Decidability of Description Logics with Transitive Closure of Roles

〈a ′ (1,1) , a′′ 〉 ∈ P21 11 I
. Therefore, a
′
(1,1)
= a, and thus
a (1,1) = a ′ (1,1) .
• Assume that k ≥ 0 or l ≥ 0. We consider the following
cases:
– Assume a (k,l) ∈ A I with (k mod 2 = 0) ∧
(l mod 2 = 0). By the axiom 10 in Definition
8 there are a (k+1,l) ∈ B I , a (k,l+1) ∈ C I such
that 〈a (k,l) , a (k+1,l) 〉 ∈ X1 1 I
, 〈a(k,l) , a (k,l+1) 〉 ∈
Y1
1 I
. Moreover, by the axioms 11, 12 in Definition
8 there are a (k+1,l+1) , a ′ (k+1,l+1)
∈ D I such that
〈a (k+1,l) , a (k+1,l+1) 〉 ∈ Y2
1 I
, 〈a(k,l+1) , a ′ (k+1,l+1) 〉 ∈
X 1 2
I
. We show that a
′
(k+1,l+1)
= a (k+1,l+1) .
By the axiom 10 in Definition 8, let a ∈ D I such that
〈a (k,l) , a〉 ∈ ε I AD . From the axiom 1 in Definition
8 we have 〈a (k,l) , a (k+1,l) 〉, 〈a (k+1,l) , a (k+1,l+1) 〉 ∈
P12 11 I
. If a(k+1,l+1) ≠ a then, by the axioms 3,
5 in Definition 8 there is an instance a ′ such that
〈a (k+1,l+1) , a ′ 〉 ∈ P12 11 I
, which contradicts the axiom
13 in Definition 8 since a (k+1,l+1) ∈ D I and
〈a (k+1,l+1) , a ′ 〉 ∈ P12 11 I
. Thus, a(k+1,l+1) = a. Analogously,
from the axiom 1 in Definition 8 we have
〈a (k,l) , a (k,l+1) 〉, 〈a (k,l+1) , a ′ (k+1,l+1) 〉 ∈ P 21 11 I
. If
a ′ (k+1,l+1) ≠ a then, by the axioms 3, 5 in Definition
8 there is an instance a ′′ such that 〈a ′ (k+1,l+1) , a′′ 〉 ∈
P 11
I
21 , which contradicts the axiom 13 in Definition 8
since a ′ (k+1,l+1) ∈ DI and 〈a ′ (k+1,l+1) , a′′ 〉 ∈ P21 11 I
.
Therefore, a ′ (k+1,l+1)
= a, and thus a (k+1,l+1) =
a ′ (k+1,l+1) .
Obviously, if (k mod 2 = 0) and (l mod 2 = 0) then
((k + 1) mod 2 = 1) and ((l + 1) mod 2 = 1)
– Assume a (k,l) ∈ D I with (k mod 2 = 1) ∧ (l mod 2 =
1). Similarly.
– Assume a (k,l) ∈ B I with (k mod 2 = 1) ∧ (l mod 2 =
0). Similarly.
– Assume a (k,l) ∈ C I with (k mod 2 = 0) ∧ (l mod 2 =
1). Similarly. □
□
We now define a mapping t : N × N → D as follows.
By the axiom 8 in Definition 8, there is D i ∈ D such that
a (0,0) ∈ D I i .
1. t(0, 0) := D i with a (0,0) ∈ D I i . From the axioms 9, 2, 6
in Definition 8 and Claim 1, there are D x (0,0) , D y
(0,0) ∈
D such that (D i , D x ) ∈ H, (D i , D y (0,0) ) ∈ V,
and 〈a (0,0) , a (1,0) 〉 ∈ X I with a (1,0) ∈ D x
(0,0)
〈a (0,0) , a (0,1) 〉 ∈ Y I with a (0,1) ∈ D y
(0,0) I
. Therefore,
we define t(1, 0) := D x (0,0) , t(0, 1) := D y (0,0) . Since
X, Y are functional and D h are disjoint for all D h ∈ D
hence such D x (0,0) , D y
(0,0) are uniquely determined from
D i .
