If there is an r-complete tuple w.r.t. S and ι, then S is

The proof of the converse direction is more involved. We assume an r-complete tuple (A_R, Q_R, Q^¬_R, R_F) to be given. Further, if φis r-simple, then we can assume R_F, and henceA_RF and N^tree_I , to be empty.

For each i, 0≤i≤n+k, we consider the canonical interpretation I_i :=I_Kⁱ

R. To distinguish the elements contained in N^aux_I , we writeaⁱ_x for the element a_x ∈N^aux_I in the domain of Ii, and define the set ∆^I_aⁱ to contain exactly those elements.

Likewise, we write aⁱ_b% for the element a_b% ∈ N^tree_I that occurs in I_i, and define

∆^I_tⁱ as above. We further write ∆^I_uⁱ for the set containing the unnamed domain elements unique to the canonical interpretation I_i, and similarly write cⁱ_%R for every element c_%R ∈∆^I_uⁱ. For any e ∈N_I(K)∪N^aux_I ∪N^tree_I , we may denote by eⁱ the corresponding element in NI(K)∪∆^I_aⁱ ∪∆^I_tⁱ, i.e., we consider aⁱ :=a for any a ∈ N_I(K). Thus, the domain of each I_i is composed of the pairwise disjoint components N_I(K), ∆^I_aⁱ, ∆^I_tⁱ, and ∆^I_uⁱ. We state this fact for future reference.

Fact 4.11. For all i, j, j⁰ ∈ {0, . . . , n+k}, the sets N_I(K), ∆^I_aⁱ, ∆^I_t^j, and ∆^Iu^j0

are pairwise disjoint.

We now construct the interpretationsJ₀, . . . ,J_n+k required for the r-satisfiability of S (whereJ₀, . . . ,J_n represent I₀, . . . ,I_n of Definition 4.2 and J_n+1, . . . ,J_n+k represent J₁, . . . ,J_k). To this end, we join the domains of the interpretations I_i and ensure that they interpret all rigid names in the same way. We first construct the common domain

∆ :=NI(K)∪

n+k

[

i=0

(∆^I_aⁱ ∪∆^I_tⁱ ∪∆^I_uⁱ) and then define the interpretations J_i, 0≤i≤n+k, as follows:

• For all a∈N_I(K), we set a^Ji :=a.

• For all rigid concept names A, we define A^Ji :=^Sn+k_j=0 A^Ij.

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

n+k

[

j=0

[

B⊆BC_R(O), O|=dBvA

(l

B)^Ij∪

n+k

[

j=0

{c^j_%R ∈∆^I_u^j | O |=∃R⁻vA,WO(c^j_%R)6=∅}.

• For all rigid role names R, we define R^Ji :=^Sn+k_j=0 R^Ij.

• For all flexible role names R, we define R^Ji :=R^Ii ∪

n+k

[

j=0

[

S∈N⁻

RR(O) O|=SvR

S^Ij ∪

n+k

[

j=0

{(e₁, e₂)∈R^Ij | ∃e∈ {e₁, e₂},WO(e)6=∅}.

In this way, we have constructed interpretations J₀, . . . ,J_n+k that have the same domain and respect rigid names.

It remains to show that the interpretations satisfy the other requirements for the r-satisfiability of S. To this end, we first provide auxiliary lemmas and begin showing a basic connection between the interpretations Ji and Ii concerning the interpretation of role names.

Lemma 4.12. For alli∈ {0, . . . , n+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 ofR^Ji. We consider the “only if”-direction.

If R is flexible, assume that either (i) (d, e) ∈ S^Ij for some rigid subrole S of R or (ii) (d, e)∈R^Ij,j 6=i, and either d ore has a witness w.r.t. O. By Fact 4.11, in both cases we must have d, e ∈ N_I(K). Hence, case (ii) is impossible since named domain elements cannot have witnesses (see Definition 4.5). In case (i), we must have S(d, e) ∈ A_R since A_R is an ABox type and I_j |= A_R, and hence (d, e)∈S^Ii ⊆R^Ii since I_i |=A_R and I_i |=O.

If R is rigid, assume again that d, e ∈ N_I(K) and (d, e) ∈ R^Ij, for some j 6= i.

Since A_R is an ABox type and I_j |=A_R, we must haveR(d, e)∈ A_R. Since also I_i |=A_R, we get (d, e)∈R^Ii.

There is a similar connection between the interpretations of concepts in Ji and I_j.

