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For Friday

For Friday. No reading Homework: Chapter 9, exercise 4 (This is VERY short – do it while you’re running your tests) Make sure you keep variables and constants straight Program 1 due. Program 1. Any questions?. Backward Chaining.

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For Friday

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  1. For Friday • No reading • Homework: • Chapter 9, exercise 4(This is VERY short – do it while you’re running your tests)Make sure you keep variables and constants straight • Program 1 due

  2. Program 1 • Any questions?

  3. Backward Chaining • Start from query or atomic sentence to be proven and look for ways to prove it. • Query can contain variables which are assumed to be existentially quantified. Sibling(x,John) ? Father(x,y) ? • Inference process should return all sets of variable bindings that satisfy the query.

  4. Method • First try to answer query by unifying it to all possible facts in the KB. • Next try to prove it using a rule whose consequent unifies with the query and then try to recursively prove all of its antecedents. • Given a conjunction of queries, first get all possible answers to the first conjunct and then for each resulting substitution try to prove all of the remaining conjuncts. • Assume variables in rules are renamed (standardized apart) before each use of a rule.

  5. Backchaining Examples KB: 1) Parent(x,y) Ù Male(x) Þ Father(x,y) 2) Father(x,y) Ù Father(x,z) Þ Sibling(y,z) 3) Parent(Tom,John) 4) Male(Tom) 7) Parent(Tom,Fred) Query: Parent(Tom,x) Answers: ( {x/John}, {x/Fred})

  6. Query: Father(Tom,s) Subgoal: Parent(Tom,s) Ù Male(Tom) {s/John} Subgoal: Male(Tom) Answer: {s/John} {s/Fred} Subgoal: Male(Tom) Answer: {s/Fred} Answers: ({s/John}, {s/Fred})

  7. Query: Father(f,s) Subgoal: Parent(f,s) Ù Male(f) {f/Tom, s/John} Subgoal: Male(Tom) Answer: {f/Tom, s/John} {f/Tom, s/Fred} Subgoal: Male(Tom) Answer: {f/Tom, s/Fred} Answers: ({f/Tom,s/John}, {f/Tom,s/Fred})

  8. Query: Sibling(a,b) Subgoal: Father(f,a) Ù Father(f,b) {f/Tom, a/John} Subgoal: Father(Tom,b) {b/John} Answer: {f/Tom, a/John, b/John} {b/Fred} Answer: {f/Tom, a/John, b/Fred} {f/Tom, a/Fred} Subgoal: Father(Tom,b) {b/John} Answer: {f/Tom, a/Fred, b/John} {b/Fred} Answer: {f/Tom, a/Fred, b/Fred} Answers: ({f/Tom, a/John, b/John},{f/Tom, a/John, b/Fred} {f/Tom, a/Fred, b/John}, {f/Tom, a/Fred, b/Fred})

  9. Incompleteness • Rule­based inference is not complete, but is reasonably efficient and useful in many circumstances. • Still can be exponential or not terminate in worst case. • Incompleteness example: P(x) Þ Q(x) ¬P(x) Þ R(x) (not Horn) Q(x) Þ S(x) R(x) Þ S(x) • Entails S(A) for any constant A but is not inferable from modus ponens

  10. Completeness • In 1930 GÖdel showed that a complete inference procedure for FOPC existed, but did not demonstrate one (non­constructive proof). • In 1965, Robinson showed a resolution inference procedure that was sound and complete for FOPC. • However, the procedure may not halt if asked to prove a thoerem that is not true, it is said to be semidecidable. • If a conclusion C is entailed by the KB then the procedure will eventually terminate with a proof. However if it is not entailed, it may never halt. • It does not follow that either C or ¬C is entailed by a KB (may be independent). Therefore trying to prove both a conjecture and its negation does not help. • Inconsistency of a KB is also semidecidable.

  11. Resolution • Propositional version. {a Ú b, ¬b Ú c} |- a Ú c OR {¬aÞ b, b Þ c} |- ¬a Þ c Reasoning by cases OR transitivity of implication • First­order form • For two literals pj and qk in two clauses • p1Ú ... pj ... Ú pm • q1Ú ... qk ... Ú qn such that q=UNIFY(pj , ¬qk), derive SUBST(q, p1Ú...pj­1Úpj+1...ÚpmÚq1Ú...qk­1 qk+1...Úqn)

  12. Implication form • Can also be viewed in implicational form where all negated literals are in a conjunctive antecedent and all positive literals in a disjunctive conclusion. ¬p1Ú...Ú¬pmÚq1Ú...ÚqnÛ p1Ù... Ù pmÞ q1Ú ...Ú qn

  13. Conjunctive Normal Form (CNF) • For resolution to apply, all sentences must be in conjunctive normal form, a conjunction of disjunctions of literals (a1Ú ...Ú am) Ù (b1Ú ... Ú bn) Ù ..... Ù (x1Ú ... Ú xv) • Representable by a set of clauses (disjunctions of literals) • Also representable as a set of implications (INF).

