1. Logics for Data and Knowledge Representation Exercises: DL

2. DL family of languages AL ::= A | B |... | P | Q |... | ⊥ | ⊤ ::= | ¬ | ⊓ | ∀ R.C | ∃ R. ⊤ ALU ⊔ ALE ∃ R.C ALN ≥nR | ≤nR ALC ¬ FL- is AL with the elimination of ⊤, ⊥ and  FL0 is FL- with the elimination of ∃ R. ⊤ SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

3. DL family of languages  Examples of formulas in each of the languages: 3 Formula ALALUALEALNALC AA  A⊔BA⊔B  ∃ R.C  ≥2R   (A ⊓ B) A⊔BA⊔B  SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

4. Formalization of a semantic network 4 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

5. 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) 5 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

6. Venn diagrams  Provide the Venn diagram for: A ⊑ B ⊓ ¬C 6 B A C SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

7. Using DPLL for reasoning tasks  DPLL solves the CNFSAT-problem by searching a truth-assignment that satisfies all clauses θ i in the input proposition P = θ 1  …  θ n  DL sentences must to be translated in PL (via TBox and ABox elimination)  Model checkingDoes 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  UnsatisfiabilityIs 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 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

8. 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 8 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

9. 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) 9 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

10. 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: 10 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}

11. 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 ))  X E’ = E Call DPLL((B  (D  E  Y ))  X → E) (Note that the formula has to be converted in CNF first) 11 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

12. 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) 12 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

13. 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 ⊑≤1 hasChild} 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 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

14. 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 ⊑⊥ 14 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

15. Drawing consequences (II) 15 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

16. 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)} ⊔ -rule 16 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS

17. 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. 17 SYNTAX :: SEMANTICS :: TBOX REASONING :: ABOX REASONING :: TABLEAU CALCULUS