Lemma 4.13. For all basic concepts B ∈ BC(O) and i, j ∈ {0, . . . , n+k}, the following hold:

a) for all a∈N_I(K), we have a∈B^Ji iff a∈B^Ii;

b) if B is rigid, then, for every e ∈ ∆^Ia^j ∪ ∆^I_t^j ∪∆^Iu^j, we have e ∈ B^Ji iff e∈B^Ij;

c) if B is flexible, then, for every e ∈∆^Ia^j ∪∆^I_t^j ∪∆^Iu^j, we have e∈B^Ji iff

• i=j and e∈B^Ii, or

• there is a B ⊆BC_R(O) with e ∈(d

B)^Ij and O |=d

B vB, or

• e∈B^Ij ∩∆^Iu^j and WO(e)6=∅.

Proof. For a), consider first the case thatB is rigid. Then,a∈B^Ii clearly implies a ∈B^Ji since s^Ii ⊆s^Ji, for s ∈N_RC ∪N_RR. On the other hand, if a ∈B^Ji, then by the definition of J_i, we must have a∈B^Ij, for at least oneI_j, 0 ≤j ≤n+k.

Since I_i and I_j are both models of the ABox type A_R, we also have a∈B^Ii. Consider now any flexible basic conceptB. By the definition ofJi on the flexible names, we have a ∈ B^Ji iff either (i) a ∈ B^Ii, or (ii) a ∈ (d

B)^Ij, for some j, 0 ≤j ≤ n+k, and B ⊆ BC_R(O) with O |= d

B vB. To see the latter w.r.t. a flexible concept of the form B =∃R, note that the definition of R^Ji implies one of the following cases (assuming that case (i) does not apply):

• a belongs to (∃S)^Ij, whereS is a rigid subrole of R. Hence, we can choose B:={∃S}since also ∃S v ∃R.

• There is an R-successor of a in J_i of the form c^j_aR2 for some R₂ ∈ N⁻_R(O) with O |= R₂ v R. By Definition 4.5, WO(a) is not defined, and hence there must exist a witness B ⊆ BC_R(O) such that O |=d

B v ∃R₂ v ∃R.

But (ii) implies (i) since a ∈ (d

B)^Ij yields B⁰(a) ∈ A_R, for all B⁰ ∈ B, which together with I_i |=A_R and I_i |=O leads to a∈B^Ii.

The claim in b) follows directly from Fact 4.11 and the definition of J_i.

For c), we first consider the case that B ∈ N_C(O) is a flexible concept name.

Then, the equivalence with one of the three cases is covered by the definition of Ji and, for the last case, also by Proposition 3.4. We now consider B to be of the form ∃R for a flexible role R ∈ N⁻_R(O). In case i = j, the claim can be restricted to the first of the three items since the other two are subsumed by it (for the second item, this holds because Ii |= O). Hence, the claim directly follows from Fact 4.11 and the definition of J_i (i.e., the interpretation of the elements in

∆^I_aⁱ∪∆^I_tⁱ∪∆^I_uⁱ is not influenced by any I_j, j 6=i). We consider the casei6=j.

• Let e be of the form d^j for some d ∈ N^aux_I ∪N^tree_I , i.e., e ∈ ∆^Ia^j ∪∆^I_t^j. We only have to regard the second item. (⇒) The definition of R^Ji yields that either (i) e ∈ (∃S)^Ji and O |= S v R for a rigid role S or (ii) there is an R-successor e⁰ of e in I_j and either e or e⁰ has a witness w.r.t. O. In case (i), we can chooseB :={∃S}. In case (ii), sinceWO is not defined for elements ofN_I(K)∪∆^Ia^j∪∆^I_t^j, we know that e⁰ is of the formc^j_dR2 for some R₂ ∈N⁻_R(O) andWO(c^j_dR2)6=∅. By Definition 3.1, we obtain O |=R₂ vR.

By the definition of W_O, there is a B ⊆ BC_R(O) with e ∈ (dB)^Ij and

O |=d

B v ∃R₂ v ∃R. (⇐) We have e∈(∃R)^Ij, and hencec^j_dR ∈∆^Iu^j and (e, c^j_dR) ∈ R^Ij, by Definition 3.1. Furthermore, B is a witness of c^j_dR, and thus we have (e, c^j_dR)∈R^Ji by the definition ofJ_i.

• Let e ∈ ∆^Iu^j. (⇒) We know that (i) e ∈ (∃S)^Ij and O |= S v R for a rigid role S (hence we can again choose B := {∃S}), or (ii) (e, d) ∈ R^Ij for some d∈∆^Ij (and hence also e∈(∃R)^Ij), and eitherWO(e) or WO(d) is non-empty. Thus, if WO(d) is undefined or empty, then the third item holds. Otherwise, we know thatd∈∆^Iu^j andWO(d)6=∅, by Definition 4.5.