  14. Example Initial CNF INF P(x) Þ Q(x) ¬P(x) Ú Q(x) P(x) Þ Q(x) ¬P(x) Þ R(x) P(x) Ú R(x) True Þ P(x) Ú R(x) Q(x) Þ S(x) ¬Q(x) Ú S(x) Q(x) Þ S(x) R(x) Þ S(x) ¬R(x) Ú S(x) R(x) Þ S(x)

  15. Resolution Proofs • INF (CNF) is more expressive than Horn clauses. • Resolution is simply a generalization of modus ponens. • As with modus ponens, chains of resolution steps can be used to construct proofs. • Factoring removes redundant literals from clauses • S(A) Ú S(A) -> S(A)

  16. Sample Proof P(w)  Q(w) Q(y)  S(y) {y/w} P(w)  S(w) True  P(x)  R(x) {w/x} True  S(x)  R(x) R(z)  S(z) {x/A, z/A} True  S(A)

  17. Refutation Proofs • Unfortunately, resolution proofs in this form are still incomplete. • For example, it cannot prove any tautology (e.g. PÚ¬P) from the empty KB since there are no clauses to resolve. • Therefore, use proof by contradiction (refutation, reductio ad absurdum). Assume the negation of the theorem P and try to derive a contradiction (False, the empty clause). • (KB Ù ¬P Þ False) Û KB Þ P

  18. Sample Proof P(w)  Q(w) Q(y)  S(y) {y/w} P(w)  S(w) True  P(x)  R(x) {w/x} True  S(x)  R(x) R(z)  S(z) {z/x} S(A)  False True  S(x) {x/A} False

  19. Resolution Theorem Proving • Convert sentences in the KB to CNF (clausal form) • Take the negation of the proposed theorem (query), convert it to CNF, and add it to the KB. • Repeatedly apply the resolution rule to derive new clauses. • If the empty clause (False) is eventually derived, stop and conclude that the proposed theorem is true.

  20. Conversion to Clausal Form • Eliminate implications and biconditionals by rewriting them. p Þ q -> ¬p Ú q p Û q ­> (¬p Ú q) Ù (p Ú ¬q) • Move ¬ inward to only be a part of literals by using deMorgan's laws and quantifier rules. ¬(p Ú q) -> ¬p Ù ¬q ¬(p Ù q) -> ¬p Ú¬q ¬"x p -> $x ¬p ¬$x p -> "x ¬p ¬¬p -> p

  21. Conversion continued • Standardize variables to avoid use of the same variable name by two different quantifiers. "x P(x) Ú$x P(x) -> "x1 P(x1) Ú $x2 P(x2) • Move quantifiers left while maintaining order. Renaming above guarantees this is a truth­preserving transformation. "x1 P(x1) Ú $x2 P(x2) -> "x1$x2 (P(x1) Ú P(x2))

  22. Conversion continued • Skolemize: Remove existential quantifiers by replacing each existentially quantified variable with a Skolem constant or Skolem function as appropriate. • If an existential variable is not within the scope of any universally quantified variable, then replace every instance of the variable with the same unique constant that does not appear anywhere else. $x (P(x) Ù Q(x)) -> P(C1) Ù Q(C1) • If it is within the scope of n universally quantified variables, then replace it with a unique n­ary function over these universally quantified variables. "x1$x2(P(x1) Ú P(x2)) -> "x1 (P(x1) Ú P(f1(x1))) "x(Person(x) Þ$y(Heart(y) Ù Has(x,y))) -> "x(Person(x) Þ Heart(HeartOf(x)) Ù Has(x,HeartOf(x))) • Afterwards, all variables can be assumed to be universally quantified, so remove all quantifiers.

  23. Conversion continued • Distribute Ù over Ú to convert to conjunctions of clauses (aÙb) Ú c -> (aÚc) Ù (bÚc) (aÙb) Ú (cÙd) -> (aÚc) Ù (bÚc) Ù (aÚd) Ù (bÚd) • Can exponentially expand size of sentence. • Flatten nested conjunctions and disjunctions to get final CNF (a Ú b) Ú c -> (a Ú b Ú c) (a Ù b) Ù c -> (a Ù b Ù c) • Convert clauses to implications if desired for readability (¬a Ú ¬b Ú c Ú d) -> a Ù b Þ c Ú d

  24. Sample Clause Conversion "x((Prof(x) Ú Student(x)) Þ($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x(¬(Prof(x) Ú Student(x)) Ú($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù$z(Book(z) Ù Has(x,z)))) "x$y$z((¬Prof(x)Ù¬Student(x))Ú ((Class(y) Ù Has(x,y)) Ù (Book(z) Ù Has(x,z)))) (¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù Book(g(x)) Ù Has(x,g(x))))

  25. Clause Conversion (¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù Book(g(x)) Ù Has(x,g(x)))) (¬Prof(x) Ú Class(f(x))) Ù (¬Prof(x) Ú Has(x,f(x))) Ù (¬Prof(x) Ú Book(g(x))) Ù (¬Prof(x) Ú Has(x,g(x))) Ù (¬Student(x) Ú Class(f(x))) Ù (¬Student(x) Ú Has(x,f(x))) Ù (¬Student(x) Ú Book(g(x))) Ù (¬Student(x) Ú Has(x,g(x))))

  26. Sample Resolution Problem • Jack owns a dog. • Every dog owner is an animal lover. • No animal lover kills an animal. • Either Jack or Curiosity killed Tuna the cat. • Did Curiosity kill the cat?

  27. In Logic Form A) $x Dog(x) Ù Owns(Jack,x) B) "x ($y Dog(y) Ù Owns(x,y)) Þ AnimalLover(x)) C) "x AnimalLover(x) Þ ("y Animal(y) Þ ¬Kills(x,y)) D) Kills(Jack,Tuna) Ú Kills(Cursiosity,Tuna) E) Cat(Tuna) F) "x(Cat(x) Þ Animal(x)) Query: Kills(Curiosity,Tuna)

  28. In Normal Form A1) Dog(D) A2) Owns(Jack,D) B) Dog(y) Ù Owns(x,y) Þ AnimalLover(x) C) AnimalLover(x) Ù Animal(y) Ù Kills(x,y) Þ False D) Kills(Jack,Tuna) Ú Kills(Curiosity,Tuna) E) Cat(Tuna) F) Cat(x) Þ Animal(x) Query: Kills(Curiosity,Tuna) Þ False

  29. Resolution Proof

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