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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>Conjunctive Query Inseparability in OWL 2 QL is ExpTime-hard</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Roman Kontchakov</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Vladislav Ryzhikov</string-name>
          <xref ref-type="aff" rid="aff1">1</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Michael Zakharyaschev</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Department of Computer Science and Information Systems</institution>
          ,
          <addr-line>Birkbeck</addr-line>
          ,
          <institution>University of London</institution>
          ,
          <country country="UK">U.K</country>
        </aff>
        <aff id="aff1">
          <label>1</label>
          <institution>Faculty of Computer Science, Free University of Bozen-Bolzano</institution>
          ,
          <country country="IT">Italy</country>
        </aff>
      </contrib-group>
      <abstract>
        <p>We settle an open question on the complexity of the following problem: given two OWL 2 QL TBoxes and a signature, decide whether these TBoxes return the same answers to any conjunctive query over any data formulated in the given signature. It has been known that the complexity of this problem is between PSpace and ExpTime. Here we show that the problem is ExpTime-complete and, in fact, deciding whether two OWL 2 QL knowledge bases (each with its own data) give the same answers to any conjunctive query is ExpTime-hard.</p>
      </abstract>
      <kwd-group>
        <kwd>Query inseparability</kwd>
        <kwd>OWL 2 QL</kwd>
        <kwd>complexity</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>Introduction</title>
      <p>In this paper, we show that the problem of deciding whether two given OWL 2 QL
ontologies return the same answers to any conjunctive query in a given signature
is ExpTime-hard. The notion of conjunctive query inseparability has recently
been used in two settings: ontology engineering and maintenance, and knowledge
base exchange.</p>
      <p>
        It has been argued [
        <xref ref-type="bibr" rid="ref4 ref5 ref6 ref8">6, 8, 4, 5</xref>
        ] that conjunctive query inseparability is
fundamental for many ontology engineering and maintenance tasks in the context of
ontology-based data access (OBDA). Indeed, suppose we want to query data via
some ontology T , and the set of concepts and roles we are interested in
comprises a signature . From our point of view, any operations with T such as
re ning T , extracting a smaller module, importing a module, updating T to a
new version, etc. should preserve answers to conjunctive queries formulated in
. In other words, all such transformations of T should result in ontologies that
are -query inseparable from T .
      </p>
      <p>
        In the knowledge exchange framework [
        <xref ref-type="bibr" rid="ref2">2</xref>
        ], the problem of inseparability by
the unions of conjunctive queries (UCQ) for two OWL 2 QL knowledge bases
(KBs) appears in the context of computing a universal UCQ-solution. Let 1
and 2 be a source and a target signature, respectively. The two signatures are
disjoint and connected by means of a declarative mapping speci cation, which is
a TBox T12 relating concepts and roles in 1 and 2. A KB K2 = (T2; A2) in the
target signature 2 is called a universal UCQ-solution for a KB K1 = (T1; A1)
in the source signature 1 under T12 if (T1 [ T12; A1) and (T2; A2) give the same
answers to any UCQ formulated in 2. Thus, in this case we are interested in
-query inseparability for OWL 2 QL KBs.
      </p>
      <p>
        The -query inseparability problem for ontologies (TBoxes) in the fragment
of OWL 2 QL without role inclusions proves to be quite simple: it is
NLogSpacecomplete and can be solved in practice using various existing reasoners [
        <xref ref-type="bibr" rid="ref4 ref6">6, 4</xref>
        ]. The
addition to this language conjunctions in the left-hand side of concept inclusions
and (unquali ed) number restrictions makes the problem coNP-complete [
        <xref ref-type="bibr" rid="ref6">6</xref>
        ].
But for the full OWL 2 QL the problem unexpectedly turned out to be much
more challenging and required new logical tools compared to the previously
analysed ontology languages. It was shown [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ] that the interaction between role
inclusions and inverse roles makes this problem PSpace-hard; on the other hand,
the established upper bound was obtained by a reduction to the emptiness
problem for alternating two-way automata, which belongs to ExpTime [
        <xref ref-type="bibr" rid="ref9">9</xref>
        ]. The proof
of PSpace-hardeness mentioned above was also adapted to query inseparability
of OWL 2 QL KBs [
        <xref ref-type="bibr" rid="ref1">1</xref>
        ].