By Definition 3.1, we then have either (i) e=c^j_% and d=c^j_%R; or (ii) d=c^j_% and e=c^j_%R−.

For (i), Definition 4.5 yields that either WO(e) is also non-empty (the third item), or there is a B ⊆BC_R(O) such that e∈(d

B)^Ij and O |=d

B v ∃R (the second item).

For (ii), we immediately have that the witness of d is also a witness of e, again by Definition 4.5.

(⇐) Let e = c^j_%. If there is a set B ∈ BCR(O) with e ∈ (d

B)^Ij and O |= d

B v ∃R, then by Definition 3.1, the element c^j_%R ∈ ∆^Iu^j exists and (e, c^j_%R)∈R^Ij. Furthermore,B is a witness ofc^j_%R, and hence (e, c^j_%R)∈R^Ji, i.e., e∈(∃R)^Ji, by the definition ofJ_i.

Ife∈(∃R)^Ij, then we also have (e, c^j_%R)∈R^Ij, by Definition 3.1. Moreover, WO(e)6=∅ yields e∈(∃R)^Ji, as in the previous case.

We finally show that J_i is in fact as intended.

Lemma 4.14. Each J_i, 0≤i≤n+k, is a model of (O,A_i).

Proof. For any assertion α ∈ A_i, we have I_i |= α, and thus Lemmas 4.12 and 4.13a) yield J_i |=α.

Consider a CIB₁u · · · uB_m vB ∈ O, where B₁, . . . , B_m are basic concepts and B is either a basic concept or ⊥. Let d ∈ B₁^Ji ∩ · · · ∩B_m^Ji. If d ∈ NI(K), we get d∈B^I₁ⁱ∩ · · · ∩B_m^Ii by Lemma 4.13a). SinceI_i |=O, this implies thatd∈B^Ii. If B =⊥, this is impossible. Otherwise, we get d∈B^Ji, again by Lemma 4.13a).

Consider now the case thatd∈∆^Ia^j∪∆^I_t^j∪∆^Iu^j. Ifi=j, then Lemma 4.13b) and Lemma 4.13c) yield the same conclusion as above since the latter collapses to the first item. Assume now that i6=j. By Lemma 4.13, for every B<sub>`</sub>, 1 ≤`≤m, we have that (i) there is a B` ⊆BCR(O) with d ∈(d

B`)^Ij and O |=d

B` v B` (in case B_{`</sub> is rigid, we can set B<sub>`} :={B_{`</sub>}); or (ii) d ∈ B<sub>`}^Ij ∩∆^Iu^j and W_O(d) 6= ∅.

Since I_j |= O, in either case, we know that d ∈ B_{`</sub><sup>Ij</sup>, and hence d ∈ B<sup>Ij</sup>. If B =⊥, this is again impossible. IfB is rigid, then Lemma 4.13b) yieldsd ∈B<sup>Ji</sup>, as required. If B is flexible and case (ii) applies for at least one B<sub>`}, then the

third item of Lemma 4.13c) yields the claim. Otherwise, it is easy to see that for B :=^Sm_{`=1</sub>B<sub>`} we haved ∈(dB)^Ij and O |=B vB₁u · · · uB_m vB. Hence, the second item of Lemma 4.13c) applies, and we also get d∈B^Ji.

We now consider role inclusions of the formR₁ vR₂and assume that (d, e)∈R^J₁ⁱ. By the definition of J_i, we must have (d, e) ∈ R^I₁^j for some j ∈ {0, . . . , n+k}.

SinceI_j |=O, we get (d, e)∈R^I₂^j. IfR₂ is rigid, we immediately get (d, e)∈R^J₂ⁱ. If R₂ is flexible, assume that (d, e) ∈/ R^J₂ⁱ. But then we must have i 6= j, R₂ cannot have a rigid subrole S with (d, e) ∈ S^Ij, and neither d nor e can have a witness w.r.t.O. This implies that alsoR₁ cannot be rigid and cannot have a rigid subroleSwith (d, e)∈S^Ij (since this would also be a rigid subrole ofR₂). Hence, the definition ofJ_iyields that (d, e)∈/ R₁^Ij, which contradicts our assumption.

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

Lemma 4.15. Each J_i, 1≤i≤n+k, is a model of χ_i.

Proof. Consider first any CQ α that occurs positively in the conjunction χ_i. Since I_i |= A_Qi and A_Qi contains an instantiation of α, we know that there is a homomorphism π of α into Ii that maps all variables to elements in ∆^I_aⁱ. By Lemmas 4.12 and 4.13, we know that π is also a homomorphism of α intoJ_i. We now consider a CQαthat occurs negatively inχ_i, and assume to the contrary that there is a homomorphism π of α into J_i. By Condition (R2), we know that Kⁱ_R 6|= α, and thus I_i |=¬α, by Proposition 3.5. Furthermore, by (R4) we know that α∈Q^¬_R.

