In this section, we give a proof of theorems 3, 4, and 5. The set Atom consists of all concepts in NC, all concepts of the form ran(r), and all concepts of the form dom(r).
The following lemma is the first main step towards a proof of the theorems. It states that it is suffient to consider inclusions of a rather simple form only; namely inclusions in which either a member of Atom is on the left hand side or a concept name on the right hand side.
Lemma 8. Let T be an ELHr-terminology andΣ a finite signature.
(i) AnELran,0-TBoxT0is a conceptΣ-interpolant for T if sig(T0) ⊆ Σ, T |= T0 andT0 |= α whenever T |= α, for all Σ-assertions of the form
r v s, C0v A, D v C, whereC0is anCran,0-concept,D ∈ Atom, and C an E L-concept.
(ii) AnELran-TBoxT0is a instanceΣ-interpolant for T if sig(T0) ⊆ Σ, T |= T0 andT0 |= α whenever T |= α, for all Σ-assertions of the form
r v s, C0v A, D v C, whereC0is anCran-concept,D ∈ Atom, and C an E L-concept.
(iii) AnELran,u,u-TBoxT0is a queryΣ-interpolant for T if sig(T0) ⊆ Σ, T |= T0 andT0 |= α whenever T |= α, for all Σ-assertions of the form
r v s, C0v A, D v C, whereC0is anCran-concept,D ∈ Atom, and C an Cu,u-concept.
Proof. First recall that it follows from Theorem 2 that it is sufficient to show that T0 with the properties stated in (i) (respectively, (ii) and (iii)) implies the same ELran,0
Σ
-inclusions as T (respectively, the same ELranΣ -inclusions and the same ELran,u,u
Σ -inclusions).
We first consider the claim for concept Σ-interpolants. Suppose T |= C1 v C2, where C1 v C2 is an ELran,0-inclusion with sig(C1 v C2) ∩ Σ = ∅. We prove by induction on the construction of C2that T0|= C1v C2.
Case 1. C2 is a concept name. Then this follows immediately from the condition that T0 |= C v A whenever T |= C v A, for all ELran,0-inclusions C v A with sig(C v A) ⊆ Σ.
Case 2. C2 = D1u D2. Then T |= C1 v D1 and T |= C1 v D2. By IH, T0|= C1v D1and T0 |= C1v D2. Hence T0|= C1v C2.
Case 3. C2 = ∃r.D. The role-depth rd(C) of an ELran,0-concept C is the number of nestings of ∃r.D in C. Now, the proof is by induction on the role-depth rd(C1) of C1. Suppose rd(C1) = 0. Then
C1= ran(s) u A1u · · · u Am
By Theorem 10, there exists a conjunct E ∈ Atom of C1such that T |= E v ∃r.D.
But then T0|= E v ∃r.D and so T0|= C1v ∃r.D.
Now assume that rd(C1) = n + 1. Let
C1= ran(s) u A1u · · · u Amu ∃r1.D1u · · · u ∃rn.Dn.
By Theorem 10 above, there are two possibilities: (i) there exists E ∈ Atom such that
|= C1v E and T |= E v ∃r.D or (ii) there exists riwith T |= riv r and T |= ran(ri) u Div D
If (i), then T0 |= E v ∃r.D, and so T0 |= C1 v ∃r.D. If (ii), then, by IH, T0 |=
ran(ri) u Div D. Moreover, we have T0|= riv r. Hence T0 |= ∃ri.Div ∃r.D and we obtain T0 |= C1v C2, as required.
Now we consider the claim for instance Σ-interpolants. Suppose T |= C1 v C2, where C1 v C2 is an ELran-inclusion with sig(C1 v C2) ∩ Σ = ∅. We prove by induction on the construction of C2that T0|= C1v C2.
Case 1. C2 is a concept name. Then this follows immediately from the condition that T0 |= C v A whenever T |= C v A, for all ELran-inclusions C v A with sig(C v A) ⊆ Σ.
