Logics for Data and Knowledge Representation

Logics for Data and Knowledge Representation Exercises: DL

DL family of languages AL <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 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • Examples of formulas in each of the languages: 3

Formalization of a semantic network SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS Person ⊑ ∃Drives.Car Person ⊑ ∃HasHobby.SportCar Person ⊑ ∃HasHobby.Opera Student ⊑ Person SportCar ⊑ Car Student(Ralf) Opera(DonCarlos) instance-of instance-of

Proofs in DL semantics SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • Verify the following equivalences hold for all interpretations (∆,I). Use definitions or Venn diagrams. • I(¬(C⊓D)) = I(¬C ⊔ ¬D) • I(¬(C ⊔ D)) = I(¬C⊓¬D) • I(¬∀R.C) = I(∃R.¬C) • 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) andC(b)} = {a ∈ ∆ | ∀b. not (R(a,b) andC(b))} = {a ∈ ∆ | ∀b. if R(a,b) then¬C(b)} = I(∀R.¬C) 5

Venn diagrams SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • Provide the Venn diagram for: A ⊑ B ⊓ ¬C C B A

Using DPLL for reasoning tasks SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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 7

Satisfiability of a TBox (a set of formulas) SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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

Satisfiability w.r.t. a TBox SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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)

Subsumption SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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: • I(XFile) = {D2} • I(Restricted) = {D1} • I(Read) = {(B, D2)} • I(Policeman) = {B} • I(FederalAgent) = {A} • I(TopSecretDocument) = {D2} • I(Access) = {(A, D1), (A, D2)}

Reduce to PL reasoning SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS The steps: T’ = Normalize(T); C’ = Expand(C, T’); D’ = Expand(D, T’); C’ = RewriteInPL(C’); D’ = RewriteInPL(D’); return DPLL(C’ → D’); • 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)) X E’ = E Call DPLL((B (D E  Y)) X → E) (Note that the formula has to be converted in CNF first)

Expansion of an ABox SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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)

ABox Reasoning services: Consistency SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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 13

Drawing consequences (I) SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • A: • Mother(Anna) • Father(Bob) • has(Cate,Anna) • has(Cate,Bob) Given the TBox and ABox below • T: • Female⊑Human • Male⊑Human • Mother⊑Female • Father⊑Male • Child≡∃has.Mother⊓ ∃has.Father • Male⊓Female⊑⊥ • Prove: • Human(Anna) • ¬Female(Bob) • Child(Cate) 14

Drawing consequences (II) SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS 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) 15

Reasoning using Tableau calculus (a) SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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)} ⊔-rule 16

Reasoning using Tableau calculus (b) SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS • 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.

Logics for Data and Knowledge Representation