Example Queries in ( kim , r 123) . KB = part of ( r 123 , cs building ) . in ( X , Y ) ← part of ( Z , Y ) ∧ in ( X , Z ) . Query Answer ? part of ( r 123 , B ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 3
Example Queries in ( kim , r 123) . KB = part of ( r 123 , cs building ) . in ( X , Y ) ← part of ( Z , Y ) ∧ in ( X , Z ) . Query Answer ? part of ( r 123 , B ) . part of ( r 123 , cs building ) ? part of ( r 023 , cs building ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 4
Example Queries in ( kim , r 123) . KB = part of ( r 123 , cs building ) . in ( X , Y ) ← part of ( Z , Y ) ∧ in ( X , Z ) . Query Answer ? part of ( r 123 , B ) . part of ( r 123 , cs building ) ? part of ( r 023 , cs building ) . no ? in ( kim , r 023) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 5
Example Queries in ( kim , r 123) . KB = part of ( r 123 , cs building ) . in ( X , Y ) ← part of ( Z , Y ) ∧ in ( X , Z ) . Query Answer ? part of ( r 123 , B ) . part of ( r 123 , cs building ) ? part of ( r 023 , cs building ) . no ? in ( kim , r 023) . no ? in ( kim , B ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 6
Example Queries in ( kim , r 123) . KB = part of ( r 123 , cs building ) . in ( X , Y ) ← part of ( Z , Y ) ∧ in ( X , Z ) . Query Answer ? part of ( r 123 , B ) . part of ( r 123 , cs building ) ? part of ( r 023 , cs building ) . no ? in ( kim , r 023) . no ? in ( kim , B ) . in ( kim , r 123) in ( kim , cs building ) � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 7
Electrical Environment outside power cb1 w5 s1 w1 circuit breaker cb2 s2 w2 w3 off s3 w0 switch on w6 w4 two-way p2 switch l1 light p1 l2 power outlet � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 8
Axiomatizing the Electrical Environment % light ( L ) is true if L is a light light ( l 1 ) . light ( l 2 ) . % down ( S ) is true if switch S is down down ( s 1 ) . up ( s 2 ) . up ( s 3 ) . % ok ( D ) is true if D is not broken ok ( l 1 ) . ok ( l 2 ) . ok ( cb 1 ) . ok ( cb 2 ) . ? light ( l 1 ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 9
Axiomatizing the Electrical Environment % light ( L ) is true if L is a light light ( l 1 ) . light ( l 2 ) . % down ( S ) is true if switch S is down down ( s 1 ) . up ( s 2 ) . up ( s 3 ) . % ok ( D ) is true if D is not broken ok ( l 1 ) . ok ( l 2 ) . ok ( cb 1 ) . ok ( cb 2 ) . ? light ( l 1 ) . = yes ⇒ ? light ( l 6 ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 10
Axiomatizing the Electrical Environment % light ( L ) is true if L is a light light ( l 1 ) . light ( l 2 ) . % down ( S ) is true if switch S is down down ( s 1 ) . up ( s 2 ) . up ( s 3 ) . % ok ( D ) is true if D is not broken ok ( l 1 ) . ok ( l 2 ) . ok ( cb 1 ) . ok ( cb 2 ) . ? light ( l 1 ) . = yes ⇒ ? light ( l 6 ) . = no ⇒ ? up ( X ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 11
Axiomatizing the Electrical Environment % light ( L ) is true if L is a light light ( l 1 ) . light ( l 2 ) . % down ( S ) is true if switch S is down down ( s 1 ) . up ( s 2 ) . up ( s 3 ) . % ok ( D ) is true if D is not broken ok ( l 1 ) . ok ( l 2 ) . ok ( cb 1 ) . ok ( cb 2 ) . ? light ( l 1 ) . = yes ⇒ ? light ( l 6 ) . = no ⇒ ? up ( X ) . = up ( s 2 ), up ( s 3 ) ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 12
connected to ( X , Y ) is true if component X is connected to Y connected to ( w 0 , w 1 ) ← up ( s 2 ) . connected to ( w 0 , w 2 ) ← down ( s 2 ) . connected to ( w 1 , w 3 ) ← up ( s 1 ) . connected to ( w 2 , w 3 ) ← down ( s 1 ) . connected to ( w 4 , w 3 ) ← up ( s 3 ) . connected to ( p 1 , w 3 ) . ? connected to ( w 0 , W ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 13
connected to ( X , Y ) is true if component X is connected to Y connected to ( w 0 , w 1 ) ← up ( s 2 ) . connected to ( w 0 , w 2 ) ← down ( s 2 ) . connected to ( w 1 , w 3 ) ← up ( s 1 ) . connected to ( w 2 , w 3 ) ← down ( s 1 ) . connected to ( w 4 , w 3 ) ← up ( s 3 ) . connected to ( p 1 , w 3 ) . ? connected to ( w 0 , W ) . = W = w 1 ⇒ ? connected to ( w 1 , W ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 14
