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DL family of languagesAL
<Atomic> ::= A | B | ... | P | Q | ... | ⊥ | ⊤
<wff> ::= <Atomic> | ¬<Atomic> | <wff> ⊓ <wff> | ∀R.C | ∃R.⊤
ALU <wff> ⊔ <wff>
ALE ∃R.C
ALN ≥nR | ≤nR
ALC ¬ <wff>
FL- is AL with the elimination of ⊤, ⊥ and FL0 is FL- with the elimination of ∃R.⊤
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
DL family of languages Examples of formulas in each of the languages:
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Formula
AL ALU ALE ALN ALC
A
A⊔B
∃R.C
≥2R
(A⊓B) A⊔B
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Formalization of a semantic network
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instance-of
instance-of
Person ⊑ ∃Drives.Car
Person ⊑ ∃HasHobby.SportCar
Person ⊑ ∃HasHobby.Opera
Student ⊑ Person
SportCar ⊑ Car
Student(Ralf)
Opera(DonCarlos)
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Proofs in DL semantics Verify the following equivalences hold for all interpretations
(∆,I). Use definitions or Venn diagrams.1. I(¬(C ⊓ D)) = I(¬C ⊔ ¬D)2. I(¬(C ⊔ D)) = I(¬C ⊓ ¬D)3. I(¬∀R.C) = I(∃R.¬C)4. I(¬∃R.C) = I(∀R.¬C)
Let us prove the last one:
I(¬∃R.C)= {a ∈ ∆ | not exists b s.t. (a,b) ∈ I(R), b ∈ I(C)}
= {a ∈ ∆ | not exists ∃b. R(a,b) and C(b)}
= {a ∈ ∆ | ∀b. not (R(a,b) and C(b))}
= {a ∈ ∆ | ∀b. if R(a,b) then ¬C(b)}
= I(∀R.¬C)
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Venn diagrams Provide the Venn diagram for: A ⊑ B ⊓ ¬C
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B
A
C
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Using DPLL for reasoning tasks DPLL solves the CNFSAT-problem by searching a truth-assignment
that satisfies all clauses θi in the input proposition P = θ1 … θn
DL sentences must to be translated in PL (via TBox and ABox elimination)
Model checking Does ν satisfy P? (ν ⊨ P?)
Check if ν(P) = true
Satisfiability Is there any ν such that ν ⊨ P?
Check that DPLL(P) succeeds and returns a ν
Unsatisfiability Is it true that there are no ν satisfying P?
Check that DPLL(P) fails
Validity Is P a tautology? (true for all ν)
Check that DPLL(P) fails
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Satisfiability of a TBox (a set of formulas) Say if the following TBox is satisfiable and provide a
model in the case: FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-
1.Policeman
I(FederalAgent) = {A} We have that I ⊨ T
I(TopSecretDocument) = {D1, D2}
I(Access) = {(A, D1), (A, D2)}
I(XFile) = {D1}
I(Restricted) = {D2}
I(Read) = {(B, D1)}
I(Policeman) = {B}
Otherwise you can either draw a Venn diagram or apply the tableaux calculus and provide the ABox built
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Satisfiability w.r.t. a TBox Consider the TBox T:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-
1.Policeman
and the formula P: TopSecretDocument ⊓ Restricted
does T ⊨ P ?
Yes, if fact these is an interpretation I (e.g. the one of the
previous exercise) such that I ⊨ T and I ⊨ P. I(TopSecretDocument) = {D1, D2}
I(Restricted) = {D1}
Notice that T does not affect P at all (i.e. we cannot further expand P w.r.t. T)
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Subsumption Consider the TBox T:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman
does T ⊨ TopSecretDocument ⊑ Restricted ?
By definition, it must be I(TopSecretDocument ) I(Restricted ) for every model I of T. Even if this is true for the I of the previous exercise, this is not true in general. It is enough to provide a counterexample:
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
I(FederalAgent) = {A}I(TopSecretDocument) = {D2}I(Access) = {(A, D1), (A, D2)}
I(XFile) = {D2}I(Restricted) = {D1}I(Read) = {(B, D2)}I(Policeman) = {B}
Reduce to PL reasoning Consider the TBox T = {A ⊑ B ⊔ C, C ⊑ D ⊓
E}. Determine if A ⊑ E by elimination of T.