Case 2. C2 = D1u D2. Then T |= C1 v D1 and T |= C1 v D2. By IH, T0|= C1v D1and T0 |= C1v D2. Hence T0|= C1v C2.
Case 3. C2 = ∃r.D. The proof is by induction on the role-depth rd(C1) of C1. Suppose rd(C1) = 0. Then
C1= ran(s1) u · · · ran(sl) u A1u · · · u Am
By Theorem 10, there exists a conjunct E ∈ Atom of C1such that T |= E v ∃r.D.
But then T0|= E v ∃r.D and so T0|= C1v ∃r.D.
Now assume that rd(C1) = n + 1. Let
C1= ran(s1) u · · · ran(sl) u A1u · · · u Amu ∃r1.D1u · · · u ∃rn.Dn. By Theorem 10 above, there are two possibilities: (i) there exists E ∈ Atom such that
|= C1v E and T |= E v ∃r.D or (ii) there exists riwith T |= riv r and T |= ran(ri) u Div D
If (i), then T0 |= E v ∃r.D. If (ii), then, by IH, T0 |= ran(ri) u Di v D. Moreover, we have T0 |= ri v r. Hence T0|= ∃ri.Div ∃r.D and we obtain T0 |= C1v C2, as required.
Now we consider the claim for query Σ-interpolants. Suppose T |= C1 v C2, where C1 v C2is an ELran,u,0-inclusion with sig(C1 v C2) ∩ Σ = ∅. We prove by induction on the construction of C2that T0|= C1v C2.
Case 1. C2 is a concept name. Then this follows immediately from the condition that T0 |= C v A whenever T |= C v A, for all ELran-inclusions C v A with sig(C v A) ⊆ Σ.
Case 2. C2 = D1u D2. Then T |= C1 v D1 and T |= C1 v D2. By IH, T0|= C1v D1and T0 |= C1v D2. Hence T0|= C1v C2.
Case 3. C2 = ∃R.D, where R = r10 u · · · u r0k. The proof is by induction on the role-depth rd(C1) of C1.
By Theorem 9, there exists s such that T |= C1v ∃s.D and T |= s v ri0for i ≤ k.
Suppose rd(C1) = 0. Then
C1= ran(s1) u · · · ran(sl) u A1u · · · u Am
By Theorem 10, there exists a conjunct E ∈ Atom of C1 such that T |= E v ∃s.D.
Therefore, T |= E v ∃R.D. But then T0|= E v ∃R.D and so T0|= C1v ∃R.D.
Now assume that rd(C1) = n + 1. Let
C1= ran(s1) u · · · ran(sl) u A1u · · · u Amu ∃r1.D1u · · · u ∃rn.Dn. By Theorem 10 above, there are two possibilities: (i) there exists E ∈ Atom such that
|= C1v E and T |= E v ∃s.D or (ii) there exists riwith T |= riv s and T |= ran(ri) u Div D
If (i), then T |= E v ∃R.D and so T0|= E v ∃R.D. If (ii), then by IH, T0|= ran(ri)u Di v D. Moreover, we have T0 |= ri v rj0 for j ≤ k. Hence T0 |= ∃ri.Di v ∃R.D and we obtain T0|= C1v C2, as required.
Case 4. C2 = ∃u.D. If T |= C1 v D, then we are done, by IH. Otherwise, by Theorem 9,
(a) there exists a subconcept ran(r) u C0 of C1 such that there exists a sequence r10, . . . , r0nsuch that T |= ∃r.C0v ∃r10. · · · ∃r0n.D or
(b) there exists a sequence r10, . . . , r0nsuch that T |= C1v ∃r01. · · · ∃rn0.D.
For (a), by Lemma 10, we have – T |= dom(r) v ∃r01. · · · ∃rn0.D, or
– T |= r v r01and T |= ran(r) u C0 v ∃r20. · · · ∃r0n.D.