      </p>
      <p>In this paper, we prove, by encoding alternating Turing machines with
polynomial tape, that the -query inseparability problem for both OWL 2 QL KBs
and ontologies is ExpTime-hard.
2</p>
      <p>-Query Entailment and Inseparability
We use the following (somewhat simpli ed) syntax of OWL 2 QL. It contains
individual names ai, concept names Ai, and role names Pi (i 1). Roles R and
basic concepts B are de ned by the grammar:</p>
      <p>R
::=</p>
      <p>Pi
j</p>
      <p>Pi ;</p>
      <p>B
::=
?
j</p>
      <p>Ai
j
9R:
A TBox, T , is a nite set of inclusions of the form</p>
      <p>B1 v B2;</p>
      <p>B1 u B2 v ?;</p>
      <p>R1 v R2;</p>
      <p>
        R1 u R2 v ?:
An ABox, A, is a nite set of atoms of the form Ak(ai) or Pk(ai; aj ). The set
of individual names in A is denoted by ind(A). T and A together form the
knowledge base (KB) K = (T ; A). The semantics for OWL 2 QL is de ned in
the usual way based on interpretations I = ( I ; I ) [
        <xref ref-type="bibr" rid="ref3">3</xref>
        ]. We use S1 S2 as
a shortcut for S1 v S2 and S2 v S1 (for both concepts and roles), and write
I j= to say that an inclusion or assertion is true in I. An interpretation I
is a model of a KB K = (T ; A) if I j= for all 2 T [ A; in this case we write
I j= K. K is consistent if it has a model. A concept B is said to be T -consistent
if (T ; fB(a)g) has a model. K j= means that I j= for all models I of K.
      </p>
      <p>A conjunctive query (CQ) q(x) is a formula 9y '(x; y), where ' is a
conjunction of atoms of the form Ak(z1) or Pk(z1; z2) with zi 2 x [ y. A tuple
a ind(A) is a certain answer to q(x) over K = (T ; A) if I j= q(a) for all
I j= K; in this case we write K j= q(a). If x = ;, the CQ q is called Boolean; a
certain answer to such a q over K is `yes' if K j= q and `no' otherwise.</p>
      <p>To de ne the main notions of this paper, consider two KBs K1 = (T1; A) and
K2 = (T2; A). For example, the Ti are di erent versions of some ontology, or one
of them is a re nement of the other by means of new axioms. The question we
are interested in is whether they give the same answers to CQs formulated in a
certain signature, say, in the common vocabulary of the Ti or in a vocabulary
relevant to an application. More precisely, by a signature, , we understand
any nite set of concept and role names. A concept (inclusion, TBox, etc.) all
concept and role names of which are in is called a -concept (inclusion, etc.).
We say that K1 -query entails K2 if, for all -queries q(x) and all a ind(A),
K2 j= q(a) implies K1 j= q(a). In other words: any certain answer to a -query
given by K2 is also given by K1. We call K1 and K2 -query inseparable if they
-query entail each other, in which case we write K1 K2.</p>
      <p>As the ABox is typically not xed or known at the ontology design stage,
we may have to compare the TBoxes over arbitrary -ABoxes rather than a
xed one, which gives the following de nition. Let T1 and T2 be TBoxes and
a signature. T1 -query entails T2 if (T1; A) -query entails (T2; A) for any
-ABox A. T1 and T2 are -query inseparable if they -query entail each other,
in which case we write T1 T2.</p>
      <p>
        In the remainder of this section, we recap the semantic criteria of -query
entailment established in [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ]. Recall rst that to compute certain answers to any
CQ q over a consistent KB K = (T ; A), it is enough to nd answers to q in the
canonical model CK of K.