We distinguish two cases.

(I) Let first πbe such that it maps no terms into N_I(K)∪^Sn+k_j=0 ∆^Ia^j∪∆^I_t^j. This in particular implies thatN_I(α) = ∅sinceαdoes not contain any names fromN^aux_I orN^tree_I . Because αis connected and by the interpretation of roles in J_i, which is based on the canonical interpretations (cf. Definition 3.1), and Fact 4.11, we have range(π)⊆∆^Ij, for a fix j. Given that I_i |=¬α, we directly get a contradiction if j =i, by Lemmas 4.12 and 4.13, and in the following assume that j 6=i.

By considering how the elements in ∆^Iu^j are connected by roles within Ij, it is easy to see that there must be a variable x∈N_V(α), for whichπ(x) =c^j_%R is such that the length of %R is minimal among all elements of range(π). Furthermore, all other π(y) for y∈NV(α) must then be of the form c^j_%R%0 for some %⁰ ∈(N⁻_R)^∗. We now define a witness query based on a tree witness f for x in α. We start with a mapping f⁰: range(π) → (N⁻_R ×2^N−R)^∗ and then set f(y) := f⁰(π(y)) for all y ∈ N_V(α). We first define f⁰(c^j_%R) := , and proceed by induction on the structure of ∆^j_u. Letc^j_%R%0R⁰ be such thatf⁰(c^j_%R%0) has already been defined. Then

f⁰(c^j_%R%0R⁰) := f⁰(c^j_%R%0)·(R⁰,C), whereC is constructed as follows: for all role atoms S(y, y⁰)∈α such that π(y) = c^j_%R%0 and π(y⁰) =c^j_%R%0R⁰, do the following:

• if there is a S⁰ ∈N⁻_RR(O) with O |=S⁰ vS and (π(y), π(y⁰))∈(S⁰)^Ji, then add S⁰ to C;

• otherwise, add S to C (since π is a homomorphism of α into J_i, we know that (π(y), π(y⁰))∈S^Ji).

It follows from Definition 3.1 that in these cases we must have O |= R⁰ v S⁰ or O |=R⁰ vS, respectively, i.e., f is indeed a tree witness for x inα.

To construct a witness queryψ forα, the next step is to find a setB ∈Con(α,f).

We first verify the last two properties of Definition 3.7, which do not depend on the particular choice ofB. First, letA(y) be an atom ofαwithf(y) =%⁰·(S,C), which implies that π(y) = c^j_%R%0|₁S ∈A^Ji, where %⁰|₁ is the projection of the sequence%⁰ on its first component. By Lemma 4.13, we know that c^j_%R%0|₁S ∈ A^Ij, and hence O |=∃S⁻ vAby Proposition 3.4. Second, consider%⁰·(S₁,C₁)·(S₂,C₂)∈range(f).

We know that (c^j_%R%0|₁S1, c^j_%R%0|₁S1S2) ∈ S₂^Ij, and hence c^j_%R%0|₁S1 ∈ (∃S₂)^Ij. Again by Proposition 3.4, we obtain O |=∃S₁⁻ v ∃S₂.

To construct B ∈ Con(α,f), we now consider all atoms of the form A(y) or S(y, z) in α such that π(y) = c^j_%R, i.e., f(y) = . For each concept atom A(y), we set B_A(y) := A and observe that c^j_%R ∈ B_A(y)^Ji since π is a homomorphism of α into J_i. For the role atoms S(y, z), we must similarly have π(z) = c^j_%RS2, (π(y), π(z)) ∈ S^Ji, fx(z) = (S₂,C), and S⁰ ∈ C for some S2, S⁰ ∈ N⁻_R(O) with O |= S₂ v S⁰ v S, where we know that if S⁰ 6= S, then S⁰ is rigid. We set B_S(y,z):=∃S₂ and consider two cases.

• If S⁰ is rigid and S₂ is flexible, we define B_S(y,z) := {∃S₂}. Notice that O |=d

B_S(y,z) vB_S(y,z) and c^j_%R ∈(d

B_S(y,z))^Ij by Proposition 3.4.

• If S⁰ is flexible, then we know that S⁰ = S and S2 is also flexible, and there can be no rigid role between S₂ and S. Hence, we obtain from the definition of J_i that either π(y) or π(z) must have a witness w.r.t. O, and hence (π(y), π(z))∈S₂^Ji.