Observe that, in the second case,
T |= C1v ∃r001. · · · ∃rm00.∃r02. · · · ∃rn0.D
for some roles r100, . . . , rm00, and so we are in case (b). In the first case, T |= dom(r) v
∃u.D and so T0 |= dom(r) v ∃u.D. Moreover |= C1 v ∃u.dom(r). Hence T0 |=
C1v C2, as required.
We now consider (b). Take a non-empty sequence r01, . . . , rk0 such that T |= C1v
∃r10. · · · ∃r0k.D. Suppose rd(C1) = 0. Then
C1= ran(s1) u · · · ran(sl) u A1u · · · u Am
By Theorem 10, there exists a conjunct E ∈ Atom of C1 such that T |= E v ∃u.D.
But then T0|= E v ∃u.D and so T0 |= C1v ∃u.D.
Now assume that rd(C1) = n + 1. Let
C1= ran(s1) u · · · ran(sl) u A1u · · · u Amu ∃r1.D1u · · · u ∃rn.Dn. By Theorem 10 above, there are two possibilities: (i) there exists E ∈ Atom such that
|= C1v E and T |= E v ∃u.D or (ii) there exists riwith T |= riv r01and T |= ran(ri) u Div ∃u.D
If (i), then T0 |= E v ∃u.D. If (ii), then, by IH, T0 |= ran(ri) u Di v ∃u.D. Hence T0|= ∃ri.Di v ∃u.D and we obtain T0 |= C1v C2, as required.
We start with restating and proving Theorem 4.
Theorem 13. Let Σ be a finite signature and T a normalized ELHr-terminology with-outΣ-loops.
Then the algorithms computingPΣ(A) and QΣ(A) (in the instance case) in Fig-ures 3 and 4 terminate for allA ∈ sig(T ). Let TΣi consist of all inclusions, whereA, r, ands range over sig(T ) ∩ Σ:
– r v s, for all r vT s;
– D v A, for all D ∈ PΣ(A);
– A v D, for all D ∈ QΣ(A);
– ran(r) v D, for all D ∈ QΣ(B) such that ran(r) v B ∈ T and B ∈ Σ;
– dom(r) v D, for all D ∈ QΣ(B) such that dom(r) v B ∈ T and B ∈ Σ.
ThenTΣi is an instanceΣ-interpolant of T .
Proof. Termination of the computation of PΣ(A) and QΣ(A) is readily checked and left to the reader. Moreover, it is straighforward to show that T |= α for every α ∈ TΣi. Thus, by Lemma 8, it is sufficient to show for all A ∈ sig(T ):
(∗) If T |= D v A for an CΣran-concept D, then there exists D0 ∈ PΣ(A) such that TΣi |= D v D0.
(∗∗) If T |= A v C for ELΣ-concepts C, then TΣi |= l
D∈QΣ(A)
D v C.
We start with the proof of (∗). The proof is by induction on the construction of D.
Suppose T |= D v A.
– D ∈ Atom. Then D ∈ PΣ(A) and so the claim follows from the definition. clearly, |= D v dom(r). Consider the first case. Assume first that A0 ∈ Σ. Then
∃r.A0 ∈ PΣ(A). We have TΣi |= ∃r.D0 v ∃r.A0, and so we are done. Assume
Proof of (∗∗) The proof is by induction on the construction of C.
Assume C is a concept name and T |= A v C. Then (∗∗) follows immediately from the definition of QΣ(A).
Assume C = ∃s.C0and T |= A v C. We prove this case by induction on the proof
because every Aiis a conjunct of the concept to the left. By considering the last conjunct of E, it follows that |= E v Aifor all Ai. assume that (a) holds. By IH and Lemma 10, there exists a conjunct D0of
l neither Point 1 not Point 2 holds, then there exists B ∈ Σ such that
– ran(s0) v B ∈ T and D0 ∈ QΣ(B).
T |= B v ∃s.C and pd∃s.C0(B) < pd∃s.C0(A). We consider the second case only and leave the first to the reader. Notice that, by IH, TΣi |=d
D∈QΣ(B)D v C.
