If S is r-satisfiable, then there is an r-complete tuple w.r.t. S . 19

Let J₁, . . . ,J_k be the interpretations over the domain ∆ that exist according to the r-satisfiability of S (cf. Definition 3.1). We define the tuple (A_R, Q^¬_R) as follows:

A_R :={A(a)|a∈N_I(φ), A∈N_RC(T), a^J1 ∈A^J1} ∪ {¬A(a)|a∈N_I(φ), A∈N_RC(T), a^J1 ∈/ A^J1};

Q^¬_R :={α_j ∈ Q_φ |p_j ∈/ ^\S}.

Obviously, A_R is an ABox type for K, and Q^¬_R satisfies Condition (R4). Further-more, it is easy to verify that eachJ_i, 1≤i≤k, can be extended to a modelJ_i⁰ of Kⁱ_R by appropriately defining the interpretations of the new individual names a_x that are introduced by A_Qi. Thus, Condition (R1) is also satisfied.

Regarding Condition (R2), assume that there are i, 1≤i≤k, and p_j ∈X_i such thatKⁱ_R |=α_j, and thusJ_i⁰ |=α_j. This means that alsoJ_i |=α_j sinceα_j does not contain any of the new individual names. But this contradicts the assumption that J_i |=χ_i.

The proof of Condition (R3) is also by contradiction. We assume that there are i, 1 ≤ i ≤ k, a tree-shaped CQ α_j ∈ Q^¬_R, and a witness B of α_j such that

Kⁱ_R |= ∃x.B(x), and thus J_i |= ∃x.B(x) as above. However, by the definition of Q^¬_R, there must be an i⁰, 1 ≤ i⁰ ≤ k, such that p_j ∈/ X_i⁰, and thus J_i⁰ 6|= α_j. Lemma 4.4 then yields that J_i⁰ 6|= ∃x.B(x), which contradicts the facts that B ⊆N_RC(T) and J_i and J_i⁰ respect the rigid names.

4.1.2 If there is an r-complete tuple w.r.t. S, then S is r-satisfiable.

The proof of the converse direction is more involved. For each i, 1 ≤ i ≤ k, we consider the canonical interpretation I_i := I_[Kⁱ

R]⁺, where [K_Rⁱ]⁺ is equal to Kⁱ_R without the negated assertions in AR. SinceKⁱ_R is consistent by Condition (R1), we know that [Kⁱ_R]⁺ 6|=A(a) for any negated assertion ¬A(a)∈ A_R. By Proposi-tion 2.11, it follows that I_i |=¬A(a), and hence I_i is a model of K_Rⁱ.

To distinguish the elements contained in N^aux_I , we define ∆^I_aⁱ := N^aux_I ∩ ∆^Ii, and write aⁱ_x instead of a_x for the elements of this set. We further write ∆^I_uⁱ for the set containing the unnamed domain elements unique to the canonical interpretation I_i, and similarly write cⁱ_%rD for every element c_%rD ∈ ∆^I_uⁱ. Thus, the domain of eachI_i is composed of the pairwise disjoint componentsN_I(φ), ∆^I_aⁱ, and ∆^I_uⁱ. We next state that as fact for future reference.

Fact 4.7. For all i, j ∈ {1, . . . , k}, the sets NI(φ), ∆^Ia^j, and ∆^I_uⁱ are pairwise disjoint.

In our construction, we make use of the subset ∆^I_uⁱ_R :=^S∞_j=0∆^i,j_uR of ∆^I_uⁱ, which is inductively defined as follows:

∆^i,0_u

R :={cⁱ_%rD | B ⊆N_RC(T), cⁱ_%∈ B^Ii ∩∆^I_uⁱ, D ∈Sub(T),T |=B v ∃r.D} ∪ {cⁱ_ai

xrD | B ⊆N_RC(T), aⁱ_x∈ B^Ii∩∆^I_aⁱ, D ∈Sub(T),T |=B v ∃r.D}

∆^i,j+1_u

R :={cⁱ_%rDsE |cⁱ_%rD ∈∆^i,j_u

R, E ∈Sub(T),T |=Dv ∃s.E}.

