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Contents
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Inference problems
Main inference problems
Let’s have a terminology T and concepts C and D.
satisfiability A concept C is satisfiable with respect to T if there exists a model I of T , such that C I is
non-empty.
subsumption A concept C is subsumed (T |= C v D) by concept D with respect to T if C I ⊆ DI for all
models I of T .
equivalence Two concepts C and D are equivalent (T |= C ≡ D) with respect to T if C I = DI
disjointness Two concepts C and D are disjoint with respect to T if C I ∩ DI = ∅).
Inference problems reductions
The following conditions assume ”with respect to terminology T ” ( T |= is missing ).
• For DL systems with bottom concept ⊥ and intersection u, all inferences can be reduced to the subsumption :
– C is unsatisfiable iff C v ⊥,
– C ≡ D iff C v D and D v C,
– C and D are disjoint iff C u D v ⊥.
For AL, upper bound of complexity is determined by subsumption.
In addition, whenever negation is available, all inferences can be reduced to the satisfiability. How ?
• For AL the unsatisfiability problem is a special case of each of the previous. The following conditions are equivalent
– C is unsatisfiable,
– C v ⊥,
– C ≡ ⊥,
– C and > are disjoint.
For AL, lower bound of complexity is determined by unsatisfiability.
Inference problems for an ABOX
Given an ABOX A and an assertion α we say that A entails α and write A |= α if all models of A are
models for α.
consistency checking An ABOX A is consistent w.r.t. the acyclic TBOX T if there is an interpretation
I, which is model of both T and A.
instance checking – Given an ABOX A and an assertion α, does it hold A |= α?
retrieval problem – Given an ABOX A and a concept C, find all individuals a, that A |= C(a).
realization problem – Given an ABOX A and an individual a, find the most specific concept C, such that
A |= C(a).
All inferences for concepts (for both ABOX and TBOX) can be reduced to the consistency
checking of an ABOX w.r.t. empty TBOX.
Inference problems for an ABOX (2)
But how to check consistency of an ABOX ? Given an acyclic TBOX T and an ABOX A, the following
procedure applies :
1. Create an expansion T1 of T
2. Create a new ABOX A1 from A in this way:
• Primitive concept assertions and role assertions is added to the A1 unchanged.
• Defined concept assertions are replaced by corresponding T1 concept descriptions.
3. The consistency of A1 (w.r.t. empty TBOX) is equivalent to the consistency of A with respect to T .
Example – CWA × OWA
Example 2.1.ABOX
hasChild(JOCAST A, OEDIP U S)
hasChild(OEDIP U S, P OLY N EIKES)
P atricide(OEDIP U S)
hasChild(JOCAST A, P OLY N EIKES)
hasChild(P OLY N EIKES, T HERSAN DROS)
¬P atricide(OEDIP U S)
Edges represent hasChild role assertions, while colors distinguish concept assertions – P atricide and ¬P atricide
JOCAST AU
UUU*
OEDIP U S
/ P OLY N EIKES
ggggg3
/ T HERSAN DROS
Query
(∃hasChild · (P atricide u ∃hasChild · ¬P atricide))(JOCAST A),
JOCAST A
/
x
/
y
Is the query entailed by the ABOX ?
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Reasoning algorithms
Reasoning algorithms
Structural subsumption algorithms solve only subsumption problem for DL languages without negation
by syntactical matching.
• efficient, but complete only for very simple languages, i.e. FL0 (polynomial).
Tableu-based algorithms ... for all other cases :-)
Sometimes, it is possible to reuse existing inference procedures for other logics. For example, decidability of
ALC has been proven due to its correspondence to L2 .
Tableau-based algorithms
1. These algorithms use negation to reduce all inference problems to unsatisfiability.
Example 3.1. (∃R · A) u (∃R · B) v ∃R · (A u B) holds iff the following concept C is unsatisfiable
(∃R · A) u (∃R · B) u ¬(∃R · (A u B))
2. A unique negation normal form C0 of the investigated concept C is constructed by ”shifting” all
negations to the ”deepest position”.