If B ∈ Σ, then B ∈ QΣ(A). By definition, TΣi |= B v l
D∈QΣ(B)
D
Thus,
TΣi |= l
D∈QΣ(A)
D v C.
If B ∈ Σ and r ∈ Σ, then QΣ(B) ⊆ QΣ(A) and the claim follows by induction. If B ∈ Σ and r ∈ Σ, then
dom(r) v l
D∈QΣ(B)
D ∈ TΣi
and, clearly,
TΣi |= l
D∈QΣ(A)
D v ∃r.>.
Thus
TΣi |= l
D∈QΣ(A)
D v l
D∈QΣ(B)
D,
and, by IH,
TΣi |= l
D∈QΣ(A)
D v C.
We now restate and show Theorem 3.
Theorem 14. Let Σ be a finite signature and T a normalized ELHr-terminology with-outΣ-loops. Define TΣCin the same way asTΣi except thatPΣ(A) is defined using the algorithm in Figure 2. ThenTΣCis a conceptΣ-interpolant of T .
Proof. Clearly, by Lemma 8, the first comments of the proof of Theorem 13 apply to this case as well. As QΣ(A) has not changed, it is sufficient to prove for all A ∈ sig(T ):
(∗) If T |= D v A for a CΣran,0-concept D, then there exists D0 ∈ PΣ0 (A) such that TΣC|= D v D0.
To prove (∗), we require the following property of PΣ(A):
(A) If ran(r) u C ∈ PΣ(A), s vT r, and s ∈ Σ, then ran(s) u C ∈ PΣ(A).
(A) is easily proved by induction on the construction of PΣ(A). Clearly, (A) holds as well if we replace PΣ(A) by PΣ0(A).
Now the proof of (∗) is by induction on the construction of D. The proof is similar to the proof of (∗) for Theorem 13 and we focus on the new steps.
– D ∈ Atom. Then D ∈ PΣ(A) and so the claim follows from the definition.
– D = D1u D2. If A has no definition of the form A ≡ B1u · · · u Bnin T , then the proof is the same as for Theorem 13 and left to the reader.
Assume now that A ≡ B1u· · ·uBn ∈ T . Then T |= D1uD2v Bi, for all Bi. We may assume that none of the Bihas a definition of the form Bi≡ B10u · · · Bk0 in T . Thus, by Lemma 10, for every i there exists Ei ∈ {D1, D2} with T |= Ei v Bi. Hence, by IH, for every Bithere exists Ei0 ∈ PΣ0 (Bi) with TΣC|= Eiv Ei0. Recall that D1u D2is a Cran,0-concept. Thus, there is at most one conjunct of the form ran(r) in D1u D2 (which might have more than one occurrence). Assume this conjunct is ran(r) (the case where there is none is simpler and left to the reader).
Let
Γ = {ran(s) | ∃i ≤ n such that ran(s) is a conjunct of Ei0}.
As TΣC|= D1u D2v Ei0for i ≤ n and by the tree-model property, we obtain that T |= r v s, for all ran(s) ∈ Γ . As r ∈ Σ, we obtain by (A) that Ei00∈ PΣ0 (Bi) for the concept Ei00obtained from Eiby replacing ran(s) by ran(r). Observe that we still have TΣC|= Eiv Ei00. But now
D0= l
Bi∈Σ
Biu l
Bi∈Σ
Ei00∈ PΣ0 (A)
as the only possible conjunct of the form ran(s) in any Ei00is ran(r). Finally, we still have TΣC|= D v D0. Thus, the claim is proved.
– Let D = ∃r.D0. Assume that A ≡ ∃s.A0∈ T ; the other cases are left to the reader.
Then, by Lemma 10, we have the following cases:
• T |= r v s and T |= ran(r) u D0 v A0.
• T |= dom(r) v A.