connected to ( X , Y ) is true if component X is connected to Y connected to ( w 0 , w 1 ) ← up ( s 2 ) . connected to ( w 0 , w 2 ) ← down ( s 2 ) . connected to ( w 1 , w 3 ) ← up ( s 1 ) . connected to ( w 2 , w 3 ) ← down ( s 1 ) . connected to ( w 4 , w 3 ) ← up ( s 3 ) . connected to ( p 1 , w 3 ) . ? connected to ( w 0 , W ) . = W = w 1 ⇒ ? connected to ( w 1 , W ) . = no ⇒ ? connected to ( Y , w 3 ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 15
connected to ( X , Y ) is true if component X is connected to Y connected to ( w 0 , w 1 ) ← up ( s 2 ) . connected to ( w 0 , w 2 ) ← down ( s 2 ) . connected to ( w 1 , w 3 ) ← up ( s 1 ) . connected to ( w 2 , w 3 ) ← down ( s 1 ) . connected to ( w 4 , w 3 ) ← up ( s 3 ) . connected to ( p 1 , w 3 ) . ? connected to ( w 0 , W ) . = W = w 1 ⇒ ? connected to ( w 1 , W ) . = no ⇒ ? connected to ( Y , w 3 ) . = Y = w 2 , Y = w 4 , Y = p 1 ⇒ ? connected to ( X , W ) . = ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 16
connected to ( X , Y ) is true if component X is connected to Y connected to ( w 0 , w 1 ) ← up ( s 2 ) . connected to ( w 0 , w 2 ) ← down ( s 2 ) . connected to ( w 1 , w 3 ) ← up ( s 1 ) . connected to ( w 2 , w 3 ) ← down ( s 1 ) . connected to ( w 4 , w 3 ) ← up ( s 3 ) . connected to ( p 1 , w 3 ) . ? connected to ( w 0 , W ) . = W = w 1 ⇒ ? connected to ( w 1 , W ) . = no ⇒ ? connected to ( Y , w 3 ) . = Y = w 2 , Y = w 4 , Y = p 1 ⇒ ? connected to ( X , W ) . = X = w 0 , W = w 1 , . . . ⇒ � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 17
% lit ( L ) is true if the light L is lit lit ( L ) ← light ( L ) ∧ ok ( L ) ∧ live ( L ) . % live ( C ) is true if there is power coming into C live ( Y ) ← connected to ( Y , Z ) ∧ live ( Z ) . live ( outside ) . This is a recursive definition of live . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 18
Recursion and Mathematical Induction above ( X , Y ) ← on ( X , Y ) . above ( X , Y ) ← on ( X , Z ) ∧ above ( Z , Y ) . This can be seen as: Recursive definition of above : prove above in terms of a base case ( on ) or a simpler instance of itself; or Way to prove above by mathematical induction: the base case is when there are no blocks between X and Y , and if you can prove above when there are n blocks between them, you can prove it when there are n + 1 blocks. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 19
Limitations Suppose you had a database using the relation: enrolled ( S , C ) which is true when student S is enrolled in course C . Can you define the relation: empty course ( C ) which is true when course C has no students enrolled in it? Why? or Why not? � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 20
Limitations Suppose you had a database using the relation: enrolled ( S , C ) which is true when student S is enrolled in course C . Can you define the relation: empty course ( C ) which is true when course C has no students enrolled in it? Why? or Why not? empty course ( C ) doesn’t logically follow from a set of enrolled relation because there are always models where someone is enrolled in a course! � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.2, Page 21
Reasoning with Variables An instance of an atom or a clause is obtained by uniformly substituting terms for variables. A substitution is a finite set of the form { V 1 / t 1 , . . . , V n / t n } , where each V i is a distinct variable and each t i is a term. The application of a substitution σ = { V 1 / t 1 , . . . , V n / t n } to an atom or clause e , written e σ , is the instance of e with every occurrence of V i replaced by t i . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 1
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( X , Y , Z , e ) σ 1 = p ( A , b , C , D ) σ 2 = p ( X , Y , Z , e ) σ 2 = p ( A , b , C , D ) σ 3 = p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 2