This definition is similar to that of ∆^I_uⁱ (cf. Definition 2.10), the only difference being that we here only consider those elements whose existence is enforced by some combination of rigid concept names at an already unnamed domain element.

Thus, there are no direct role connections between elements of NI(φ) and ∆^I_uⁱ

R. Fact 4.8. For all i, 1≤i≤k, we have ∆^I_uⁱ

R ⊆∆^I_uⁱ.

We now construct the interpretationsJ1, . . . ,Jkas required for the r-satisfiability ofS, that is, they share the same domain and respect rigid names, and eachJ_i is a model of T and χ_i =^V_pj∈Xiα_j ∧^V_p

j∈X_i¬α_j. Recall that we then do not need to specifically define an interpretation for time point 0, since any J_ι(0) will be a model of A₀ = ∅ and χ_ι(0). To obtain interpretations J₁, . . . ,J_k as required, we join the domains of the interpretations I_i and ensure that they interpret all rigid

concept names in the same way. We first construct the common domain and then define the interpretations J_i, 1≤i≤k, as follows:

• For all a∈N_I(φ), we set a^Ji :=a.

• For all flexible concept names A, we define A^Ji :=A^Ii∪

• For all (flexible) role namesr, we define r^Ji :=r^Ii ∪

In this way, we have constructed interpretations J₁, . . . ,J_k that have the same domain and respect the rigid concept names since, for all A ∈ N_RC, the defini-tion of A^Ji is independent of i. It remains to show that they satisfy the other requirements for the r-satisfiability of S as described above.

We start by showing some facts about the elements in the sets ∆^Iu^jR. Lemma 4.9. For all i, j ∈ {1, . . . , k} and c^j_%rD ∈∆^IuR^j, the following hold:

a) For all concepts C ∈Sub(T), we have c^j_%rD ∈C^Ji iff T |=DvC.

b) There is a witness B of ∃r.D w.r.t. T such that B^Ij is non-empty.

Proof. We begin with the proof of a), for which we use induction on the shape of C. For the induction start, let c^j_%rD ∈ ∆^Iu^jR and C = A∈ N_C(T). We consider (⇒) and the definition ofJ_i. Ifc^j_%rD ∈A^Ij, then we immediately haveT |=DvA by Definition 2.10. If c^j_%rD ∈ B^Ij for some B ⊆ NRC(T) with T |= B v A, then we also get c^j_%rD ∈A^Ij since I_j |=T, and thus T |=DvA as above. The other direction, (⇐), immediately follows from Definition 2.10 and the definition ofJ_i. The claim for C => holds because of the interpretation of >.

ForC =C₁uC₂, we havec^j_%rD ∈C₁^Ji∩C₂^Ji iffT |=DvC₁ and T |=DvC₂ by the induction hypothesis. This is equivalent to T |=DvC₁uC₂.

Let now C = ∃s.C₁. If c^j_%rD ∈ (∃s.C₁)^Ji, then there exists c^j_%rDsE ∈ ∆^Iu^jR with (c^j_%rD, c^j_%rDsE) ∈ s^Ji and c^j_%rDsE ∈ C₁^Ji. By the induction hypothesis, we get T |= E v C₁. Moreover, we have T |= D v ∃s.E by Definition 2.10, which implies that T |= D v ∃s.C₁. On the other hand, if T |= D v ∃s.C₁, then we have c^j_%rDsC1 ∈ ∆^IuR^j. Since T |= C1 v C1, by the induction hypothesis we get c^j_%rDsC1 ∈ C₁^Ji. By the definition of J_i, we also have (c^j_%rD, c^j_%rDsC1) ∈ s^Ji, and thus c^j_%rD ∈(∃s.C₁)^Ji.

For the proof of b), we proceed by induction on the construction of ∆^Iu^jR. For elements c^j_%rD ∈∆^j,0_u

R, the definition of ∆^j,0_u

R directly yields the claim.