As in the proof of Theorem 13, the second case and the case A0 ∈ Σ are straigh-forward. Consider the first case under the assumption A0 ∈ Σ. Then, if A0 has no definition of the form A0 ≡ B1u · · · u Bn, then we have (a) T |= ran(r) v A0 or (b) T0 |= D0v A0. Let (a) hold. Then ran(r) ∈ PΣ0(A0). Thus, ∃r.> ∈ PΣ0(A) and TΣC |= ∃r.D0 v ∃r.>, and we are done. If (b), then, by IH, there exists D00 ∈ PΣ0 (A0) such that TΣC |= D0 v D00. As, by definition, D0 does not can-tain any conjunct of the form ran(r), if follows that D00cannot contain any such conjunct. But then ∃r.D00∈ PΣ0 (A) and TΣC|= ∃r.D0v ∃r.D00, and we are done.
Now let A0 ≡ B1u · · · u Bn ∈ T . Then, for each Bi, we may assume that T |=
ran(r) v Bior T |= D0v Bi, by Lemma 10. Now the proof is similar to the proof above and left to the reader.
Extend the relation ≺Σto ≺uΣby adding (A, B) to ≺Σif there are A ./ ∃r.A0 ∈ T such that CTΣ(r) = ∅ and ran(r) v B ∈ T . T contains Σ-u-loops if ≺uΣ contains a cycle.
Now, we extend the notion of a proof depth to Cu-concepts. Let A ∈ NCand C be a Cu-concept of the form C = ∃S.C1, where S is either a role name or a conjunction of roles, such that for a ELHr-terminology T we have T |= A v C. By Theorem 9, whenever ∃(t1u· · ·utl).D is a sub-concept of C for some ti∈ NRand D a Cu-concept, there exists r ∈ NRsuch that T |= r v ti, 1 ≤ i ≤ l and T |= A v C0, where C0 is obtained from C by replacing ∃(t1u · · · u tl).D with ∃r.D. Let C∗be the result of
recursive replacing of all expressions of the form ∃(t1u · · · u tl).D with ∃r.D for an appropriate r. Then we define pdC(A) = pdC∗(A).
Finally, we restate and proof Theorem 5.
Theorem 15. Let Σ be a finite signature and T a normalized ELHr-terminology with-out Σ-u-loops. Define TΣq in the same way as TΣi with the exception thatQΣ(A) is defined using the algorithm in Figure 5. ThenTΣqis a queryΣ-interpolant of T . Proof. Termination of the computation of PΣ(A) and QΣ(A) is readily checked and left to the reader. Moreover, it is straighforward to show that T |= α for every α ∈ TΣq. only. Again, the proof is similar to (∗∗) for Theorem 13.
The proof is by induction on the construction of C.
Assume C is a concept name and T |= A v C. Then (∗∗) follows immediately from the definition of QΣ(A).
Σ-concept. We prove this case by induction on the proof depth of A wrt. ∃S.C0.
Assume C0= A1u · · · u Aku ∃r1.C1u · · · ∃rm.Cm. Then
|= l
B∈Σ,T |=ran(r)uA0vB
B v A1u · · · u Ak
because every Aiis a conjunct of the concept to the left. By considering the last conjunct of E, it follows that |= E v Aifor all Ai.
For each ∃ri.Ciwe have
– (a) T |= ran(r) v ∃ri.Cior (b) T |= A0v ∃ri.Ci, by Lemma 10.
Fix an ∃ri.Ci. Assume first that (b) holds. If A0∈ Σ, then A0is a conjunct ofd
B∈Σ,T |=ran(r)uA0vBB and, therefore, |= E v ∃ri.Ci. If A06∈ Σ, then by IH,
T |= l
D∈QΣ(A0)
D v ∃ri.Ci.
By considering the second conjunct of E, we have |= E v ∃ri.Ci. Now assume that (a) holds. By IH and Lemma 10, there exists a conjunct D0of
l
B∈Σ,ran(r)vB∈T ,D∈QΣ(B)
D u l
ran(r)vB∈T ,B∈Σ
B
such that TΣq |= D0 v ∃ri.Ci. Similarly to the proof of Theorem 13, we consider the following cases. If
– ran(r) v D0∈ T or
– D0 ∈ QΣ(B) with ran(r) v B ∈ T and ran(s0) v B 6∈ T for all s0∈ S0,
then D0 is a conjunct of E and, therefore, TΣq |= E v ∃ri.Ci. On the other hand, if neither Point 1 not Point 2 holds, then there exists B ∈ Σ such that
– ran(s0) v B ∈ T for some s0∈ S0and D0 ∈ QΣ(B).
