Decidability of Description Logics with Transitive Closure of Roles

• E(R) = {〈[x], [y]〉 | R ∈ L(〈[x], [y]〉)} ∪ {〈[y], [x]〉 |
Inv(R) ∈ L(〈[x], [y]〉)}
It holds that T is a tableau of D w.r.t. (T , R).
Proof of Lemma 6. Let T = (S, L, E). We will prove that T
satisfies all the properties from Definition 9.
• D ∈ L([x 0 ]) since D ∈ L(x 0 ) according to Definition 4;
• Property P1. Let [x] ∈ S. According to the definition of
neighborhood (tbox-rule) we have nnf(¬C ⊔ D) ∈ L(x)
for all x ∈ V with C ⊑ D ∈ T . This implies that
nnf(¬C ⊔ D) ∈ L(x) for all C ⊑ D ∈ T .
• Property P2 holds since the nodes x ∈ V are built
from valid neighborhoods (i.e. satisfying clash-rule) and
L([x]) = L(x).
• Properties P3, P4 hold thanks to the ⊓-rule and ⊔-rule in
the definition of neighborhoods (Definition 3);
• Property P5. Assume ∀S.C ∈ L([x]) and 〈[x], [y]〉 ∈
E(S). According to the definition of E, we consider the
following cases:
1. S ∈ L(〈[x], [y]〉). By the construction of G, there
are x ′ ∈ [x], y ′ ∈ [y] such that S ∈ L(〈x ′ , y ′ 〉) or
Inv(S) ∈ L(〈y ′ , x ′ 〉). By the construction of T it follows
that x ′ , y ′ are respectively a core and neighbor
node of a neighborhood (x ′ , N, l) with S ∈ l(x ′ , y ′ ),
y ′ ∈ N. By ∀-rule we have C ∈ l(y ′ ). Moreover, by
the construction of T it follows C ∈ L(y ′ ). By the construction
of G, it holds C ∈ L([y]) since L(y) = L(y ′ ),
and thus C ∈ L([y]).
2. Inv(S) ∈ L(〈[y], [x]〉). By the construction of G, there
are x ′ ∈ [x], y ′ ∈ [y] such that Inv(S) ∈ L(〈y ′ , x ′ 〉) or
S ∈ L(〈x ′ , y ′ 〉). By the construction of T it follows
that x ′ , y ′ are respectively a core and neighbor node
of a neighborhood (y ′ , N, l) with Inv(S) ∈ l(y ′ , x ′ ),
x ′ ∈ N. By ∀-rule we have C ∈ l(y ′ ). Moreover, by
the construction of T it follows C ∈ L(y ′ ). By the construction
of G, it holds C ∈ L([y]) since L(y) = L(y ′ )
and thus C ∈ L([y]).
• Property P6. Assume ∃R.C ∈ L([x]). We will show that
there exists [y] ∈ S such that C ∈ L([y]) and 〈[x], [y]〉 ∈
E(R).
By the construction of G, we have ∃R.C ∈ L(x). By
the construction of T, x is a core node of a neighborhood
(x, N, l). By ∃-rule there is a neighbor y ∈ N such that
C ∈ l(y) and R ∈ l(〈x, y〉). Again, by the construction
of T, x has a neighbor y in T such that C ∈ L(y) and
R ∈ L(〈x, y〉) or Inv(R) ∈ L(〈y, x〉). We consider the
following cases:
– R ∈ L(〈x, y〉). By the construction of G, we have C ∈
L([y]) and R ∈ L(〈[x], [y]〉). By the construction of the
tableau T , it holds 〈[x], [y]〉 ∈ E(R) and C ∈ L([y]).
– Inv(R) ∈ L(〈y, x〉). By the construction of G, we have
C ∈ L([y]) and Inv(R) ∈ L(〈[y], [x]〉). By the construction
of the tableau T , it holds 〈[y], [x]〉 ∈ E(R)
and C ∈ L([y]).