For the induction step, we considerc^j_%sErD ∈∆^j,l+1_uR ,l≥0, i.e., we havec^j_%sE ∈∆^j,l_uR and T |= E v ∃r.D. By the induction hypotheses, there are r₁, . . . , r_` ∈ N_R,

` ≥ 0, and a set B ⊆ N<sub>RC</sub>(T) such that T |= B v ∃r<sub>1</sub>. . . r<sub>`</sub>s.E and B^Ij is non-empty. This implies thatT |=B v ∃r₁. . . r_`sr.D, which concludes the proof.

We now state a basic connection between the interpretationsJi andIi concerning the interpretation of role names.

Lemma 4.10. For all i ∈ {1, . . . , k}, role names r ∈ N_R, and d, e ∈ ∆^Ii, we have (d, e)∈r^Ji iff (d, e)∈r^Ii.

Proof. The “if”-direction follows directly from the definition of r^Ji. For the

“only if”-direction, Facts 4.7 and 4.8 and the definition of J_i either directly yield (d, e) ∈ r^Ii, or e ∈∆^I_uⁱ

R is of the form c^j_%rD and either d =c^j_% or d =a^j_x = %. By Definition 2.10, the latter two options also imply that (d, e)∈r^Ii.

There is a similar connection between the interpretations of concepts inI_j andJ_i. Lemma 4.11. For all i, j ∈ {1, . . . , k} and all concepts C ∈Sub(T), the follow-ing hold:

a) For all e ∈N_I(φ), we have e∈C^Ji iff e∈C^Ii. b) For all e ∈∆^Ia^j ∪(∆^Iu^j \∆^Iu^jR), we have e∈C^Ji iff

• i=j and e∈C^Ii, or

• there is aB ⊆ N_RC(T) such that e∈ B^Ij and T |=B v C.

c) For all e ∈∆^Iu^jR, we have e∈C^Ji iff e∈C^Ij.

Proof. Item c) is a direct consequence of Lemmata 2.12, and 4.9a) and Fact 4.8.

We now prove the other two items simultaneously by induction on the structure of C.

For the induction start, we begin with a). For rigid concept names, it follows from the fact that each I_j, 1 ≤ j ≤ k, is a model of A_R and from the definition of J_i. For flexible concept names C, by Fact 4.7 we have e ∈ C^Ji iff e ∈C^Ii or e ∈ B^Ij for some j, 1≤j ≤k and B ⊆N_RC(T) with T |=B vC. Since both I_i and I_j are models of A_R, in the latter case we also have e ∈ B^Ii. Since I_i |= T, this implies that e∈C^Ii.

We consider b). Because of Fact 4.7, for C ∈ N_RC, the definition of J_i directly yields e∈C^Ji iff e∈C^Ij. For C ∈N_C\N_RC, we obtain the claim from Fact 4.7, the definition of J_i, and the fact that all I_j are models ofT.

The claim for C => is again trivial by the interpretation of >.

We consider C = C₁ uC₂ for a) and b) under the assumption that i = j. The induction hypothesis directly yields the equivalence between e ∈ C₁^Ji ∩C₂^Ji and e ∈ C₁^Ii ∩C₂^Ii. Moreover, any B ⊆ N_RC(T) with e ∈ B^Ii and T |= B v C₁ uC₂ also yields that e∈(C₁uC₂)^Ii, and thus e∈(C₁uC₂)^Ji as above.

For case b) with i6=j, by the induction hypothesis e∈ (C₁uC₂)^Ji implies that there areB₁,B₂ ⊆N_RC(T) withe∈(B₁u B₂)^Ij,T |=B₁ vC₁, andT |=B₂ vC₂. But then it also holds that T |= B1 u B2 v C1 uC2, and thus e ∈ (C₁uC2)^Ij. On the other hand, if e∈ B^Ij and T |=B vC₁uC₂ for some B ⊆N_RC(T), then also T |=B vC₁ andT |=B v C₂. Together with the induction hypothesis, this leads to e∈(C₁uC2)^Ji.