T’ = {A ≡ (B ⊔ C) ⊓ X, C ≡ D ⊓ E ⊓ Y}
A’ = (B ⊔ (D ⊓ E ⊓ Y)) ⊓ X
E’ = E
A’ = (B (D E Y)) XE’ = E
Call DPLL((B (D E Y)) X → E)
(Note that the formula has to be converted in CNF first)
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The steps:
T’ = Normalize(T);
C’ = Expand(C, T’);
D’ = Expand(D, T’);
C’ = RewriteInPL(C’);
D’ = RewriteInPL(D’);
return DPLL(C’ → D’);
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Expansion of an ABox Provide the expansion of A w.r.t. T (without
normalization), where:
TBox T = {Student ⊑ Faculty, Professor ⊑ Faculty ⊓ Teach}
ABox A = {Professor(Bob), Faculty(Rui)}
Professor(Bob), Faculty(Bob), Teach(Bob)
Faculty(Rui)
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
ABox Reasoning services: Consistency An ABox A is consistent with respect to a TBox T if there is an
interpretation I which is a model of both A and T.
T = {Parent⊑≤1hasChild}
A = {hasChild(mary, bob), hasChild(mary, cate), Parent(mary)}
A is consistent ALONE but is not consistent with respect T.
In fact, from A mary has two children while T imposes maximum one
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Drawing consequences (I)Given the TBox and ABox below T:
Female⊑Human Male⊑Human Mother⊑Female Father⊑Male Child≡∃has.Mother⊓ ∃has.Father Male⊓Female⊑⊥
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Prove:1. Human(Anna)2. ¬Female(Bob)3. Child(Cate)
A: Mother(Anna) Father(Bob) has(Cate,Anna) has(Cate,Bob)
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Drawing consequences (II)
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Expand A w.r.t. T:
A: Mother(Anna) Female(Anna) Human(Anna) Father(Bob) Male(Bob) Human(Bob) , ¬Female(Bob) has(Cate,Anna) Child(Cate) has(Cate,Bob) Child(Cate)
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Reasoning using Tableau calculus (a) Given the ABox A = {hasParent(Speedy, Furia)},
prove with Tableau algorithm the satisfiability of the following formula:
∃Parent.Horse ⊓ (Horse ⊓ Mule)
Both ∃Parent.Horse and (Horse ⊓ Mule) have to be satisfied ⊓-rule
(i) ∃hasParent.Horse A’’ = A’ ∪ {Horse(Furia)} ∃-rule
(ii) (Horse ⊓ Mule) Horse ⊔ Mule
It is not consistent for A’ = A ∪ { Horse(Furia)} ⊔-rule
(backtracking)
It is consistent for A’ = A ∪ { Mule(Furia)} ⊔-rule16
SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS
Reasoning using Tableau calculus (b) Using the tableau calculus, say whether the formula (∀R.A ⊓ ∃R.¬A)
⊔ ∀R.(A ⊓ B) is satisfiable given the TBox T = {A ⊑ B}. Motivate your response with a proof.
By ⊔-rule we split into (i) (∀R.A ⊓ ∃R.¬A) (x) and (ii) ∀R.(A ⊓ B) (x).
We need one of the two to be satisfiable.
(i) By ⊓-rule we put into the ABox both ∀R.A (x) and ∃R.¬A (x)
(i.1) For ∀R.A (x), by ∀-rule we add into the ABox both R(x, y) and A(y)
(i.2) For ∃R.¬A (x), by ∃-rule we add into the ABox both R(x, y) and ¬A(y)
i.1. and i.2 clearly contradict each other(backtracking)
(ii) It is clearly in contradiction with the axiom in T. In fact A and B are disjoint
Since none of the two is satisfiable, then the formula is NOT satisfiable.
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SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS