Decidability of Description Logics with Transitive Closure of Roles

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9. (Y 2 1 ) I = {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 1)∧(l mod 2 = 1)} 10. (P 11 12 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 0)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 0)} 11. (P 11 21 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 1)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 0)} 12. (P 22 12 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 1)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 1)} 13. (P 22 21 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 0)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 1)} 14. (P 21 21 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 0)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 0)} 15. (P 21 12 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 1)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 0)} 16. (P 12 21 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 1)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 1)} 17. (P 12 12 ) I = {〈a (k,l) , a (k+1,l) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 0)}∪ {〈a (k,l) , a (k,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 1)} 18. (ε AD ) I = {〈a (k,l) , a (k+1,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 0)} 19. (ε DA ) I = {〈a (k,l) , a (k+1,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 1)} 20. (ε BC ) I = {〈a (k,l) , a (k+1,l+1) 〉 | (k mod 2 = 1) ∧ (l mod 2 = 0)} 21. (ε BC ) I = {〈a (k,l) , a (k+1,l+1) 〉 | (k mod 2 = 0) ∧ (l mod 2 = 1)} 22. D i I = {a (k,l) | t(k, l) = D i } for each D i ∈ D 23. A I = {a (k,l) | (k mod 2 = 0) ∧ (l mod 2 = 0)} 24. D I = {a (k,l) | (k mod 2 = 1) ∧ (l mod 2 = 1)} 25. B I = {a (k,l) | (k mod 2 = 1) ∧ (l mod 2 = 0)} 26. C I = {a (k,l) | (k mod 2 = 0) ∧ (l mod 2 = 1)} 27. X I = X1 1 I ∪ X 1I 2 ∪ X 2I 1 ∪ X 2I 2 28. Y I = Y1 1 I ∪ Y 1I 2 ∪ Y 2I 1 ∪ Y 2I 2 We now check that I satisfies all axioms in Definition 8. 1. X i r ⊑ P ij ⊑ Prs ij for all i, j, r, s ∈ {1, 2}, r ≠ s. For each k, l ≥ 0, we consider the following cases: rs, Ys j • Assume (k mod 2 = 0) ∧ (l mod 2 = 0). From the assertions 2, 6, we have 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I , and 〈a (k,l) , a (k,l+1) 〉 ∈ Y1 1 I . From the assertions 10, 11 we have 〈a (k,l) , a (k+1,l) 〉 ∈ P12 11 I and 〈a(k,l) , a (k,l+1) 〉 ∈ P21 11 I . • Assume (k mod 2 = 1) ∧ (l mod 2 = 0). Similarly. • Assume (k mod 2 = 0) ∧ (l mod 2 = 1). Similarly. • Assume (k mod 2 = 1) ∧ (l mod 2 = 1). Similarly. 