Finally, we consider the case of an existential restrictionC =∃r.C₁. For a) and b) with i = j, by the definition of J_i, Lemma 4.10, Item c), and the induction hypothesis the existence of a d ∈ (C₁)^Ji with (e, d) ∈ r^Ji is equivalent to the existence of a d ∈ (C₁)^Ii with (e, d) ∈ r^Ii. As before, the second option of b) is subsumed by the first one, in this case.

For case b) with i 6=j, e ∈(∃r.C₁)^Ji implies that there is a c^j_%rD ∈(C₁)^Ji ∩∆^IuR^j

such that either e = c^j_% or e = a^j_x = %. Since e /∈ ∆^Iu^jR, in both case we must have c^j_%rD ∈ ∆^j,0_u

R, and thus there exists a B ⊆ N_RC(T) such that e ∈ B^Ij and T |= B v ∃r.D. Furthermore, the fact that c^j_%rD ∈ (C₁)^Ji implies that T |= D v C₁ by Lemma 4.9, and thus T |= B v ∃r.C₁. On the other hand, if there exists a B ⊆ N_RC(T) with e ∈ B^Ij and T |=B v ∃r.C₁, then by the defini-tion of ∆^j,0_u

R we havec^j_%rC1 ∈∆^j_u

R, where again either e =c^j_% or e=a^j_x =%. Thus, in particular it holds that (e, c^j_%rC

1)∈r^Ji. SinceT |=C₁ vC₁, by Lemma 4.9 we have c^j_%rC1 ∈(C₁)^Ji, and thus e∈(∃r.C₁)^Ji, as required.

We finally show that J_i is in fact as intended.

Lemma 4.12. Each J_i, 1≤i≤k, is a model of T.

Proof. Consider a GCI C vD ∈ T and an element d∈ C^Ji. If d ∈N_I(φ)∪∆^Iu_R^j

for some j, 1≤j ≤k, we get d∈C^Ij by Lemma 4.11. SinceI_j |=T, we obtain

d∈D^Ij, and thus d∈D^Ji again by Lemma 4.11. A similar argument applies in case that d ∈∆^I_aⁱ ∪(∆^I_uⁱ\∆^I_uⁱ

R).

For d ∈ ∆^Ia^j ∪(∆^Iu^j \∆^Iu^jR) with i 6= j, by Lemma 4.11b) we know that there is a set B ⊆ N_RC(T) such that d ∈ B^Ij and T |= B v C. But then we also have T |=B vD, which leads to d∈D^Ji by another application of Lemma 4.11.

We now provide the final missing piece to show r-satisfiability of S.

Lemma 4.13. Each J_i, 1≤i≤k, is a model of χ_i.

Proof. Consider first any CQ α that occurs positively in the conjunction χ_i. Since Ii |= AQi and AQi contains an instantiation of α, we know that there is a homomorphism π of α into I_i that maps all variables to elements in ∆^I_aⁱ. By Lemmata 4.10 and 4.11b), we know that π is also a homomorphism ofα into J_i. We now consider a CQ α that occurs negatively in χ_i. By Condition (R2), we know that [Kⁱ_R]⁺ 6|=α, and thusI_i |=¬α by Proposition 2.11. We now assume to the contrary that there is a homomorphism π of α into J_i. Since α is connected and domain elements d, e ∈ ∆ can only be connected by r^Ji if they belong to the same domain ∆^Ij (cf. Fact 4.7), we can assume that there is an index j, 1≤j ≤k, such thatπ maps all terms of α into ∆^Ij.

Assume first that π maps all terms into ∆^Ii, which in particular includes N_I(φ).

Then by Lemmata 4.10 and 4.11, π is also a homomorphism of α intoI_i, which contradicts the fact that I_i |=¬α.