• If S₂ is rigid, then we immediately get (π(y), π(z))∈S₂^Ij ⊆S₂^Ji.

To summarize, in the cases for which B_β is not (yet) defined, we know that π(y) = c^j_%R ∈ B_β^Ji. From Lemma 4.13, we obtain that then either (i) there is a B_β ⊆ BC_R(O) with c^j_%R ∈ (d

B_β)^Ij and O |= d

B_β v B_β, or (ii) c^j_%R ∈ B_β^Ij and WO(c^j_%R)6=∅.

• If WO(c^j_%R) 6= ∅, by Definitions 4.5 and 3.1 and Proposition 3.3, there must be a set B ⊆ BC_R(O) that is a witness of ∃R⁻ w.r.t. O, and fur-thermore an element c^j_%0 ∈ ∆^j_u that satisfies d

B in I_j. It remains to show that {∃R⁻} ∈ Con(α,f). But for all A(y) ∈ α with f_x(y) = , we have O |= ∃R⁻ v A by Proposition 3.4, because c^j_%R ∈ B_A(y)^Ij = A^Ij (this follows in both cases (i) and (ii) above). Likewise, for an element (R⁰,C) ∈ (N⁻_R × 2^N−R) ∩range(f_x), we know that c^j_%RR0 ∈ ∆^j_u, and hence c^j_%R ∈ (∃R⁰)^Ij and O |= ∃R⁻ v ∃R⁰ by Proposition 3.4. Hence, we have found a witness query ψ :=∃x.B(x) for α. Moreover, we have I_j |=ψ via the homomorphism that maps x to c^j_%0.

• If WO(c^j_%R) = ∅, then (i) must hold for all concepts B_β above (except in the cases where we have directly defined B_S(y,z) := {∃S2}), and we define B as the union of all B_β. In particular, we obtain that c^j_%R ∈ (d

B)^Ij. To show the two remaining conditions of Definition 3.7, we start again with the concept atoms A(y)∈ψ with f(y) =. We know that

O |=l

B v l

B_A(y) vB_A(y) =A,

as required. Likewise, for any (S₂,C)∈(N⁻_R ×2^N−R)∩range(f) we know that there must be a role atomS(y, z) such thatf(z) = (S₂,C) and O |=S⁰ vS for some S⁰ ∈ C since α is connected. But then we get

O |=l

B vl

BS(y,z) vBS(y,z) =∃S2, as before.

It remains to construct a tree witness query ψ as in Definition 4.6. In this construction, we maintain the invariant that I_j |= ψ via the homomor-phism that maps all variables y_%⁰ with %⁰ ∈range(f) to c^j_%R%0|1. We start by including in ψ all atoms from B|_R(y), hence satisfying the first condition of Definition 4.6 and our invariant since we know that c^j_%R ∈ (d

B)^Ij. The second condition is immediately satisfied by the same arguments as above, by observing that, for any A(y)∈ α with f(y) = , the set B_A(y) ⊆ B con-tains only rigid basic concepts. Likewise, for all (S₂,C)∈range(f) for which there is a role atomS(y, z) withf(y) =for which case (i) above applies, we know that B_S(y,z)⊆ B contains only rigid basic concepts whose conjunction implies ∃S₂. Hence, in this case (S₂,C) is rigidly witnessed in ψ.

However, if (S₂,C) ∈range(f) is such that S₂ is flexible and all role atoms of the form S(y, z) for which f(y) = , O |= S⁰ v S, and S⁰ ∈ C are such thatS⁰ is rigid, thenB contains only∃S₂ andC consists of all these rolesS⁰. If there is an alternative set B_∃S2 ⊆ BC_R(O) with O |= dB_∃S2 v ∃S₂ and c^j_%R ∈ (d

B∃S₂)^Ij, then we can add B∃S₂(y) to ψ. Otherwise, we know that (S₂,C) cannot be rigidly witnessed inψ, and we have to add all atoms S⁰(y, y_(S2,C)) to ψ in order to fulfill the last condition for (S₂,C). Since we

haveO |=S₂ vS⁰ for all these S⁰, the above atoms can be mapped into I_j as required for our invariant. Since (S₂,C) is not rigidly witnessed inψ, we further need to consider the atoms of α that are mapped below c^j_%RS2. We continue the construction of ψ by induction on the structure of f, until all remaining paths are already rigidly witnessed in ψ or we have reached a leaf of the tree described by f.

Assume hence that we have already defined ψ up to a variable of the form y_%⁰, but that some %⁰ ·(S₂,C) ∈ range(f) is not rigidly witnessed.

