Inference/Reasoning problems in ALC Tableaux reasoning in ALC Logics for Data and Knowledge Representation 5. Reasoning in ALC Luciano Serafini FBK-irst, Trento, Italy October 14, 2012 L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Basic Inference Problems The basic inference problems on concepts and T-boxes are the following: Concept subsumption C is subsumed by D , or equivalently, D subsumes C , in symbols | = C ⊑ D , if and only if C I ⊆ D I in all interpretations I L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Basic Inference Problems The basic inference problems on concepts and T-boxes are the following: Concept subsumption C is subsumed by D , or equivalently, D subsumes C , in symbols | = C ⊑ D , if and only if C I ⊆ D I in all interpretations I Concept Subsumption w.r.t. T-Box C is subsumed by D w.r.t., T-box T , or equivalently, D subsumes C in T , in symbols | = C ⊑ T D , (an alternative notation T | = C ⊑ D ) if and only if C I ⊆ D I in all interpretations I that satisfies T . L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Basic Inference Problems The basic inference problems on concepts and T-boxes are the following: Concept subsumption Concept consistency C is subsumed by D , or equivalently, D C is consistent if and only if there exists an interpretation I such that C I � = ∅ . subsumes C , in symbols | = C ⊑ D , if and only if C I ⊆ D I in all interpretations I Concept Subsumption w.r.t. T-Box C is subsumed by D w.r.t., T-box T , or equivalently, D subsumes C in T , in symbols | = C ⊑ T D , (an alternative notation T | = C ⊑ D ) if and only if C I ⊆ D I in all interpretations I that satisfies T . L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Basic Inference Problems The basic inference problems on concepts and T-boxes are the following: Concept subsumption Concept consistency C is subsumed by D , or equivalently, D C is consistent if and only if there exists an interpretation I such that C I � = ∅ . subsumes C , in symbols | = C ⊑ D , if and only if C I ⊆ D I in all interpretations I Concept consistency w.r.t a Tbox C is consistent w.r.t. T if and only if there a model I of T with C I � = ∅ Concept Subsumption w.r.t. T-Box C is subsumed by D w.r.t., T-box T , or equivalently, D subsumes C in T , in symbols | = C ⊑ T D , (an alternative notation T | = C ⊑ D ) if and only if C I ⊆ D I in all interpretations I that satisfies T . L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Basic Inference Problems The basic inference problems on concepts and T-boxes are the following: Concept subsumption Concept consistency C is subsumed by D , or equivalently, D C is consistent if and only if there exists an interpretation I such that C I � = ∅ . subsumes C , in symbols | = C ⊑ D , if and only if C I ⊆ D I in all interpretations I Concept consistency w.r.t a Tbox C is consistent w.r.t. T if and only if there a model I of T with C I � = ∅ Concept Subsumption w.r.t. T-Box C is subsumed by D w.r.t., T-box T , or Consistency of a T-box equivalently, D subsumes C in T , in symbols | = C ⊑ T D , (an alternative A T-box T is consistent, if there is an notation T | = C ⊑ D ) if and only if interpretation I that satisfies T , i.e., C I ⊆ D I in all interpretations I that I | = T . satisfies T . L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC Dependencies between basic inference problems Concept subsumption ⇔ concept consistency | = C ⊑ D ⇐ ⇒ C ⊓ ¬ D is not consistent (1) T | = C ⊑ D ⇐ ⇒ C ⊓ ¬ D is not consistent w.r.t., T (2) Proof. We prove property (2). Indeed (1) is a special case of (2) with T = ∅ . = T , C I ⊆ D I T | = C ⊑ D ⇐ ⇒ for all I such that I | = T , ( C ⊓ ¬ D ) I = ∅ ⇐ ⇒ for all I s.t. I | = T , ( C ⊓ ¬ D ) I � = ∅ ⇐ ⇒ there is no I | ⇐ ⇒ C ⊓ ¬ D is not satisfiable in T L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC Dependencies between basic inference problems Concept consistency w.r.t., T-box ⇔ T-box consistency C is consistent w.r.t. T ⇐ ⇒ T ∪ {∃ P new . C } is consistent (3) Where P new is a “fresh” role, i.e., a role symbol not appearing in T Proof. ⇒ If C is consistent w.r.t. T , there is an