k+1)
jbj . Hence,Iencodesck+1 at momentk+ 1. The case ofδ(q, bi) = (q0, b0i, R)is analogous.
It follows that if M accepts~athen(TM,A~a)is unsatisfiable. Indeed, ifM accepts~athen the
computation is a sequence of configurationsc0, . . . ,cmsuch thatcm=hb1. . . bs(n), i, qfi. Suppose
(TM,A~a)is satisfied in a modelI. By Lemma C.1,dI∈H
I(m)
iqf , which contradicts (13).
Conversely, if M rejects~athen(TM,A~a)is satisfiable. Letc0, . . . ,cm be a sequence of con-
figurations representing the rejecting computation of M on~a,ck = hb1,k, . . . , bs(n),k, ik, qki, for
0 ≤k≤m. We define an interpretationIwith∆I ={d},dI =dand, for everya∈Γ,q ∈Q,
1≤i≤s(n)andk≥0, we set (note thatqmis a rejecting state and so,δ(qf, a)is undefined):
HiqI(k)= ∆I, if k≤m, i=i kandq=qk, ∅, otherwise, SiaI(k)= ∆I, if k≤manda=bi,k, ∆I, if k=m+ 1anda=bi,m, ∅, otherwise, DIi(k)= ∆I, if 0< k≤m+ 1andi=ik−1, ∆I, if k=m+ 2 ∅, otherwise.
It can be easily verified thatI |= (TM,A~a).
D. PROOF OF THEOREM 6.3
LEMMA6.4. LetKbe aTU∗DL-Lite
N
boolKB andqK = max(QT ∪QA) + 1. IfKis satisfiable then it can be satisfied in an interpretationIsuch that(≥qK2∗R)I=∅, for eachR∈roleK.
PROOF. LetI |=K. Without loss of generality, we will assume that the domain∆I is at most countable. Construct a new interpretationI∗as follows. We take∆I×
Nas the domain ofI∗and
setaI∗= (aI,0), for alla∈obA. For eachn∈Z, we set
AI∗(n)={(u, i)|u∈AI(n), i∈N}, for every concept nameA,
SI∗(n)={((u, i),(v, i))|(u, v)∈SI(n), i∈N}, for every role nameS.
It should be clear thatI∗|=K.
Suppose thatu∈∆Ihas at leastqK-many2∗R-successors inIand assume that the pairs
(u, i),(u1, i) , . . . , (u, i),(uqK−1, i) , (u, i),(uqK, i) , . . .
are all in(2∗R)I∗. We can also assume that if(uj,0) =aI∗, for somea∈obA, thenj < qK. We
then rearrange some of theR-arrows of the form((u, i),(uj, i0)), simultaneouslyat all moments of
time, in the following manner. We remove((u, i),(uj, i))from(2∗R)I
∗
, for alljandisuch that
j ≥qKori >0. Note that this operation does not affect the2∗R-arrows to the ABox individuals. To preserve the extension of concepts of the form≥q2∗R, we then add new2∗R-arrows of the form
((u, i),(uj, i0)), fori > i0 ≥0, to(2∗R)I
∗
in such a way that the following conditions are satisfied: – for every(uj, i0), there is precisely one2∗R-arrow of the form((u, i),(uj, i0)),
– for every(u, i), there are precisely(qK−1)-many2∗R-arrows of the form((u, i),(uj, i0)).
Such a rearrangement is possible becauseI∗contains countably infinitely many copies ofI. We leave it to the reader to check that the resulting interpretation is still a model ofK.
The rearrangement process is then repeated for each otheru ∈∆I with at leastqK-many2∗R- successors.