We consider first all concept atoms A(y) ∈ α with f(y) = %⁰ · (S₂,C).

Since c^j_%R%0|₁S2 ∈ A^Ji, by Lemma 4.13 we know that either (i’) there is a set B_A ⊆ BC_R(O) with O |= d

B_A v A and c^j_%R%0|₁S2 ∈ (d

B_A)^Ij, or (ii’) c^j_%R%0|1S2 ∈ A^Ij and WO(c^j_%R%0|1S2) 6= ∅. In case (i’), we add the atoms BA(y_%⁰_·(S2,C)) to ψ to satisfy the corresponding condition of Definition 4.6, while maintaining our invariant. In case (ii’), there must be a prefix%⁰⁰R⁰ of

%R%⁰|₁S₂ and a set B_R⁰ ⊆ BC_R(O) such that O |= d

B_R⁰ v ∃R⁰ and either c^j_%⁰⁰ ∈(d

BR⁰)^Ij or %⁰⁰∈NI(K)∪N^aux_I and %⁰⁰ ∈(d

BR⁰)^Ij. If %⁰⁰R⁰ is a prefix of %R, then this contradicts the fact that W_O(c^j_%R) = ∅. But if %R is a prefix of %⁰⁰R⁰, we would have added B_R⁰(y_%⁰⁰⁰) to ψ earlier, where %⁰⁰⁰ is the path in range(f) corresponding to %⁰⁰, and hence%⁰·(S₂,C) would be rigidly witnessed in ψ. This shows that case (ii’) is impossible.

Consider now any successor%⁰·(S₂,C)·(S₃,C⁰)∈range(f) and all role atoms S(y, z)∈αwithf(y) =%⁰·(S₂,C),f(z) = %⁰·(S₂,C)·(S₃,C⁰), andO |=S⁰ vS for someS⁰ ∈ C. If S₃ is rigid or there is one such atom where S⁰ is flexible (and hence S⁰ = S), then we obtain as above that c^j_%R%0|1S2 ∈ (∃S₃)^Ji. By Lemma 4.13 and the same arguments as above, we know that there is a set B_∃S3 ⊆ BC_R(O) with O |= dB_∃S3 v ∃S₃ and c^j_%R%0|1S2 ∈ (dB_∃S3)^Ij. Hence, we can add the atoms B∃S₃(y_%⁰·(S₂,C)) to ψ in order to ensure that

%⁰ ·(S₂,C)·(S₃,C⁰) is rigidly witnessed in ψ. If S₃ is flexible and there is no such atom and no set B_∃S3 as above, then we again have to add all the atoms S⁰(y_%⁰·(S₂,C), y_%⁰·(S₂,C)·(S₃,C⁰)) to ψ and continue the construction with

%⁰·(S₂,C)·(S₃,C⁰), which cannot be rigidly witnessed.

The construction of ψ terminates since f is finite, and when it does we have added enough rigid atoms to ψ in order to satisfy Definition 4.6, and furthermore know that I_j |=ψ.

In both cases, we have found a witness query ψ for αsuch that I_j |=ψ, and thus we obtain a contradiction from (R5) and Proposition 3.5 (recall that α ∈Q^¬_R).

(II) In the remainder of the proof, let π be such that it maps at least one term into ∆_n := N_I(K) ∪ ^Sn+k_j=0 ∆^Ia^j ∪ ∆^I_t^j. In this case, we directly define a homomorphism π⁰ of α intoI_i in order to obtain a contradiction to the fact that I_i |=¬α. We start defining π⁰ for all terms t∈N_V(α)∪N_I(α) for which we have

π(t) ∈∆_n: if π(t) = e^j for e ∈ N_I(K)∪N^aux_I ∪N^tree_I , then we set π⁰(t) := eⁱ. We first show that

for all B ∈BC(O) and all t ∈N_V(α)∪N_I(α) withπ(t)∈∆_n,

we have π⁰(t)∈B^Ii whenever π(t)∈B^Ji. (1) By Lemma 4.13, π(t)∈ B^Ji implies that (i) π(t)∈B^Ii, or (ii) π(t) ∈∆^Ia^j∪∆^I_t^j and there is a set B ⊆BCR(O) such that π(t) ∈(d

B)^Ij and O |=d

B v B. In case (i), we haveπ⁰(t) =π(t), and hence the claim holds. In case (ii), ifi=j, the claim follows as in case (i); otherwise, we further distinguish the following two cases.