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , D ) σ 2 = p ( X , Y , Z , e ) σ 2 = p ( A , b , C , D ) σ 3 = p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 3
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , e ) p ( A , b , C , D ) σ 2 = p ( X , Y , Z , e ) σ 2 = p ( A , b , C , D ) σ 3 = p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 4
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , e ) p ( A , b , C , D ) σ 2 = p ( X , b , Z , e ) p ( X , Y , Z , e ) σ 2 = p ( A , b , C , D ) σ 3 = p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 5
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , e ) p ( A , b , C , D ) σ 2 = p ( X , b , Z , e ) p ( X , Y , Z , e ) σ 2 = p ( X , b , Z , e ) p ( A , b , C , D ) σ 3 = p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 6
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , e ) p ( A , b , C , D ) σ 2 = p ( X , b , Z , e ) p ( X , Y , Z , e ) σ 2 = p ( X , b , Z , e ) p ( A , b , C , D ) σ 3 = p ( V , b , W , e ) p ( X , Y , Z , e ) σ 3 = � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 7
Application Examples The following are substitutions: σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { A / X , Y / b , C / Z , D / e } σ 3 = { A / V , X / V , Y / b , C / W , Z / W , D / e } The following shows some applications: p ( A , b , C , D ) σ 1 = p ( A , b , C , e ) p ( X , Y , Z , e ) σ 1 = p ( A , b , C , e ) p ( A , b , C , D ) σ 2 = p ( X , b , Z , e ) p ( X , Y , Z , e ) σ 2 = p ( X , b , Z , e ) p ( A , b , C , D ) σ 3 = p ( V , b , W , e ) p ( X , Y , Z , e ) σ 3 = p ( V , b , W , e ) � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 8
Unifiers Substitution σ is a unifier of e 1 and e 2 if e 1 σ = e 2 σ . Substitution σ is a most general unifier (mgu) of e 1 and e 2 if ◮ σ is a unifier of e 1 and e 2 ; and ◮ if substitution σ ′ also unifies e 1 and e 2 , then e σ ′ is an instance of e σ for all atoms e . If two atoms have a unifier, they have a most general unifier. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 9
Unification Example Which of the following are unifiers of p ( A , b , C , D ) and p ( X , Y , Z , e ): σ 1 = { X / A , Y / b , Z / C , D / e } σ 2 = { Y / b , D / e } σ 3 = { X / A , Y / b , Z / C , D / e , W / a } σ 4 = { A / X , Y / b , C / Z , D / e } σ 5 = { X / a , Y / b , Z / c , D / e } σ 6 = { A / a , X / a , Y / b , C / c , Z / c , D / e } σ 7 = { A / V , X / V , Y / b , C / W , Z / W , D / e } σ 8 = { X / A , Y / b , Z / A , C / A , D / e } Which are most general unifiers? � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 10
Unification Example p ( A , b , C , D ) and p ( X , Y , Z , e ) have as unifiers: σ 1 = { X / A , Y / b , Z / C , D / e } σ 4 = { A / X , Y / b , C / Z , D / e } σ 7 = { A / V , X / V , Y / b , C / W , Z / W , D / e } σ 6 = { A / a , X / a , Y / b , C / c , Z / c , D / e } σ 8 = { X / A , Y / b , Z / A , C / A , D / e } σ 3 = { X / A , Y / b , Z / C , D / e , W / a } The first three are most general unifiers. The following substitutions are not unifiers: σ 2 = { Y / b , D / e } σ 5 = { X / a , Y / b , Z / c , D / e } � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 11
1: procedure unify ( t 1 , t 2 ) ⊲ Returns mgu of t 1 and t 2 or ⊥ . E ← { t 1 = t 2 } ⊲ Set of equality statements 2: S ← {} ⊲ Substitution 3: while E � = {} do 4: select and remove x = y from E 5: if y is not identical to x then 6: if x is a variable then 7: replace x with y in E and S 8: S ← { x / y } ∪ S 9: else if y is a variable then 10: replace y with x in E and S 11: S ← { y / x } ∪ S 12: else if x is p ( x 1 , . . . , x n ) and y is 13: p ( y 1 , . . . , y n ) then E ← E ∪ { x 1 = y 1 , . . . , x n = y n } 14: else 15: return ⊥ ⊲ t 1 and t 2 do not unify 16: return S ⊲ S is mgu of t 1 and t 2 17: � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 12