2. Xr i ⊑ X, Yr i ⊑ Y for all i, r ∈ {1, 2}. From assertions 27 and 28. 3. ε AD ⊑ (P 11 12 ) + , ε AD ⊑ (P 11 21 ) + . For each k, l ≥ 0, we consider the following cases: Y 1 2 • Assume (k mod 2 = 0) ∧ (l mod 2 = 0). From the assertion 18, we have 〈a (k,l) , a (k+1,l+1) 〉 ∈ ε I AD . From the assertions 2 and 8 it follows that 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I , 〈a(k+1,l) , a (k+1,l+1) 〉 ∈ I (note that (k + 1 mod 2 = 1)). By the assertion 10 we have 〈a (k,l) , a (k+1,l) 〉 ∈ P 11 12 I and 〈a (k+1,l) , a (k+1,l+1) 〉 ∈ P12 11 I . This implies that 〈a (k,l) , a (k+1,l+1) 〉 ∈ (P12 11 ) +I . On the other hand, from the assertions 6 and 4 we have 〈a (k,l) , a (k,l+1) 〉 ∈ Y1 1 I , 〈a(k,l+1) , a (k+1,l+1) 〉 ∈ X2 1 I (note that (l + 1 mod 2 = 1)). By the assertion 11 we have 〈a (k,l) , a (k,l+1) 〉 ∈ P21 11 I , 〈a (k,l+1) , a (k+1,l+1) 〉 ∈ P21 11 I . This implies that 〈a (k,l) , a (k+1,l+1) 〉 ∈ (P21 11 ) +I . • Assume (k mod 2 = 1) ∧ (l mod 2 = 0). Similarly. • Assume (k mod 2 = 0) ∧ (l mod 2 = 1). Similarly. • Assume (k mod 2 = 1) ∧ (l mod 2 = 1). Similarly. 4. ε DA ⊑ (P12 22 ) + , ε DA ⊑ (P21 22 ) + . Similarly. 5. ε BC ⊑ (P21 21 ) + , ε BC ⊑ (P12 21 ) + . Similarly. 6. ε CB ⊑ (P21 12 ) + , ε CB ⊑ (P12 12 ) + . Similarly. 7. ⊤ ⊑ ≤ 1Pr,s ij for all i, j, r, s ∈ {1, 2}, r ≠ s. For each k, l ≥ 0, we consider the following cases: • Assume (k mod 2 = 0) ∧ (l mod 2 = 0). From the assertions 10, 11, 14, 17 we have 〈a (k,l) , a (k+1,l) 〉 ∈ P12 11 I , 〈a(k,l) , a (k,l+1) 〉 ∈ P21 11 I , 〈a(k,l) , a (k,l+1) 〉 ∈ I and 〈a(k,l) , a (k+1,l) 〉 ∈ P 12 P 21 21 12 I . • Assume (k mod 2 = 1) ∧ (l mod 2 = 0). Similarly. • Assume (k mod 2 = 0) ∧ (l mod 2 = 1). Similarly. • Assume (k mod 2 = 1) ∧ (l mod 2 = 1). Similarly. 8. ⊤ ⊑ ≤ 1X, ⊤ ⊑ ≤ 1Y . For each k, l ≥ 0, we consider the following cases: • Assume (k mod 2 = 0) ∧ (l mod 2 = 0). From the assertions 2, 6 we have 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I , 〈a (k,l) , a (k,l+1) 〉 ∈ Y1 1 I . From the assertion 28, we have 〈a (k,l) , a (k+1,l) 〉 ∈ X I , 〈a (k,l) , a (k,l+1) 〉 ∈ Y I .
• Assume (k mod 2 = 1) ∧ (l mod 2 = 0). Similarly. • Assume (k mod 2 = 0) ∧ (l mod 2 = 1). Similarly. • Assume (k mod 2 = 1) ∧ (l mod 2 = 1). Similarly. 9. ⊤ ⊑ ≤ 1ε AD . It is obvious from the assertion 18 for each k, l ≥ 0. 10. ⊤ ⊑ ≤ 1ε DA . It is obvious from the assertion 19 for each k, l ≥ 0. 11. ⊤ ⊑ ≤ 1ε BC . It is obvious from the assertion 20 for each k, l ≥ 0. 