      </p>
      <p>Let [R] = fS j T j= R Sg. We write [R] T [S] if T j= R v S; thus, T
is a partial order on the set f[R] j R a role in T g. For each [R], we introduce a
witness w[R] and de ne a generating relation K on the set of these witnesses
together with ind(A) by taking:
a
w[S]</p>
      <p>K w[R] if a 2 ind(A) and [R] is
K 6j= R(a; b) for any b 2 ind(A);</p>
      <p>K w[R] if [R] is T -minimal with T j= 9S</p>
      <p>T -minimal such that K j= 9R(a) and
v 9R and [S ] 6= [R].</p>
      <p>Clearly, K can be computed in polynomial time in jKj. A K-path is a nite
sequence aw[R1] w[Rn], n 0, such that a 2 ind(A) and, if n &gt; 0, then
a K w[R1] and w[Ri] K w[Ri+1], for i &lt; n. Denote by tail( ) the last element
in the path . The canonical model CK of K is de ned by taking CK to be the
set of all K-paths and setting:
aCK = a; for all a 2 ind(A);
ACK = fa 2 ind(A) j K j= A(a)g [</p>
      <p>w[R] j T j= 9R
f
P CK = f(a; b) 2 ind(A) ind(A) j R(a; b) 2 A with [R]
v Ag; for all concept names A;</p>
      <p>T [P ]g [
f( ;
f(</p>
      <p>w[R]) j tail( )
w[R]; ) j tail( )</p>
      <p>K w[R]; [R]
K w[R]; [R]</p>
      <p>T [P ]g [
T [P ]g; for all role names P:
Theorem 1. For all consistent OWL 2 QL KBs K = (T ; A), CQs q(x) and
tuples a ind(A), we have K j= q(a) i CK j= q(a).</p>
      <p>Thus, to decide -query entailment between KBs K1 and K2, it su ces to
check whether CK2 j= q(a) implies CK1 j= q(a) for all -queries q(x) and tuples
a. This relationship between CK2 and CK1 can be characterised semantically in
terms of nite -homomorphisms.</p>
      <p>For an interpretation I and a signature , the -types tI (x) and rI (x; y),
for x; y 2 I , are given by:
tI (x) = f -concept B j x 2 BI g;
rI (x; y) = f -role R j (x; y) 2 RI g:
A -homomorphism from an interpretation I to I0 is a function h :
such that h(aI ) = aI0 , for all individual names a interpreted in I,
I !</p>
      <p>I0
tI (x)
tI0 (h(x))
and
rI (x; y)
rI0 (h(x); h(y));
for all x; y 2</p>
      <p>
        I :
It is known that answers to -CQs are preserved under -homomorphisms.
Thus, if there is a -homomorphism from CK2 to CK1 , then K1 -query entails
K2. However, it was observed [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ] that the converse does not hold in general and
a more subtle notion of homomorphism is required. We say that I is nitely
-homomorphically embeddable into I0 if, for every nite sub-interpretation I1
of I, there exists a -homomorphism from I1 to I0.
      </p>
      <p>
        Theorem 2 ([
        <xref ref-type="bibr" rid="ref4">4</xref>
        ]). Let K1 and K2 be consistent OWL 2 QL KBs. Then K1
query entails K2 i CK2 is nitely -homomorphically embeddable into CK1 .
      </p>
      <p>To obtain a similar criterion for TBoxes (rather than KBs), we use the fact
that inclusions in OWL 2 QL, di erent from disjointness axioms, involve only
one concept or role in the left-hand side and making sure that the TBoxes entail
the same -inclusions, one can show that it is enough to consider singleton
-ABoxes of the form fB(a)g.</p>
      <p>
        Theorem 3 ([
        <xref ref-type="bibr" rid="ref4">4</xref>
        ]). Let T1 and T2 be OWL 2 QL TBoxes. Then T1
tails T2 i the following conditions hold :
-query
en(p) T2 j=
(h) C(T2;fB(a)g) is
all T1-consistent
implies T1 j= , for all -inclusions ;
nitely -homomorphically embeddable into C(T1;fB(a)g), for
      </p>
      <p>-concepts B.</p>
      <p>
        It was shown [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ] that checking -query entailment for OWL 2 QL TBoxes is
PSpace-hard and can be done in ExpTime.