Otherwise, we have j 6=i and π maps at least one term into ∆^Ij \N_I(φ). By the interpretation of roles in Ji and since α is connected, this means that no term of αcan be mapped into N_I(φ), (i.e., α contains no individual names andπ maps all variables into ∆^Ia^j∪∆^Iu^j). Furthermore, for all role atomsr(y, z)∈ At(α), we either have (i) π(y)∈∆^Ia^j∪(∆^Iu^j\∆^IuR^j) andπ(z)∈∆^IuR^j, or (ii) π(y), π(z)∈∆^IuR^j. Since α is connected, there is at most one variable in α that is mapped into

∆^Ia^j∪(∆^Iu^j\∆^IuR^j) byπ. Thus, in α, there are only role connections starting from this variable and role connections between other variables (mapped into ∆^IuR^j) via a single role and in one direction. This means that α is tree-shaped.

We now show that there is a witness B of α w.r.t. T such that B^Ij is non-empty. For this, let x be the root variable of α. By our assumption that π is a homomorphism of α into J_i, we know that π(x)∈(Con(α))^Ji.

• If π(x) is contained in ∆^Ia^j ∪(∆^Iu^j \∆^IuR^j), then Lemma 4.11b) yields that there is a witness B of α w.r.t. T such that π(x)∈ B^Ij.⁵

5Recall that we assumed thatCon(α) occurs inT.

• Ifπ(x) is of the form c^j_%rD ∈∆^Iu^jR, then by Lemma 4.9b) there is a witnessB of ∃r.D such thatB^Ij is non-empty. Since c^j_%rD ∈(Con(α))^Ji, Lemma 4.9a) implies that B is also a witness of α.

However, by Condition (R4) we know that α ∈ Q^¬_R. Hence, by Condition (R3) we have [K_R^j]⁺ 6|= ∃x.B(x), and thus Ij |= ¬∃x.B(x) by Proposition 2.11. This contradicts the fact that B^Ij is non-empty.

This finishes the proof of Lemma 4.6. We next use this characterization to solve TCQ satisfiability (and entailment) in polynomial space.

4.1.3 The Upper Bound ctd.

The key insight of the previous section is that we do not need to store the expo-nentially large set S in order to check the conditions of Definition 4.5. It suffices to guess an ABox typeAR and a set Q^¬_R in advance, and then check, in each step of an LTL-satisfiability test for φ^p, if there is a world X_i ⊆ {p₁, . . . , p_m} that satisfies the requirements specified in Definition 4.5.

For this purpose, we use the polynomial-space-bounded Turing machines for LTL-satisfiability constructed in [SC85]. Given the propositional LTL-formulaφ^p, the machine M_φ^p iteratively guesses complete sets of (negated) subformulae of φ^p specifying which subformulae are satisfied at each point in time. Every such set induces a unique world X_i ⊆ {p₁, . . . , p_m} containing the propositional variables that are true.

In [SC85, Theorem 4.7], it is shown that if φ^p is satisfiable, then there must be a periodic model of φ^p with a period that is exponential in the size of φ^p. Hence, Mφ^p first guesses two polynomial-sized indices specifying the beginning and end of the first period. Then it continuously increments a (polynomial-sized) counter and in each step guesses a complete set of (negated) subformulae of φ^p. It then checks Boolean consistency of this set and consistency with the set of the previous time point according to the temporal operators. For example, if the previous set contains the formula p₁Up₂, then either it also contains p₂ or it must contain p₁ and the current set must contain p₁Up₂. In this way, the satisfaction of the U-formula is deferred to the next time point.

In each step, the oldest set is discarded and replaced by the next one. When the counter reaches the beginning of the period, it stores the current set and contin-ues until it reaches the end of the period. At that point, instead of gcontin-uessing the next set of subformulae, the set stored at the beginning of the period is used and checked for consistency with the previous set as described above. M_φ^p addition-ally has to ensure that all U-subformulae are satisfied within the period. Thus, the Turing machine never has to remember more than three sets of polynomial size.

Note that [SC85] do not directly regard past operators, which are considered by us. However, we can certainly adapt the complete sets of subformulae guessed by M_φ^p to also include the past operators. This does not affect the space require-ments of the Turing machines; in particular, the period that has to be guessed is still exponential in the size of φ^p. We now modify this procedure to prove the desired PSpace upper bound.