Logical Consequence Atom g is a logical consequence of KB if and only if: g is an instance of a fact in KB , or there is an instance of a rule g ← b 1 ∧ . . . ∧ b k in KB such that each b i is a logical consequence of KB . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 13
Aside: Debugging false conclusions To debug answer g that is false in the intended interpretation: If g is a fact in KB , this fact is wrong. Otherwise, suppose g was proved using the rule: g ← b 1 ∧ . . . ∧ b k where each b i is a logical consequence of KB . ◮ If each b i is true in the intended interpretation, this clause is false in the intended interpretation. ◮ If some b i is false in the intended interpretation, debug b i . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 14
Proofs A proof is a mechanically derivable demonstration that a formula logically follows from a knowledge base. Given a proof procedure, KB ⊢ g means g can be derived from knowledge base KB . Recall KB | = g means g is true in all models of KB . A proof procedure is sound if KB ⊢ g implies KB | = g . A proof procedure is complete if KB | = g implies KB ⊢ g . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 15
Bottom-up proof procedure KB ⊢ g if there is g ′ added to C in this procedure where g = g ′ θ : C := {} ; repeat select clause “ h ← b 1 ∧ . . . ∧ b m ” in KB such that there is a substitution θ such that for all i , there exists b ′ i ∈ C and θ ′ i where b i θ = b ′ i θ ′ i and there is no h ′ ∈ C and θ ′ such that h ′ θ ′ = h θ C := C ∪ { h θ } until no more clauses can be selected. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 16
Example live ( Y ) ← connected to ( Y , Z ) ∧ live ( Z ) . live ( outside ) . connected to ( w 6 , w 5 ) . connected to ( w 5 , outside ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 17
Example live ( Y ) ← connected to ( Y , Z ) ∧ live ( Z ) . live ( outside ) . connected to ( w 6 , w 5 ) . connected to ( w 5 , outside ) . C = { live ( outside ) , connected to ( w 6 , w 5 ) , connected to ( w 5 , outside ) , live ( w 5 ) , live ( w 6 ) } � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 18
Soundness of bottom-up proof procedure If KB ⊢ g then KB | = g . Suppose there is a g such that KB ⊢ g and KB �| = g . Then there must be a first atom added to C that has an instance that isn’t true in every model of KB . Call it h . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 19
Soundness of bottom-up proof procedure If KB ⊢ g then KB | = g . Suppose there is a g such that KB ⊢ g and KB �| = g . Then there must be a first atom added to C that has an instance that isn’t true in every model of KB . Call it h . Suppose h isn’t true in model I of KB . There must be an instance of clause in KB of form h ′ ← b 1 ∧ . . . ∧ b m where � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 20
Soundness of bottom-up proof procedure If KB ⊢ g then KB | = g . Suppose there is a g such that KB ⊢ g and KB �| = g . Then there must be a first atom added to C that has an instance that isn’t true in every model of KB . Call it h . Suppose h isn’t true in model I of KB . There must be an instance of clause in KB of form h ′ ← b 1 ∧ . . . ∧ b m where h = h ′ θ and b i θ is an instance of an element of C . ◮ Each b i θ is true in I . ◮ h is false in I . ◮ So an instance of this clause is false in I . ◮ Therefore I isn’t a model of KB . ◮ Contradiction. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 21
Fixed Point The C generated by the bottom-up algorithm is called a fixed point. C can be infinite; we require the selection to be fair. Herbrand interpretation: The domain is the set of constants. We invent a constant if the KB or query doesn’t contain one. Each constant denotes itself. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 22