• Ifπ(t)∈∆^Ia^j, thenπ(t) must be of the forma^j_x. Letβ ∈ Q_φbe the (unique) CQ containing the variable x. By (R3), the existence of the element a^j_x implies that β ∈ Q_R. Hence, the element of the form aⁱ_x must also exist, i.e., π⁰(t) = aⁱ_x is well-defined. Since only A_QR, A_Qι(j), and A_RF contain assertions abouta_x, by Definition 3.1 and Proposition 3.3 we know that the elements of B are implied by the conjunction of

– all elements of BC⁻(a_x, β), and

– all rigid concepts∃R for which there is∃S(a_x)∈R_F withO |=SvR.

But for all concepts of the latter form, (R6) implies that ∃R(a_x) is already implied by some hO,A_R∪ A_QR ∪ A_Q

ι(i0) ∪ A_i⁰i, and hence must follow also from d

BC⁻(a_x, β). This shows that O |=d

BC⁻(a_x, β)v d

B, and hence B⁰(a_x) ∈ A_QR for all B⁰ ∈ B. Since I_i |= A_QR and I_i |=O, we obtain that aⁱ_x ∈(d

B)^Ii ⊆B^Ii, as required.

• Ifπ(t)∈∆^I_t^j, then π(t) is of the form a^j_b%. In this case, the elementaⁱ_b% also exists since A_RF is the same for all time points. By Proposition 3.3 and the definition ofA_RF,π(t)∈B^Ij implies thatB subsumes the conjunction of all rigid basic concepts satisfied by c_b% in some IhO,{∃S(b)}i where ∃S(b)∈ R_F. Another application of Proposition 3.3 yields that B is also satisfied by π⁰(t) = aⁱ_b% in I_i.

This concludes the proof of (1).

In particular, it follows that all concept atoms in α that are of the form A(t) with π(t) ∈ ∆_n are satisfied by π⁰ in I_i. We proceed by showing that this is also the case for all role atoms in α involving only terms of the above form.

Let hence R(t, t⁰) ∈ α be such that π(t), π(t⁰) ∈ ∆n. If π(t) and π(t⁰) are both contained in ∆^Ii, the claim follows immediately from Lemma 4.12 and the fact that π⁰(t) =π(t) andπ⁰(t⁰) =π(t⁰). If this is not the case, then since in J_i there are no role connections between elements of different sets ∆^Ia^j∪∆^I_t^j,π(t) andπ(t⁰) must both belong to some N_I(K)∪∆^Ia^j∪∆^I_t^j for a fixed j 6=i, and it remains to consider the following cases:

• R is rigid and at least one ofπ(t) orπ(t⁰) is contained in ∆^Ia^j, but neither is contained in ∆^I_t^j. Since (π(t), π(t⁰))∈R^Ji, we know that (π(t), π(t⁰))∈R^Ij. By Definition 3.1, there must be an assertion S(τ(t), τ(t⁰)) ∈ A_QR ∪ A_Qι(j) such that O |= S v R, where τ(t) :=e if π(t) = e^j. By Definition 4.4, we get R(τ(t), τ(t⁰))∈ A_QR. SinceI_i |=A_QR, we conclude (π⁰(t), π⁰(t⁰))∈R^Ii.

• R is rigid and at least one of π(t) or π(t⁰) is contained in ∆^I_t^j. We again have (π(t), π(t⁰)) ∈ R^Ij. By Definition 3.1 and the definition of A_RF, we know thatR(τ(t), τ(t⁰))∈ A_RF. SinceI_i |=A_RF, we get (π⁰(t), π⁰(t⁰))∈R^Ii.

• R is flexible. Since WO is not defined for elements of ∆n, there must be a rigid role S such that (π(t), π(t⁰)) ∈ S^Ij and O |= S v R. As in the two previous cases, it follows that (π⁰(t), π⁰(t⁰))∈S^Ii. Since I |=O, we obtain (π⁰(t), π⁰(t⁰))∈R^Ii.

It remains to defineπ⁰ for the variables ofαthat are mapped byπ into^Sn+k_j=0 ∆^Iu^j. Consider any such variable that is mapped to c^j_eRR1...Rm for some individual name e ∈ NI(K)∪N^aux_I ∪N^tree_I . Since α is connected, there must be a variable y with π(y) =c^j_eR and an atom S(t, y)∈α such that O |=Rv S and π(t) =e^j. Recall that for all such atoms we have π⁰(t) = eⁱ. To determine the value of π⁰(y), we consider all atoms of the above form. If i = j, then by Lemmas 4.12 and 4.13 all these atoms can be satisfied by setting π⁰(y) := π(y) = cⁱ_eR. Otherwise, we distinguish the following two cases.