Lemma 4.14. IfNRC 6=∅butNRR =∅, then TCQ entailment inELis inPSpace w.r.t. combined complexity.

Proof. Let K = hT,∅i be a TKB and φ be a TCQ. We analyze the complexity of the satisfiability problem by showing how an r-satisfiable set S can be found.

By Lemma 4.6, it suffices to find a tuple (AR, Q^¬_R) satisfying conditions (R1)–

(R4). All these conditions are such that it is not necessary to actually construct the whole set S—it is enough to show that each world X_i we encounter when checking φ^p (not φ^p_S) for satisfiability induces a knowledge baseKⁱ_R that satisfies all requirements.

We can thus run a modified version of the Turing machineM_φ^p that first guesses the setsA_R and Q^¬_R required by Definition 4.5, which can clearly be done in poly-nomial space, and then proceeds as before, but additionally executes the following checks for the worldX induced by each guessed complete set of propositional sub-formulae:

(R1) Check the KB K_R = hT,A_R∪ A_QXi for consistency, where A_QX is formed by instantiating all CQs α_j with p_j ∈X.

This consistency test can be done in polynomial time in the (polynomial) size of K_R [BBL05] and thus needs only polynomial space.

(R2) Check, for each p_j ∈X, whether K_R 6|=α_j holds.

Note that we have K_R 6|= α_j iff [K_R]⁺ 6|= α_j. For (⇐), we have that, if [K_R]⁺6|=α_j, thenI[K_R]⁺ 6|=α_j by Proposition 2.11. Furthermore, the canon-ical modelI_[K

R]⁺ is also a model ofK_R (cf. the beginning of Section 4.1.2), and thus [K_R]⁺ 6|= α_j. For (⇒), we directly have that every model of K_R that does not satisfy α_j is also a model of [K_R]⁺. Hence, it suffices to check the non-entailment [K_R]⁺ 6|=α_j, which can be done in (deterministic) exponential time and polynomial space by Lemma 3.5.

(R3) Check, for eachα ∈Q^¬_R and every witnessB ofα w.r.t.T, whether it holds that K_R 6|=∃x.B(x).

Since each B is of polynomial size, the actual non-entailment test can be done in polynomial space by the same arguments as above. However, while we can easily enumerate all α ∈Q^¬_R and B ⊆N_RC(T) in polynomial space, we still have to determine whether B is actually a witness ofα w.r.t. T.

In [BBM11, Lemma 12], it is shown that, for two concept namesA, B, it can be decided in polynomial time whether there are role namesr₁, . . . , r_{`</sub> such thatT |=Av ∃r<sub>1</sub>. . . r<sub>`}.B. Essentially, it suffices to check reachability ofB fromAin an appropriate graph derived fromT. This idea is also implicitly used in the form of the reachability relation ; in [BBL05, KKS12].

We can use this approach for our problem by introducing two new concept names AB and Aα and then checking in polynomial time whether

T ∪ {AB v B,Con(α)vAα} |=AB v ∃r1. . . r`.Aα

holds for some role names r₁, . . . , r`, which is equivalent to the fact that B is a witness of α w.r.t. T.

(R4) Check, for each pj ∈X, whether αj ∈Q^¬_R.

The set S required for Lemma 3.2 corresponds to the set of all worlds X encoun-tered during a run of this modified Turing machine. Under this definition of S, it is easy to see that the above checks are actually equivalent to (R1)–(R4) from Definition 4.5. By Lemmata 3.2 and 4.6, the described Turing machine accepts the input K and φ iff φ has a model w.r.t. K (recall that we can disregard the mapping ι due to our assumptions). Since we do not have to store S explic-itly and all checks can be done with a nondeterministic Turing machine using only polynomial space, according to [Sav70], TCQ entailment can be decided in PSpace.

This finishes the proof of Theorem 4.1.

Im Dokument Temporal Query Answering in EL (Seite 20-28)