Fixed Point The C generated by the bottom-up algorithm is called a fixed point. C can be infinite; we require the selection to be fair. Herbrand interpretation: The domain is the set of constants. We invent a constant if the KB or query doesn’t contain one. Each constant denotes itself. Let I be the Herbrand interpretation in which every ground instance of every element of the fixed point is true and every other atom is false. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 23
Fixed Point The C generated by the bottom-up algorithm is called a fixed point. C can be infinite; we require the selection to be fair. Herbrand interpretation: The domain is the set of constants. We invent a constant if the KB or query doesn’t contain one. Each constant denotes itself. Let I be the Herbrand interpretation in which every ground instance of every element of the fixed point is true and every other atom is false. I is a model of KB . Proof: � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 24
Fixed Point The C generated by the bottom-up algorithm is called a fixed point. C can be infinite; we require the selection to be fair. Herbrand interpretation: The domain is the set of constants. We invent a constant if the KB or query doesn’t contain one. Each constant denotes itself. Let I be the Herbrand interpretation in which every ground instance of every element of the fixed point is true and every other atom is false. I is a model of KB . Proof: suppose h ← b 1 ∧ . . . ∧ b m in KB is false in I . Then h is false and each b i is true in I . Thus h can be added to C . Contradiction to C being the fixed point. I is called a Minimal Model. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 25
Completeness If KB | = g then KB ⊢ g . Suppose KB | = g . Then g is true in all models of KB . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 26
Completeness If KB | = g then KB ⊢ g . Suppose KB | = g . Then g is true in all models of KB . Thus g is true in the minimal model. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 27
Completeness If KB | = g then KB ⊢ g . Suppose KB | = g . Then g is true in all models of KB . Thus g is true in the minimal model. Thus g is in the fixed point. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 28
Completeness If KB | = g then KB ⊢ g . Suppose KB | = g . Then g is true in all models of KB . Thus g is true in the minimal model. Thus g is in the fixed point. Thus g is generated by the bottom up algorithm. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 29
Completeness If KB | = g then KB ⊢ g . Suppose KB | = g . Then g is true in all models of KB . Thus g is true in the minimal model. Thus g is in the fixed point. Thus g is generated by the bottom up algorithm. Thus KB ⊢ g . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 30
Top-down Proof procedure A generalized answer clause is of the form yes ( t 1 , . . . , t k ) ← a 1 ∧ a 2 ∧ . . . ∧ a m , where t 1 , . . . , t k are terms and a 1 , . . . , a m are atoms. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 31
Top-down Proof procedure A generalized answer clause is of the form yes ( t 1 , . . . , t k ) ← a 1 ∧ a 2 ∧ . . . ∧ a m , where t 1 , . . . , t k are terms and a 1 , . . . , a m are atoms. The SLD resolution of this generalized answer clause on a i with the clause a ← b 1 ∧ . . . ∧ b p , where a i and a have most general unifier θ , is ( yes ( t 1 , . . . , t k ) ← a 1 ∧ . . . ∧ a i − 1 ∧ b 1 ∧ . . . ∧ b p ∧ a i +1 ∧ . . . ∧ a m ) θ. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 32
Top-down Proof Procedure To solve query ? B with variables V 1 , . . . , V k : Set ac to generalized answer clause yes ( V 1 , . . . , V k ) ← B while ac is not an answer do Suppose ac is yes ( t 1 , . . . , t k ) ← a 1 ∧ a 2 ∧ . . . ∧ a m select atom a i in the body of ac choose clause a ← b 1 ∧ . . . ∧ b p in KB Rename all variables in a ← b 1 ∧ . . . ∧ b p Let θ be the most general unifier of a i and a . Fail if they don’t unify Set ac to ( yes ( t 1 , . . . , t k ) ← a 1 ∧ . . . ∧ a i − 1 ∧ b 1 ∧ . . . ∧ b p ∧ a i +1 ∧ . . . ∧ a m ) θ end while . Answer is V 1 = t 1 , . . . , V k = t k where ac is yes ( t 1 , . . . , t k ) ← � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 33