We have
ran(s0) v l
D00∈QΣ(B)
D00∈ TΣq,
for some s0∈ S0and, therefore, TΣq |= ∃S0.> v ∃S.∃ri.Ci.
Summarising we obtain that that TΣq|= ∃S0.E v ∃S.C0, as required.
The proofs of the case when pd∃S.C0(A) > 0 is exactly the same as in the proof of Theorem 13 and left to the reader.
Assume C = ∃u.C0and T |= A v C. Then, by Theorem 9, there exists r10, . . . , r0n such that T |= A v ∃r10. . . ∃rn0.C0.
We prove by induction on the sequence length k that whenever T |= A v ∃t1. . . ∃tkC0 for some A ∈ NC∩sig(T ) and ti∈ NR∩sig(T ) we have TΣq |=d
D∈QΣ(A)D v ∃u.C0. If k = 0 the claim is trivial. Assume now we proved the claim for all k < n. Let T |= A v ∃r01. . . ∃rn0.C0. We proceed by induction on the proof depth of A wrt.
∃r10. . . ∃rn0.C0.
Assume pd∃r0
1...∃rn0.C0(A) = 0. Then A ./ ∃r.A0 ∈ T for some A ∈ NC and T |= r v r10 such that T |= ran(r) u A0 v ∃r02. . . ∃r0n.C0.
Let S0=d
s0∈CTΣ(r)s0, if CTΣ(r) 6= ∅, or u otherwise. Note that if r10 ∈ Σ we have CTΣ(r) 6= ∅ and for some s0 ∈ S0we have T |= s0v r01.
Now QΣ(A) contains ∃S0.E, where
E = l
B∈Σ, ran(r)vB∈T
∀s0∈S0(ran(s0)vB6∈T ), D∈QΣ(B)
D u l
D∈QΣ(A0) A0∈Σ
D u l
B∈Σ T |=ran(r)uA0vB
B.
Assume first that n ≥ 2. Then we have
– (a) T |= ran(r) v ∃r20. . . ∃rn0.C0or (b) T |= A0 v ∃r02. . . ∃r0n.C0, by Lemma 10.
Assume first that (b) holds. By IH TΣq |= d
D∈QΣ(A0)D v ∃u.C0. Thus TΣq |= E v
∃u.C0. Now assume that (a) holds. By IH and Lemma 10, there exists a conjunct D0of l
B∈Σ,ran(r)vB∈T ,D∈QΣ(B)
D u l
ran(r)vB∈T ,B∈Σ
B
such that TΣq |= D0v ∃u.C0. Again, similarly to the proof of Theorem 13, if – ran(r) v D0∈ T or
– D0 ∈ QΣ(B) with ran(r) v B ∈ T and either S0 = u or ran(s0) v B 6∈ T for all s0∈ S0,
then D0 is a conjunct of E and, therefore, TΣq |= E v ∃u.C0. On the other hand, if neither Point 1 not Point 2 holds, then there exists B ∈ Σ such that
– ran(s0) v B ∈ T for some s0∈ S0and D0 ∈ QΣ(B).
We have
ran(s0) v l
D00∈QΣ(B)
D00∈ TΣq,
for some s0∈ S0and, therefore, TΣq |= ∃S0.> v ∃S.∃u.C0.
Summarising we obtain that that TΣq|= ∃S0.E v ∃u.C0, as required.
The case of n = 1 can be proved by combining the reasoning above and the one needed for proving the case C = ∃s01u · · · u s0k.
The proofs of the case when pd∃s.C0(A) > 0 is exactly the same as in the proof of Theorem 13 and left to the reader.
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