Example 3.2. (∃R · A) u (∃R · B) u (∀R · (¬A t ¬B))
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Tableau-based algorithms (2)
3. The (un)satisfiability is proven by constructing (Henkin-like) model. It starts with one object, say
b ∈ ∆I , which is assumed to belong to the interpretation of investigated concept C0 .
Example 3.3. b ∈ (∃R · A)I and b ∈ (∃R · B)I and b ∈ (∀R · (¬A t ¬B))I
4. For each construct similar sound rules are introduced :
existential restriction introduces a new individual as role filler conforming the constraints.
value restriction together with roles relationships imposes new constraints on individuals.
union branches the execution – with possible backtracking.
number restrictions may force identifing individuals when violated.
Tableau-based satisfiability algorithm for ALCN
.
• To store intermediate states of model construction, an ABox enriched with (in)equality assertions = can be used.
• A set S of ABoxes Ai is evolved. Furthermore, S is satisfiable, iff at least one ABox is satisfiable.
• Investigating satisfiability of concept C0 in negation normal form, we start with S0 = {{C0 (x0 )}} and apply the following rules
until no more can be applied :
→u −rule
condition: A contains (C1 u C2 )(x), but it does not contain both C1 (x) and C2 (x).
action: A1 = A ∪ {C1 (x), C2 (x)}
→t −rule
condition: A contains (C1 t C2 )(x), but neither C1 (x) nor C2 (x).
action: A1 = A ∪ {C1 (x)} and A2 = A ∪ {C2 (x)}
→∃ −rule
condition: A contains ∃R · C(x) but there is no individual name z such that C(z) and R(x, z) in A
action: A1 = A ∪ {C(z), R(x, z)}
... and other
Tableau-based satisfiability algorithm for ALCN (2)
... Rules above apply until not possible – in the end, S is a set of complete ABoxes.
Example 3.4.
S0
=
→u
→∃
→∀
→t
{{(∃R · A u ∃R · B u ∀R · (¬A t ¬B))(b)}}
{{. . . , (∃R · A)(b), (∃R · B)(b), (∀R · (¬A t ¬B))(b)}}
{{. . . , R(b, c), A(c), R(b, d), B(d), (∀R · (¬A t ¬B))(b)}}
{{. . . , R(b, c), A(c), R(b, d), B(d), (¬A t ¬B)(c), (¬A t ¬B)(d)}}
{{. . . , R(b, c), A(c), R(b, d), B(d), ¬A(c), ¬A(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬B(c), ¬A(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬A(c), ¬B(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬B(c), ¬B(d)}}
Tableau-based satisfiability algorithm for ALCN (3)
If no ABox contains clash, i.e. a concept and its negation, bottom concept, or number restrictions violation, then the ABox is
consistent.
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Example 3.5.
...S
=
{{. . . , R(b, c), A(c), R(b, d), B(d), ¬A(c), ¬A(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬B(c), ¬A(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬A(c), ¬B(d)},
{. . . , R(b, c), A(c), R(b, d), B(d), ¬B(c), ¬B(d)}}
The resulting set of ABOXes is satisfiable (second ABOX), hence the subsumption does not hold. The following model has been
constructed :
b D
zz
}zz
R
c(A)
DDR
D"
d(B)
terminating, sound, complete
Extension to the consistency checking of ABoxes
1. Let’s have an ABox in negation normal form,
2. Add to the ABox inequality assertions for all different individual names occuring in the ABox to obtain
a new ABox, say A0 .
3. Starting with S = {{A0 }}, we can use the algorithm above.
4. !!! BUT !!! In this case the algorithm need not terminate. In order to regain termination, the rules
must be chosen in an appropriate way : the generating rules (f.e.→∃ −rule) cannot be used while some
other rule is applicable.
• Consistency of ALCN ABoxes is decidable (but PSpace-C, as well).
Resources
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