Example live ( Y ) ← connected to ( Y , Z ) ∧ live ( Z ) . live ( outside ) . connected to ( w 6 , w 5 ) . connected to ( w 5 , outside ) . ? live ( A ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 34
Example live ( Y ) ← connected to ( Y , Z ) ∧ live ( Z ) . live ( outside ) . connected to ( w 6 , w 5 ) . connected to ( w 5 , outside ) . ? live ( A ) . yes ( A ) ← live ( A ) . yes ( A ) ← connected to ( A , Z 1 ) ∧ live ( Z 1 ) . yes ( w 6 ) ← live ( w 5 ) . yes ( w 6 ) ← connected to ( w 5 , Z 2 ) ∧ live ( Z 2 ) . yes ( w 6 ) ← live ( outside ) . yes ( w 6 ) ← . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 35
Function Symbols Often we want to refer to individuals in terms of components. Examples: 4:55 p.m. English sentences. A classlist. We extend the notion of term. So that a term can be f ( t 1 , . . . , t n ) where f is a function symbol and the t i are terms. In an interpretation and with a variable assignment, term f ( t 1 , . . . , t n ) denotes an individual in the domain. One function symbol and one constant can refer to infinitely many individuals. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 36
Lists A list is an ordered sequence of elements. Let’s use the constant nil to denote the empty list, and the function cons ( H , T ) to denote the list with first element H and rest-of-list T . These are not built-in. The list containing sue , kim and randy is cons ( sue , cons ( kim , cons ( randy , nil ))) append ( X , Y , Z ) is true if list Z contains the elements of X followed by the elements of Y append ( nil , Z , Z ) . append ( cons ( A , X ) , Y , cons ( A , Z )) ← append ( X , Y , Z ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 37
Unification with function symbols Consider a knowledge base consisting of one fact: lt ( X , s ( X )) . Should the following query succeed? ask lt ( Y , Y ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 38
Unification with function symbols Consider a knowledge base consisting of one fact: lt ( X , s ( X )) . Should the following query succeed? ask lt ( Y , Y ) . What does the top-down proof procedure give? � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 39
Unification with function symbols Consider a knowledge base consisting of one fact: lt ( X , s ( X )) . Should the following query succeed? ask lt ( Y , Y ) . What does the top-down proof procedure give? Solution: variable X should not unify with a term that contains X inside. E.g., X should not unify with s ( X ). Simple modification of the unification algorithm, which Prolog does not do! � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.3, Page 40
Complete Knowledge Assumption Often you want to assume that your knowledge is complete. Example: assume that a database of what students are enrolled in a course is complete. We don’t want to have to state all negative enrolment facts! The definite clause language is monotonic: adding clauses can’t invalidate a previous conclusion. Under the complete knowledge assumption, the system is non-monotonic: adding clauses can invalidate a previous conclusion. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 1
Equality Equality is a special predicate symbol with a standard domain-independent intended interpretation. Suppose interpretation I = � D , φ, π � . t 1 and t 2 are ground terms then t 1 = t 2 is true in interpretation I if t 1 and t 2 denote the same individual. That is, t 1 = t 2 if φ ( t 1 ) is the same as φ ( t 2 ). t 1 � = t 2 when t 1 and t 2 denote different individuals. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 2
Equality Equality is a special predicate symbol with a standard domain-independent intended interpretation. Suppose interpretation I = � D , φ, π � . t 1 and t 2 are ground terms then t 1 = t 2 is true in interpretation I if t 1 and t 2 denote the same individual. That is, t 1 = t 2 if φ ( t 1 ) is the same as φ ( t 2 ). t 1 � = t 2 when t 1 and t 2 denote different individuals. Example: D = { ✂ , ☎ , ✎ } . φ ( phone ) = ☎ , φ ( pencil ) = ✎ , φ ( telephone ) = ☎ What equalities and inequalities hold? � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 3