• If W_O(c^j_eR) 6= ∅, then it follows from (e^j, c^j_eR) ∈ R^Ij that (e^j, c^j_eR) ∈ R^Ji. By (1), we infer that eⁱ = π⁰(t) ∈ (∃R)^Ii, and hence the element cⁱ_eR also exists and the pair (eⁱ, cⁱ_eR) satisfies all role atoms S(t, y) that are mapped to (e^j, c^j_eR) by π. Hence, we can set π⁰(y) := cⁱ_eR for all such variablesy.

• If WO(c^j_eR) = ∅, then by the definition of J_i we know that for all atoms S(t, y) as above there is a rigid role S⁰ such that O |= R v S⁰ v S and (e^j, c^j_eR) ∈ (S⁰)^Ji. Note that R must be flexible since otherwise ∃R would be a witness forc^j_eR. In particular, we know thatφis not r-simple. Further-more, since (e^j, c^j_eR) ∈ R^Ij, Definition 3.1 and Proposition 3.3 yield that the assertion ∃R(e) is implied by the basic concepts obtained from the as-sertions involving e inKⁱ_R. We now show by a case distinction on the form of e that this is already the case if we ignore the assertions in A_RF.

– If e ∈ N_I(K), then the basic concepts obtained from the assertions about e in A_RF are of the form ∃R⁰ for rigid roles R⁰. Since A_R is an ABox type and Ij is a model of both AR and AR_F, we must have

∃R⁰(e) ∈ A_R, and hence the assertions about e in A_RF are subsumed byA_R. Hence, ∃R(e) follows fromKⁱ_R already without A_RF.

– If e = a_x ∈ N^aux_I and β is the CQ in which x appears, then we know that ∃R is implied by the conjunction of all elements of BC⁻(a_x, β) and all rigid concepts ∃R⁰ for which there is a ∃R⁰⁰(a_x) ∈ R_F with O |= R⁰⁰ v R⁰. By (R6) and Proposition 3.3, all concepts of the latter form are implied by d

BC⁻(a_x, β), and hence ∃R⁰(a_x) must be contained in A_QR (see Definition 4.4). This shows again that ∃R(e) follows from K_Rⁱ without A_RF.

– If e ∈ N^tree_I , then ∃R(e) must follow exclusively from A_RF (and O).

Since A_RF contains only rigid assertions, the corresponding rigid ba-sic concepts thus constitute a witness for c^j_eR, which contradicts our assumption.

We have thus shown the entailment required to apply (R6), and infer that

∃R(e) ∈ A_RF. Since I_i |= A_RF, this means that (eⁱ, aⁱ_eR) ∈ (S⁰)^Ii holds for all rigid roles S⁰ identified above. This shows tat (eⁱ, aⁱ_eR) satisfies all role atoms S(t, y) that are mapped to (e^j, c^j_eR) by π. We can thus define π⁰(y) :=aⁱ_eR.

We have thus definedπ⁰for all variables that are directly connected to some termt withπ(t)∈∆_n. We proceed to defineπ⁰ by induction on the tree structure of the homomorphism π into J_i below these variables. It is easy to see that we do not have to change π on the variables y with π(y)∈∆^I_uⁱ; hence, in the following we consider only the case that π(y)∈∆^Iu^j with j 6= i. In the construction of π⁰, we maintain the invariant that whenever π(y) = c^j_%R, then either (i) WO(c^j_%R) = ∅ and π⁰(y) = aⁱ_%R or (ii)WO(c^j_%R)6=∅ and π⁰(y) is of the form cⁱ_%0R.

Hence, assume that π(y) = c^j_%R ∈ ∆^Iu^j and π⁰(y) has already been defined. We first show that all concept atomsA(y)∈αare satisfied byπ⁰. Sincec^j_%R ∈A^Ji, we know by Lemma 4.13 that either (i’) there is a B ⊆BC_R(O) with c^j_%R ∈(dB)^Ij and O |= d

B v A; or (ii’) c^j_%R ∈ A^Ij and WO(c^j_%R) 6= ∅. If case (ii) from above applies, i.e., we have π⁰(y) =cⁱ_%⁰_R, then two applications of Proposition 3.4 yield that O |= ∃R⁻ v A and π⁰(y) = cⁱ_%0R ∈ A^Ii. If case (i) applies, then (i’) must hold, and we know by Proposition 3.4 that O |= ∃R⁻ v d

B, and hence π⁰(y) =aⁱ_%R ∈ (d

B)^Ii by the definition of A_RF. Since I_i |=O, we conclude that π⁰(y)∈A^Ii.

To continue the definition of π⁰, consider a fixed π⁰(y⁰) that has already been

To continue the definition of π⁰, consider a fixed π⁰(y⁰) that has already been

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