I
,
Moreover, from the axiom 9, 2, 6 in Definition 8, there
are D y (1,0) , D x (0,1) ∈ D such that (D x (0,0) , D y (1,0) ) ∈ H,
(D y (0,0) , D x (0,1) ) ∈ V, and 〈a (1,0) , a (1,1) 〉 ∈ Y I with
a (1,1) ∈ D y
(1,0) I
, 〈a(0,1) , a ′ (1,1) 〉 ∈ XI with a ′ (1,1)
∈
D x
(0,1) I
. By the axioms 11, 12, 2, 6 in Definition 8
we have 〈a (1,0) , a (1,1) 〉 ∈ Y2
1 I
, 〈a(0,1) , a ′ (1,1) 〉 ∈ X1 I
2 .
From Claim 1 we have a (1,1) = a ′ (1,1). This implies
that D y (1,0) = D x (0,1) since D y (1,0) , D x
(0,1) are disjoint by
the axiom 8 in Definition 8. Therefore we can define
t(1, 1) := D y (1,0) = D (0,1)
x .
2. Assume that t(i, j) := D i ′ with a(i, j) ∈ D I i ′ . From
the axiom 9, 2, 6 in Definition 8 and Claim 1, there
are D x
(i,j) , D y (i,j) ∈ D such that (D x
(i,j) , D y (i,j) ) ∈
H, (D x
(i,j) , D y
(i,j) ) ∈ V, and 〈a (i,j) , a (i+1,j) 〉 ∈ X I
with a (i+1,j) ∈ D (i,j) I
, 〈a(i,j) , a (i,j+1) 〉 ∈ Y I
with a (i,j+1) ∈ D (i,j)
y
D (i,j)
x
x
, t(i, j+1) := D (i,j)
y
I
. Therefore, t(i + 1, j) :=
. Since X, Y are functional and
, D y
(i,j)
D h are disjoint for all D h ∈ D hence such D x
(i,j)
are uniquely determined from D i ′.
Moreover, from the axiom 9, 2, 6 in Definition 8, there
are D y
(i+1,j) , D x
(i,j+1) ∈ D such that (D x
(i,j) , D y
(i+1,j) ) ∈
H, (D y
(i,j) , D x
(i,j+1) ) ∈ V, and 〈a (i+1,j) , a (i+1,j+1) 〉 ∈
Y I with a (i+1,j+1) ∈ D y
(i+1,j) I
, 〈a(i,j+1) , a ′ (i+1,j+1) 〉 ∈
X I with a ′ (i+1,j+1) ∈ I D(i,j+1) x . We now distinguish the
following cases:
(a) Assume that a (i,j) ∈ A I . From Claim 1 and the axiom
8 in Definition 8 we can show D y
(i+1,j) = D x (i,j+1) .
Therefore we can define t(i + 1, j + 1) := D y (i+1,j) =
D x (i,j+1) .
(b) Assume that a (i,j) ∈ B I . Similarly.
(c) Assume that a (i,j) ∈ C I . Similarly.
(d) Assume that a (i,j) ∈ D I . Similarly.
It remains to be shown that (1) t is well defined, (2) the
horizontal and vertical matching conditions are satisfied.
(1) is obvious from the construction of the mapping t.
(2) From the definition of t, for each a (k,l) there is a D i ∈ D
such that t(k, l) = D i and a (k,l) ∈ D I i . Again, by
the construction of t, there are D j , D k ∈ D such that
t(k + 1, l) = D j , t(k, l + 1) = D j and a (k+1,l) ∈ D I j ,
a (k,l+1) ∈ D I k . By the axioms 2 and 9, we have
〈D i , D j 〉 ∈ H and 〈D i , D k 〉 ∈ V.