3
      </p>
    </sec>
    <sec id="sec-2">
      <title>ExpTime-hardness</title>
      <p>
        Our main result is the following theorem:
Theorem 4. The -query entailment and
ExpTime-hard for OWL 2 QL KBs.
-query inseparability problems are
Proof. The proof is by encoding alternating Turing machines with polynomial
tape and using the well-known fact that APSpace = ExpTime. For more details
on alternating Turing machines the reader is referred to [
        <xref ref-type="bibr" rid="ref7">7</xref>
        ].
      </p>
      <p>Suppose we are given an alternating Turing machine M = ( ; Q; q0; q1; ),
where is a tape alphabet, Q a set of states partitioned into existential Q9 and
universal Q8 states, q0 2 Q9 an initial state, q1 2 Q an accepting state, and
: (Q n fq1g)
! (Q
f 1; 0; +1g)2
is a transition function that, for any q 2 Q and a 2 , gives two alternative
transitions. We assume that existential and universal states strictly alternate,
that is, any transition from an existential state results in a universal state, and
vice versa. We also assume that M terminates on every input and there is a
polynomial function f such that M uses at most f (m) tape cells on any input
of length m. It will be convenient to extend the transition function by the
instructions (q1; a) = ((q1; a; 0); (q1; a; 0)), for all a 2 , which have an e ect
of going into an in nite loop if M reaches the accepting state. Thus, M accepts
an input w i there is a run of the modi ed machine M 0 on w such that all
branches of the run are in nite.</p>
      <p>Our aim is to construct, given M and an input w, two TBoxes, T1 and T2, such
that M accepts w i the canonical model C2 of K2 = (T2; fA(c)g) is nitely
homomorphically embeddable into the canonical model C1 of K1 = (T1; fA(c)g),
where comprises the concept and role names in T2 (in fact, can be any set
of concept and role names including those from T2).</p>
      <p>In the de nition of the TBox T1 (but not T2), we use inclusions of the form
B v 9R:(C1 u u Ck) as an abbreviation for B v 9R0, R0 v R and 9R0 v Ci,
for 1 i k, where R0 is a fresh role name. If Ci is a complex concept then
9R0 v Ci is also treated as an abbreviation for the respective concept and role
inclusions.</p>
      <p>We de ne T1 and T2 in three steps. Let n = f (jwj).</p>
      <p>Step 1. First we encode con gurations and transitions of M 0 using T1. We
represent a con guration (that is, the contents of every cell on the tape, the
state and the position of the head) by a sequence of (n + 1) R-arrows, for some
roles R, which will be called a block. More precisely, the rst arrow in each block
is auxiliary; it is used to distinguish the type of the block. The range of the
remaining n arrows belongs to one of the concepts Ca, for a 2 . For example,
if the range of the (i + 1)th arrow in the block belongs to Ca then the ith cell of
the tape contains a in the con guration de ned by the block.</p>
      <p>C1</p>
      <p>A</p>
      <p>P</p>
      <p>P</p>
      <p>P
Ca1</p>
      <p>Ca2</p>
      <p>Can 1</p>
      <p>Can
P</p>
      <p>Qqn0;a1;1</p>
      <p>Fig. 1. Encoding a con guration by a P -block.</p>
      <p>The initial block of (n+1)-many P -arrows represents the initial con guration,