12. ⊤ ⊑ ≤ 1ε CB . It is obvious from the assertion 21 for each k, l ≥ 0. 13. ⊤ ⊑ ⊔ (D i ⊓ ( ¬D j )). Since t is a tiling, 1≤i≤l 1≤j≤l,j≠i each (k, l) has a unique D i ∈ D such that t(k, l) = D i . Thus, from the assertion 22, each a (k,l) has a unique D i ∈ D such that a (k,l) ∈ D I i . ⊔ ⊔ 14. D i ⊑ ∀X. D j ⊓ ∀Y. for each (D i,D j)∈H (D i,D k )∈V D k D i ∈ D. From the assertion 22, if a (k,l) ∈ D i I then t(k, l) = D i . Since t is a tiling, according to Definition 7 we have 〈D i , D j 〉 ∈ H and 〈D i , D k 〉 ∈ V with t(k + 1, l) = D j and t(k, l + 1) = D k . From the assertions 28 and 2-9 we have 〈a (k,l) , a (k+1,l) 〉 ∈ X I and 〈a (k,l) , a (k,l+1) 〉 ∈ Y I . From the assertion 22, we have a (k+1,l) ∈ D j I and a (k,l+1) ∈ D k I . 15. A ⊑ ¬B ⊓ ¬C ⊓ ¬D ⊓ ∃X1 1 .B ⊓ ∃Y1 1 .C ⊓ ∃ε AD .D ⊓ ∀P12 22 .⊥ ⊓ ∀P21 22 .⊥. For each k, l ≥ 0, we consider the following cases: • Assume (k mod 2 = 0) ∧ (l mod 2 = 0). From the assertions 23 we have a (k,l) ∈ A I . From the assertions 24, 25, 26, we have a (k,l) /∈ B I , a (k,l) /∈ C I , a (k,l) /∈ D I . Moreover, from the assertions 2, 6 we have 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I , 〈a(k,l) , a (k,l+1) 〉 ∈ Y1 1 I . By the assertions 18 and 24 we have 〈a (k,l) , a (k+1,l+1) 〉 ∈ ε I AD and a (k+1,l+1) ∈ D I . Additionally, according to the assertions 12, 13, 〈a (k,l) , a (k+1,l) 〉, 〈a (k,l) , a (k,l+1) 〉 /∈ P12 11 I and 〈a (k,l) , a (k+1,l) 〉, 〈a (k,l) , a (k,l+1) 〉 /∈ P21 11 I . • Assume (k mod 2 = 1) ∧ (l mod 2 = 0). From the assertion 23, it follows a (k,l) /∈ A I . • Assume (k mod 2 = 0) ∧ (l mod 2 = 1). Similarly. • Assume (k mod 2 = 1) ∧ (l mod 2 = 1). Similarly. 16. B ⊑ ¬A ⊓ ¬C ⊓ ¬D ⊓ ∃X2 2 .A ⊓ ∃Y2 1 .D ⊓ ∃ε BC .C ⊓ ∀P21 12 .⊥ ⊓ ∀P12 12 .⊥. Similarly. 17. C ⊑ ¬A ⊓ ¬B ⊓ ¬D ⊓ ∃X2 1 .D ⊓ ∃Y2 2 .A ⊓ ∃ε CB .B ⊓ ∀P21 21 .⊥ ⊓ ∀P12 21 .⊥. Similarly. 18. D ⊑ ¬A ⊓ ¬B ⊓ ¬C ⊓ ∃X1 2 .C ⊓ ∃Y1 2 .B ⊓ ∃ε DA .A ⊓ ∀P12 11 .⊥ ⊓ ∀P21 11 .⊥. Similarly. • ”Only-If-direction”. On the other hand, assume that the concept A is satisfiable w.r.t. the axioms in Definition 8 , and let I = 〈∆ I , . I 〉 be an interpretation such that A I ≠ ∅. Assume that a (0,0) ∈ A I . This interpretation can be used to find a compatible tiling for D. First, we show the following claim: Claim 1 There are individuals a (k,l) ∈ ∆ I with k, l ≥ 0 such that • If (k mod 2 = 0) ∧ (l mod 2 = 0) then a (k,l) ∈ A I . Additionally, there are a (k+1,l) ∈ B I , a (k,l+1) ∈ C I , a (k+1,l+1) ∈ D I such that 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I , 〈a (k+1,l) , a (k+1,l+1) 〉 ∈ Y2 1 I , 〈a(k,l) , a (k,l+1) 〉 ∈ Y1 1 I and 〈a (k,l+1) , a (k+1,l+1) 〉 ∈ X2 1 I . • If (k mod 2 = 1) ∧ (l mod 2 = 1) then a (k,l) ∈ D I . Additionally, there are a (k+1,l) ∈ C I , a (k,l+1) ∈ B I , a (k+1,l+1) ∈ A I such that 〈a (k,l) , a (k+1,l) 〉 ∈ X1 2 I , 〈a (k+1,l) , a (k+1,l+1) 〉 ∈ Y2 2 I , 〈a(k,l) , a (k,l+1) 〉 ∈ Y1 2 I and 〈a (k,l+1) , a (k+1,l+1) 〉 ∈ X2 2 I . • If (k mod 2 = 1) ∧ (l mod 2 = 0) then a (k,l) ∈ B I . Additionally, there are a (k+1,l) ∈ A I , a (k,l+1) ∈ D I , a (k+1,l+1) ∈ C I such that 〈a (k,l) , a (k+1,l) 〉 ∈ X2 2 I , 〈a (k+1,l) , a (k+1,l+1) 〉 ∈ Y1 1 I , 〈a(k,l) , a (k,l+1) 〉 ∈ Y2 1 I and 〈a (k,l+1) , a (k+1,l+1) 〉 ∈ X1 2 I . • If (k mod 2 = 0) ∧ (l mod 2 = 1) then a (k,l) ∈ C I . Additionally, there are a (k+1,l) ∈ D I , a (k,l+1) ∈ A I , a (k+1,l+1) ∈ B I such that 〈a (k,l) , a (k+1,l) 〉 ∈ X2 1 I , 〈a (k+1,l) , a (k+1,l+1) 〉 ∈ Y1 2 I , 〈a(k,l) , a (k,l+1) 〉 ∈ Y2 2 I and 〈a (k,l+1) , a (k+1,l+1) 〉 ∈ X1 1 I . Proof:[Proof of the claim 1] • Assume k = 0, l = 0. We have a (0,0) ∈ A I . By the axiom 10 in Definition 8 there are a (1,0) ∈ B I , a (0,1) ∈ C I such that 〈a (0,0) , a (1,0) 〉 ∈ X1 1 I , 〈a(0,0) , a (0,1) 〉 ∈ Y1 1 I . Moreover, by the axioms 11, 12 in Definition 8 there are a (1,1) , a ′ (1,1) ∈ DI such that 〈a (1,0) , a (1,1) 〉 ∈ Y2 1 I , 〈a (0,1) , a ′ (1,1) 〉 ∈ X1 2 I . We show that a ′ (1,1) = a (1,1) . By the axiom 10 in Definition 8, let a ∈ D I such that 〈a (0,0) , a〉 ∈ ε I AD . From the axiom 1 in Definition 8 we have 〈a (0,0) , a (1,0) 〉, 〈a (1,0) , a (1,1) 〉 ∈ P12 11 I . If a (1,1) ≠ a then, by the axioms 3, 5 in Definition 8 there is an instance a ′ such that 〈a (1,1) , a ′ 〉 ∈ P12 11 I , which contradicts the axiom 13 in Definition 8 since a (1,1) ∈ D I and 〈a (1,1) , a ′ 〉 ∈ P12 11 I . Thus, a(1,1) = a. Analogously, from the axiom 1 in Definition 8 we have 〈a (0,0) , a (0,1) 〉, 〈a (0,1) , a ′ (1,1) 〉 ∈ P 21 11 I . If a ′ (1,1) ≠ a then, by the axioms 3, 5 in Definition 8 there is an instance a ′′ such that 〈a ′ (1,1) , a′′ 〉 ∈ P21 11 I , which contradicts the axiom 13 in Definition 8 since a ′ (1,1) ∈ DI and
Page 1 and 2: Decidability of Description Logics
Page 3 and 4: ⊤ I = ∆ I , (C ⊓ D) I = C I
Page 5 and 6: of x ′ or x ′ is a Inv(R)-succe
Page 7 and 8: B A X, X 2 2 , P 22 21 C D C D Y, Y
Page 9 and 10: 1. I is an interpretation. Indeed,
Page 11: A tree T = (V, E, L) will be obtain

Decidability of Description Logics with Transitive Closure of Roles

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