interpretation I that satisfies C and such = that C I � = ∅ . Let I ′ be the extension of I where ( P new ) = ∆ × C I . Since C I is not empty we have that for all d ∈ ∆ I there is a d ′ ∈ C I such that ( d , d ′ ) ∈ ( P new ) I , this implies that d ∈ ( ∃ P new . C . Since this holds for every d ∈ ∆ I , we have that I | = ⊤ ⊑ ∃ P new . C , and therefore I is a model for T ∪ {⊤ ⊑ ∃ P new . C } . ⇐ = If T ∪ {⊤ ⊑ ∃ P new . C } is consistent then there is a model I that satisfies ⊤ ⊑ ∃ P new . C . Since ⊤ I is not empty, this implies that there is a d ∈ ∃ P new . C , which implies that there is a d ′ , with ( d , d ′ ) ∈ P new and d ′ ∈ C I , i.e., C is consistent. L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC Dependencies between basic inference problems T | = C ⊑ D T ∪ C ⊓ ¬ D is not {⊤ ⊑ ∃ P new ( C ⊓ ¬ D ) } consistent in T is not consistent L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC (un)satisfiability general properties - exercises Exercise Show that | = C ⊑ D implies | = ∃ R . C ⊑ ∃ R . D L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC (un)satisfiability general properties - exercises Exercise Show that | = C ⊑ D implies | = ∃ R . C ⊑ ∃ R . D Solution We have to prove that for all I , ( ∃ R . C ) I ⊆ ( ∃ R . C ) I under the hypothesis that for all I , C I ⊆ D I . Let x ∈ ( ∃ R . C ) I , we want to show that x is also in ( ∃ R . D ) I . If x ∈ ( ∃ R . C ) I , then by the interpretation of ∃ R there must be an y with ( x , y ) ∈ R I such that y ∈ C I . By the hypothesis that C I ⊆ D I for all I , we have thaty ∈ D I . The fact that ( x , y ) ∈ R I and y ∈ D I implies that x ∈ ( ∃ R . D ) I . L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC (un)satisfiability and validity - exercises Exercise For each of the following formula say if it is valid, satisfiable or unsatisfiable. If it is not valid provide a model that falsify it. ∀ R ( A ⊓ B ) ≡ ∀ RA ⊓ ∀ RB ∀ R ( A ⊔ B ) ≡ ∀ RA ⊔ ∀ RB ∃ R ( A ⊓ B ) ≡ ∃ RA ⊓ ∃ RB ∃ R ( A ⊔ B ) ≡ ∃ RA ⊔ ∃ RB L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC (un)satisfiability and validity - exercises Exercise For each of the following formula say if it is valid, satisfiable or unsatisfiable. If it is not valid provide a model that falsify it. ∀ R ( A ⊓ B ) ≡ ∀ RA ⊓ ∀ RB ∀ R ( A ⊔ B ) ≡ ∀ RA ⊔ ∀ RB ∃ R ( A ⊓ B ) ≡ ∃ RA ⊓ ∃ RB ∃ R ( A ⊔ B ) ≡ ∃ RA ⊔ ∃ RB Solution ∀ R ( A ⊓ B ) ≡ ∀ RA ⊔ ∀ RB is valid and we can prove that ( ∀ R ( A ⊓ B )) I = ( ∀ R . A ⊓ ∀ R . B ) I for all interpretations I . ( ∀ R ( A ⊓ B )) I = { ( x , y ) ∈ R I | y ∈ ( A ⊓ B ) I } = { ( x , y ) ∈ R I | y ∈ A I ∩ B I } = { ( x , y ) ∈ R I | y ∈ A I } ∩ { ( x , y ) ∈ R I | y ∈ B I } = ( ∀ R . A ) I ∩ ( ∀ R . B ) I = ( ∀ R . A ⊓ ∀ R . B ) I L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC (un)satisfiability and validity - exercises Exercise For each of the following formula say if it is valid, satisfiable or unsatisfiable. If it is not valid provide a model that falsify it. ∀ R ( A ⊓ B ) ≡ ∀ RA ⊓ ∀ RB ∀ R ( A ⊔ B ) ≡ ∀ RA ⊔ ∀ RB ∃ R ( A ⊓ B ) ≡ ∃ RA ⊓ ∃ RB ∃ R ( A ⊔ B ) ≡ ∃ RA ⊔ ∃ RB Solution ∀ R ( A ⊔ B ) ≡ ∀ RA ⊔ ∀ RB is not valid. The following model is such that ( ∀ R ( A ⊔ B )) I � = ( ∀ RA ⊔ ∀ RB ) I s 0 s 0 ∈ ( ∀ R ( A ⊔ B )) I but R R s 0 �∈ ( ∀ RA ) and A , ¬ B s 1 s 2 ¬ A , B s 0 �∈ ( ∀ RB ) I However notice that the containment: ∀ R . A ⊔ ∀ R . B ⊑ ∀ R . ( A ⊔ B ) is valid L. Serafini LDKR
Inference/Reasoning problems in ALC Tableaux reasoning in ALC ALC (un)satisfiability and validity - exercises Exercise For each of the following formula say if it is valid, satisfiable or unsatisfiable. If it is not valid provide a model that falsify it. ∀ R ( A ⊓ B ) ≡ ∀ RA ⊓ ∀ RB ∀ R ( A ⊔ B ) ≡ ∀ RA ⊔ ∀ RB ∃ R ( A ⊓ B ) ≡ ∃ RA ⊓ ∃ RB ∃ R ( A ⊔ B ) ≡ ∃ RA ⊔ ∃ RB Solution ∃ R ( A ⊓ B ) ≡ ∃ RA ⊓ ∃ RB is not valid. The following model is such that ( ∃ R ( A ⊓ B )) I � = ( ∃ RA ⊓ ∀ RB ) I s 0 s 0 ∈ ( ∃ RA ) I and R R s 0 ∈ ( ∃ RB ) I but A , ¬ B s 1 s 2 ¬ A , B s 0 �∈ ( ∃ R ( A ⊓ B )) I However notice that the containment: ∃ R ( A ⊓ B ) ⊑ ∃ RA ⊓ ∃ RB is valid L. Serafini LDKR
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