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Logics for Data and Knowledge Representation

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Logics for Data and Knowledge Representation. Exercises: DL. DL family of languages. AL ::= A | B | ... | P | Q | ... | ⊥ | ⊤ ::= | ¬ | ⊓ | ∀ R.C | ∃ R. ⊤ ALU ⊔ ALE ∃R. C ALN ≥nR | ≤nR ALC ¬ - PowerPoint PPT Presentation
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Logics for Data and Knowledge Representation Exercises: DL
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Logics for Data and Knowledge Representation

Exercises: DL

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


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