that is, symbols a1; : : : ; an written in the n cells of the tape (comprising the input
w in the rst m cells padded with the blanks) and the initial state q0 (see Fig. 1):
A v 9P:9P:(Ca1 u 9P:(Ca2 u 9P:(: : : 9P:(Can u Qqn0;a1;1) : : : ))):
(T1-1)</p>
      <p>The concepts Qjq;a;k, for 0 j n, de ne the remaining components of a
con guration: q 2 Q is the current state and k, 1 k n, is the current position
of the head on the tape. In addition, a 2 speci es the contents of the cell
scanned by the head (this duplicates the information given by Ca but simpli es
the construction). Finally, the index j is required to propagate the current state
and head position along the points encoding the con guration. At the end of the
tape, the concept Qqn;a;k starts a separate branch for each of the two transitions:
we take, for q 2 Q and a 2 with 0(q; a) = ((q1; a1; d1); (q2; a2; d2)), where
d1; d2 2 f+1; 0; 1g, the inclusions</p>
      <p>Qqn;a;k v 9P:(X1 u Bq01;a1;k;d1 ) u 9P:(X2 u Bq02;a2;k;d2 ); for 1
k
n; (T1-2)
where X1 and X2 are two fresh concept names (distinguishing the two branches)
j
and the Bq0;a0;k;d are fresh concepts encoding the next state q0 and the symbol
a0 to be placed to the kth cell in the next con guration with k being the current
position of the head and d indicating the direction in which the head moves.</p>
      <p>For all cells before the head, that is, for all j with j + 1 &lt; k + d and j + 1 &lt; k,
we use the following inclusions to pass the information along the sequences of
arrows encoding the con guration:</p>
      <p>Bq;a;k;d v 9P:(Cb u Bqj;+a1;k;d);
j
(these inclusions, in fact, generate, for each b 2 , a branch in C1 to represent
the same cell but with a di erent symbol, b, tentatively assigned to the cell|we
shall see later how the correct branch and the symbol are selected to match the
cell contents in the preceding con guration). If j +1 = k or j +1 = k +d then the
j
cell contents is changed according to the subscript of Bq;a;k;d and the contents
of the cell scanned by the head is then recorded in the subscript of Qjq;a;k:
Bqk;a;2k; 1 v 9P:(Cb u 9P:(Ca u 9Sa;0 u Qqk;b;k 1));</p>
      <p>Bqk;a;k;0 v 9P:(Ca u 9Sa;0 u Qqk;a;k);</p>
      <p>1
Bqk;a;1k;+1 v 9P:(Ca u 9Sa;0 u 9P:(Cb u Qqk;+b;1k+1));
for b 2 ;
The three situations are depicted in Fig. 2, where the Sa;0 arrows are not shown,
and all arrows are labelled by P . The lled nodes represent the beginning of
the pattern in each of the three inclusions (T1-4){(T1-6) and the subscript of the
concept Ca in the label of the hatched nodes determines the respective subscript
j
of the sequence of roles Qq;a;k that follows the pattern.</p>
      <p>Note that there is only one branch for the cell whose contents is modi ed: the
branch corresponds to the new symbol, a, in that cell (see explanations below).
(a)
(b)
(c)</p>
      <p>Bqk;a;2k; 1</p>
      <p>Bqk;a;2k;0
Bqk;a;2k;+1
cell (k</p>
      <p>1)
Cb
Cb0</p>
      <p>Cb00
Cb; Bqk;a;1k;0</p>
      <p>Cb0 ; Bqk;a;1k;0</p>
      <p>Cb00 ; Bqk;a;1k;0
Cb; Bqk;a;1k;+1</p>
      <p>Cb0 ; Bqk;a;1k;+1
Cb00 ; Bqk;a;1k;+1
Ca; Qqk;b;k 1</p>
      <p>Ca; Qqk;b0;k 1
Ca; Qqk;b00;k 1</p>
      <p>Ca; Qqk;a;k</p>
      <p>Ca; Qqk;a;k
Ca; Qqk;a;k
Ca
Ca; Qqk;a;k
Ca; Qqk;a;k
cell (k + 1)</p>
      <p>Cb; Qqk;+b;1k 1
Cb0 ; Qqk;+b;1k 1
Cb00 ; Qqk;+b;1k 1
Cb; Qqk;+a1;k
Cb0 ; Qqk;+a1;k
Cb00 ; Qqk;+a1;k
Cb; Qqk;+b;1k+1
Cb0 ; Qqk;+b01;k+1
Cb00 ; Qqk;+b010;k+1
Finally, if k j &lt; n then the current state and head position with the symbol
scanned by the head are simply propagated along the tape:</p>
      <p>Qq;a;k v 9P:(Cb u Qjq+;a1;k);
j
for b 2
(T1-7)
(generating a separate branch for each symbol tentatively assigned to the cell;
see cases (a) and (b) in Fig. 2).</p>
      <p>Step 2. The axioms (T1-3){(T1-7) generate a separate P -successor for each
symbol b in the alphabet . The correct one will be chosen by a -homomorphism
from the canonical model C2 of K2 into the canonical model C1 of K1. To
exclude wrong choices, we add the following concept and role inclusions to T1 and
T2. Each element d1 of C1 that represents a possible cell contents, say a, also
generates a block of arrows Sb;0; : : : ; Sb;n, for all symbols b di erent from a:
Ca v D;</p>
      <p>Ca v 9Sb;0;
9Sb;j 1 v 9Sb;j ;
9Sb;n v Cb;
On the other hand, T2 makes sure that every element d2 of a concept D in C2
generates a block Za;0; : : : ; Za;n, for each a 2 :</p>
      <p>D v 9Za;0;
9Za;j 1 v 9Za;j ;
9Za;n v Ca:
for 0 &lt; j
n;
(T2-1)
(T2-2)
(T2-3)
Suppose a -homomorphism h : C2 ! C1 maps d2 to d1. The following role
inclusions in T1 ensure then that all the Za;0; : : : ; Za;n blocks but one can be
mapped by h to the respective blocks Sa;0; : : : ; Sa;n in C1 (see Fig. 3):
Sa;j v Za;j ;
for a 2
and 0
j
n:
(T1-12)
So, the only remaining block is Za;0; : : : ; Za;n such that d1 belongs to Ca in C1
(that is, the symbol a tentatively contained in this cell). Then the following role
inclusions allow the sequence of Za;j to be mapped along the inverses of P in
C1, but only if the same cell contains a in the preceding con guration (that is,
the element n + 1 steps closer to the root in C1 belongs to Ca; see Fig. 3):
P
v Za;j ;
for a 2
and 0
j
n:
(T1-13)
Thus, the blocks Sb;0; : : : ; Sb;n act as `sinks' for the respective blocks Zb;0; : : : ; Zb;n
from C2.</p>
      <p>C2
C1</p>
      <p>Note that the cell whose contents is changed generates the additional blocks
Sa;0; : : : ; Sa;n to allow the respective blocks of Za;0; : : : ; Za;n from C2 to be sunk
there (because there can be no match along the P -edges).</p>
      <p>Step 3. It remains to encode the acceptance condition for the Turing machine
M 0. To this end, we extend T2 with four blocks of roles: for each i = 0; : : : ; 3, we
take role names Pi;0; : : : ; Pi;n and add to T2 the inclusions
9Pi;j 1 v 9Pi;j ; for 0
i
The four blocks are arranged into an in nite tree-like structure by the following
concept inclusions:</p>
      <p>A v 9P0;0;
9P0;n v 9P1;0;
9P1;n v 9P3;0;
9P3;n v 9P1;0;
9P0;n v 9P2;0;
9P2;n v 9P3;0;
9P3;n v 9P2;0:
More precisely, the P0;j block can be thought of as the root, from which two
branches start with P1;j and P2;j blocks, respectively. Each of these blocks is
followed by a P3;j block. Then we again have the P1;j and P2;j blocks, and so
on. Such a pattern is required to represent con gurations of M 0 with alternating
universal and existential states.