Equality Equality is a special predicate symbol with a standard domain-independent intended interpretation. Suppose interpretation I = � D , φ, π � . t 1 and t 2 are ground terms then t 1 = t 2 is true in interpretation I if t 1 and t 2 denote the same individual. That is, t 1 = t 2 if φ ( t 1 ) is the same as φ ( t 2 ). t 1 � = t 2 when t 1 and t 2 denote different individuals. Example: D = { ✂ , ☎ , ✎ } . φ ( phone ) = ☎ , φ ( pencil ) = ✎ , φ ( telephone ) = ☎ What equalities and inequalities hold? phone = telephone , phone = phone , pencil = pencil , telephone = telephone pencil � = phone , pencil � = telephone � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 4
Equality Equality is a special predicate symbol with a standard domain-independent intended interpretation. Suppose interpretation I = � D , φ, π � . t 1 and t 2 are ground terms then t 1 = t 2 is true in interpretation I if t 1 and t 2 denote the same individual. That is, t 1 = t 2 if φ ( t 1 ) is the same as φ ( t 2 ). t 1 � = t 2 when t 1 and t 2 denote different individuals. Example: D = { ✂ , ☎ , ✎ } . φ ( phone ) = ☎ , φ ( pencil ) = ✎ , φ ( telephone ) = ☎ What equalities and inequalities hold? phone = telephone , phone = phone , pencil = pencil , telephone = telephone pencil � = phone , pencil � = telephone Equality does not mean similarity! � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 5
Properties of Equality Equality is: Reflexive: X = X Symmetric: if X = Y then Y = X Transitive: if X = Y and Y = Z then X = Z For each n -ary function symbol f f ( X 1 , . . . , X n ) = f ( Y 1 , . . . , Y n ) if X 1 = Y 1 and · · · and X n = Y n . For each n -ary predicate symbol p p ( X 1 , . . . , X n ) if p ( Y 1 , . . . , Y n ) and X 1 = Y 1 and · · · and X n = Y n . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 6
Unique Names Assumption Suppose the only clauses for enrolled are enrolled ( sam , cs 222) enrolled ( chris , cs 222) enrolled ( sam , cs 873) To conclude ¬ enrolled ( chris , cs 873), what do we need to assume? � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 7
Unique Names Assumption Suppose the only clauses for enrolled are enrolled ( sam , cs 222) enrolled ( chris , cs 222) enrolled ( sam , cs 873) To conclude ¬ enrolled ( chris , cs 873), what do we need to assume? ◮ All other enrolled facts are false ◮ Inequalities: sam � = chris ∧ cs 873 � = cs 222 The unique names assumption (UNA) is the assumption that distinct ground terms denote different individuals. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 8
Completion of a knowledge base: propositional case Suppose the rules for atom a are a ← b 1 . . . . a ← b n . equivalently a ← b 1 ∨ . . . ∨ b n . Under the Complete Knowledge Assumption, if a is true, one of the b i must be true: a → b 1 ∨ . . . ∨ b n . Thus, the clauses for a mean a ↔ b 1 ∨ . . . ∨ b n � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 9
Clark Normal Form The Clark normal form of the clause p ( t 1 , . . . , t k ) ← B . is the clause p ( V 1 , . . . , V k ) ← ∃ W 1 . . . ∃ W m V 1 = t 1 ∧ . . . ∧ V k = t k ∧ B . where V 1 , . . . , V k are k variables that did not appear in the original clause W 1 , . . . , W m are the original variables in the clause. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 10
Clark Normal Form The Clark normal form of the clause p ( t 1 , . . . , t k ) ← B . is the clause p ( V 1 , . . . , V k ) ← ∃ W 1 . . . ∃ W m V 1 = t 1 ∧ . . . ∧ V k = t k ∧ B . where V 1 , . . . , V k are k variables that did not appear in the original clause W 1 , . . . , W m are the original variables in the clause. When the clause is an atomic clause, B is true . Often can be simplified by replacing ∃ W V = W ∧ p ( W ) with P ( V ). � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 11