• Property P7 is satisfied due to the bidirectional definition
of E.
• Property P8. Assume that 〈[x], [y]〉 ∈ E(Q + ) with Q ∈
R ∪ {Inv(P ) | P ∈ R} and 〈[x], [y]〉 /∈ E(Q). By the
construction of G, there are x ′ ∈ [x], y ′ ∈ [y] such that
Q + ∈ L(〈x ′ , y ′ 〉) and Q /∈ L(〈x ′ , y ′ 〉).
By the construction of T, there is ϕ =
〈x 0 , x 1 , x 2 , · · · , x n , x n+1 〉 with x ′ = x 2 , and y ′ = w
where w = x 1 if x 1 is not blocked or w = z if x 1
is blocked by z. Furthermore, these nodes satisfy the
following conditions
– L(x 1 ) = L(x n+1 ), L(x 0 ) = L(x n ) .
– Q ∈ L ′ (〈x i , x i+1 〉) for all i ∈ {2, · · · , n} where
L ′ (〈x i , x i+1 〉) = L(〈x i , x i+1 〉) if 〈x i , x i+1 〉 ∈ E;
L ′ (〈x i , x i+1 〉) = Inv( L(〈x i+1 , x i 〉) ) if 〈x i+1 , x i 〉 ∈
E; L ′ (〈x i , x i+1 〉) = L(〈x i , z〉) ) if 〈x i , z〉 ∈ E and
x i+1 blocks z.
By the definition of ∼, we have y ′ , x n+1 ∈ [x 1 ] and x 0 ∈
[x n ]. From the definition of G, we consider the following
cases for all i ∈ {2, · · · , n − 1}:
– If Q ∈ L(〈x i , x i+1 〉) then Q ∈ L(〈[x i ], [x i+1 ]〉),
– If Q ∈ Inv( L(〈x i+1 , x i 〉) ) then Q ∈ L(〈[x i ], [x i+1 ]〉),
– If Q ∈ L(〈x i , z〉) where x i+1 blocks z then Q ∈
L(〈[x i ], [x i+1 ]〉).
– If Q ∈ L(〈x n , x n+1 〉) then Q ∈ L(〈[x n ], [x n+1 ]〉).
This implies that Q ∈ L(〈[x i ], [x i+1 ]〉) for all i ∈
{2, · · · , n − 1} and Q ∈ L(〈[x n ], [x 1 ]〉).
• Property P9. From the construction of neighborhoods. □
Lemma (3). Let D be an SHI + -concept. Let T and R
be a terminology and role hierarchy. If D has a model w.r.t.
(T , R) then there exists a completion tree with cyclic paths.
Proof of Lemma 3.
According to Lemma 5 there is a tableau T = (S, L, E) for
D. A tree T = (V, E, L) can be inductively built from T
together with a function π from V to S. This construction
is quite intuitive since we can define a valid neighborhood
from each individual s ∈ S as follows:
• We define l(v) := L(s). v is valid since any node whose
label is included in the label of a node in the tableau T is
always valid.
• Let S ′ (s) ⊆ S such that s ′ ∈ S ′ (s) iff L(〈s, s ′ 〉) ≠ ∅
where
L(〈s, s ′ 〉) := {R | 〈s, s ′ 〉 ∈ E(R) for some R ∈
R (T ,R,D) }.
Let S(s) ⊆ S ′ sub(T ,R,D)
(s) such that for each C ∈ 2
and R ∈ 2 R if there is a t ∈ S ′ (s) with L(t) = C and
L(〈s, t〉) = R then there is a unique node t ′ ∈ S(s) with
L(t ′ ) = L(t) and L(〈s, t〉) = L(〈s, t ′ 〉). This implies
that S(s) is finite.
• For each t ∈ S(s) we define a node u ∈ N 0 such that
l(u) = L(t) and l(〈v, u〉) = L(〈s, s ′ 〉). From the construction,
(v, N 0 , l) is valid.