(T2-4)
(T2-5)
(T2-6)
(T2-7)
P0;0</p>
      <p>P0;1
0</p>
      <p>P0;n
C2
A
C1
A</p>
      <p>P1;0
P2;0</p>
      <p>X1P1;1
X2P2;1
1
2</p>
      <p>P1;n</p>
      <p>P3;0 P3;1
P2;n</p>
      <p>P3;0 P3;1
3
3</p>
      <p>P3;n
P3;n</p>
      <p>P1;0
P2;0
P1;0
P2;0
X1
X2</p>
      <p>X1
X2
X1</p>
      <p>X2</p>
      <p>We extend T1 with the following role inclusions:</p>
      <p>P v Pi;j ;
for 0
i
3 and 0
j
n:
(T1-14)
Thus, each the initial P0;i block in C2 can be -homomorphically mapped into
the initial P -block in C1 (see Fig. 4). Its successor must be a universal state,
which is re ected by the two blocks, P1;j and P2;j , in C2. Both of these blocks
should be -homomorphically mapped into C1. The following concept inclusions
in T2 ensure that the P1;j block is mapped to the P -block that begins with X1
and the P2;j to the P -block that begins with X2:
9P1;0 v X1;
9P2;0 v X2;
(T2-8)
Each of the P1;j and P2;j blocks in C2 is followed by a P3;j block, which
corresponds to an existential state: the P3;j block can be mapped to either of
the two P -blocks (beginning with X1 or X2); see Fig. 4, where possible
homomorphisms are shown by thick grey dashed arrows.</p>
      <p>It remains to add to T2 concept inclusions making sure that each of the
domain elements representing cells of the tape in a non-initial con guration
belongs to the concept D, thus enforcing synchronisation of the cell contents:
9Pi;j v D;
for 1
i</p>
      <p>One can show now that T1 and T2 are as required: M accepts w i the
canonical model C2 of K2 = (T2; fA(c)g) is nitely -homomorphically
embeddable into the canonical model C1 of K1 = (T1; fA(c)g), where contains the
concept and role names in T2. It remains to use Theorem 2 and the fact that
APSpace = ExpTime.</p>
      <p>Next, we show that -query inseparability is also ExpTime-hard. To this
end, let be the signature of T2. Then we take T11 and T22 be to the `localised'
TBoxes T1 and T2, respectively: namely, Tkk is the result of attaching the
superscript k to each symbol (concept or role name) in the TBox Tk (such symbols
are local for the respective TBox, that is, the signatures of T11 and T22 do not
intersect). Then we take Tk to link those local symbols to the signature : each
TTkk0 =coTnkksi[stTskof. aCllo ninscidluesriothnes porfotbhlee mforomf S-kquverSy,infosrepsayrmabbiollistySofin(T10;. fNAo1w(a,)lget)
and (T10 [ T20; fA1(a); A2(a)g). Evidently, the canonical model of (T10; fA1(a)g) is
contained in the canonical model of (T10 [ T20; fA1(a); A2(a)g). The converse
query entailment is only possible if there is a -homomorphism from the
canonical model of (T20; fA2(a)g) into the canonical model of (T10; fA1(a)g), which
amounts to checking whether T2 is -query entailed by T1. q</p>
      <p>The ExpTime lower bound for -query entailment in the case of OWL 2 QL
TBoxes can be proved, using Theorem 3, by a minor modi cation of T1 and T2
above: for each of the role names R involved in T2 (and so in T1), we take a fresh
role R0 and add an extra role inclusion R v R0 to the each of the two TBoxes;
and take to include the concepts A, X1, X2 and the Ca, for a 2 , together
with the newly introduced role names R0. Thus we obtain:
Theorem 5. The -query entailment and -query inseparability problems for
OWL 2 QL TBoxes are ExpTime-complete.</p>
    </sec>
    <sec id="sec-3">
      <title>Conclusion</title>
      <p>
        In this paper, we have proved that the -query entailment and inseparability
problems for both OWL 2 QL KBs and TBoxes are ExpTime-hard. This lower
bound is tight for OWL 2 QL TBoxes matching the upper bound previously
established in [
        <xref ref-type="bibr" rid="ref4">4</xref>
        ]. We are working on a game-theoretic proof showing that it is
tight for the case of KBs as well.
      </p>
    </sec>
  </body>
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