Clark normal form For the clauses student ( mary ) . student ( sam ) . student ( X ) ← undergrad ( X ) . the Clark normal form is student ( V ) ← V = mary . student ( V ) ← V = sam . student ( V ) ← ∃ X V = X ∧ undergrad ( X ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 12
Clark’s Completion Suppose all of the clauses for p are put into Clark normal form, with the same set of introduced variables, giving p ( V 1 , . . . , V k ) ← B 1 . . . . p ( V 1 , . . . , V k ) ← B n . which is equivalent to p ( V 1 , . . . , V k ) ← B 1 ∨ . . . ∨ B n . Clark’s completion of predicate p is the equivalence ∀ V 1 . . . ∀ V k p ( V 1 , . . . , V k ) ↔ B 1 ∨ . . . ∨ B n If there are no clauses for p , � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 13
Clark’s Completion Suppose all of the clauses for p are put into Clark normal form, with the same set of introduced variables, giving p ( V 1 , . . . , V k ) ← B 1 . . . . p ( V 1 , . . . , V k ) ← B n . which is equivalent to p ( V 1 , . . . , V k ) ← B 1 ∨ . . . ∨ B n . Clark’s completion of predicate p is the equivalence ∀ V 1 . . . ∀ V k p ( V 1 , . . . , V k ) ↔ B 1 ∨ . . . ∨ B n If there are no clauses for p , the completion results in ∀ V 1 . . . ∀ V k p ( V 1 , . . . , V k ) ↔ false Clark’s completion of a knowledge base consists of the completion of every predicate symbol along the unique names assumption. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 14
Completion example p ← q ∧ ∼ r . p ← s . q ← ∼ s . r ← ∼ t . t . s ← w . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 15
Completion Example Consider the recursive definition: passed each ([ ] , St , MinPass ) . passed each ([ C | R ] , St , MinPass ) ← passed ( St , C , MinPass ) ∧ passed each ( R , St , MinPass ) . In Clark normal form, this can be written as � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 16
Completion Example Consider the recursive definition: passed each ([ ] , St , MinPass ) . passed each ([ C | R ] , St , MinPass ) ← passed ( St , C , MinPass ) ∧ passed each ( R , St , MinPass ) . In Clark normal form, this can be written as passed each ( L , S , M ) ← L = [ ] . passed each ( L , S , M ) ← ∃ C ∃ R L = [ C | R ] ∧ passed ( S , C , M ) ∧ passed each ( R , S , M ) . Here we renamed the variables as appropriate. Thus, Clark’s completion of passed each is � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 17
Completion Example Consider the recursive definition: passed each ([ ] , St , MinPass ) . passed each ([ C | R ] , St , MinPass ) ← passed ( St , C , MinPass ) ∧ passed each ( R , St , MinPass ) . In Clark normal form, this can be written as passed each ( L , S , M ) ← L = [ ] . passed each ( L , S , M ) ← ∃ C ∃ R L = [ C | R ] ∧ passed ( S , C , M ) ∧ passed each ( R , S , M ) . Here we renamed the variables as appropriate. Thus, Clark’s completion of passed each is ∀ L ∀ S ∀ M passed each ( L , S , M ) ↔ L = [ ] ∨ ∃ C ∃ R L = [ C | R ] ∧ passed ( S , C , M ) ∧ passed each ( R , S , M ) . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 18
Clark’s Completion of a KB Clark’s completion of a knowledge base consists of the completion of every predicate. The completion of an n -ary predicate p with no clauses is p ( V 1 , . . . , V n ) ↔ false . You can interpret negations in the body of clauses. ∼ a means a is false under the complete knowledge assumption. ∼ a is replaced by ¬ a in the completion. This is negation as failure. � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 19
Defining empty course Given database of: course ( C ) that is true if C is a course enrolled ( S , C ) that is true if student S is enrolled in course C . Define empty course ( C ) that is true if there are no students enrolled in course C . � D. Poole and A. Mackworth 2016 c Artificial Intelligence, Lecture 13.4, Page 20
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