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Resolution and Refutation Proofs

Resolution and Refutation Proofs. Introduction to Artificial Intelligence CS440/ECE448 Lecture 13 Homework due March 2. Last lecture. Substitutions and unification Generalized Modus Ponens Resolution definition This lecture Refutation proofs True-or-false questions

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Resolution and Refutation Proofs

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  1. Resolution and Refutation Proofs Introduction to Artificial Intelligence CS440/ECE448 Lecture 13 Homework due March 2

  2. Last lecture • Substitutions and unification • Generalized Modus Ponens • Resolution definition This lecture • Refutation proofs • True-or-false questions • Fill-in-the-blanks questions • Resolution properties Reading • Chapter 9

  3. Generalized Modus Ponens (GMP) p1’, p2’,…, pn’, (p1p2… pn  q) wherepi’ = pi q for all i For example, let p1’ = Faster (Bob, Pat) p2’ = Faster (Pat, Steve) Faster (x, y)  Faster (y, z)  Faster (x, z) Unify p1’ and p2’ with the premise  = { x/Bob, y/Pat, z/Steve } Apply substitution to the conclusion q = Faster(Bob, Steve) pi and q atomic sentences Universally quantified variables

  4. Forward Chaining Example • White: Facts added in turn • Yellow: The result of implication of rules. • Buffalo(x) Pig(y)  Faster(x, y) • Pig(y) Slug(z)  Faster(y, z) • Faster(x, y) Faster(y, z)  Faster(x, z) • Buffalo(Bob) [ Unifies with 1-a ] • Pig(Pat) [ Unifies with 1-b, GMP Fires ] [ Unifies with 2-a ] • Faster(Bob,Pat) [ Unifies with 3-a, 3-b ] • Slug(Steve) [ Unifies with 2-b, GMP Fires ] • Faster(Pat, Steve) [ Unifies with 3-b and with 6,GMP Fires ] • Faster (Bob, Steve) • …

  5. Pig(y) Slug(z)  Faster (y, z) Slimy(a) Creeps(a)  Slug(a) Pig(Pat) Slimy(Steve) Creeps(Steve) Backward Chaining Example

  6. Refutation Proofs • Given • a knowledge base KB (collection of true sentences), • a proposition P, we wish to prove that P is true. • Proof by contradiction (refutation): • Assume that P is FALSE (i.e., that P is TRUE). • Show that a contradiction arises. • A complete approach to refutation can be obtained using a single inference rule: resolution.

  7. Resolution Inference Rule • Idea: If  is true or  is true and  is false or  is true then  or  must be true • Basic resolution rule from propositional logic:          • Can be expressed in terms of implications   ,       • Note that Resolution rule is a generalization of Modus Ponens ,    is equivalent to TRUE  ,     TRUE  

  8. Generalized Resolution Generalized resolution rule for first order logic (with variables) If pj can be unified with qk, then we can apply the resolution rule: p1  …  pj …  pm q1  …  qk …  qn Subst(, (p1  …  pj-1  pj+1 …  pm q1  …  qk-1  qk+1 …  qn)) where  = Unify (pj, qk) • Example: KB:  Rich(x)  Unhappy(x) Rich(Me) Substitution:  = { x/Me } Conclusion: Unhappy(Me)

  9. Canonical Form • For generalized Modus Ponens, entire knowledge base is represented as Horn Sentences. • For resolution, entire database will be represented using Conjunctive Normal Form (CNF) • Any first order logic sentence can be converted to a Canonical CNF form. • Note: Can also do resolution with implicative form, but let’s stick to CNF.

  10. Converting any FOL to CNF • Literal = (possibly negated) atomic sentence, e.g.,  Rich(Me) • Clause = disjunction of literals, e.g.,  Rich(Me)  Unhappy(Me) • The KB is a conjunction of clauses • Any FOL sentence can be converted to CNF as follows: • Replace P  QbyP  Q • Move inwards to literals, e.g., x P becomes x P • Standardize variables, e.g., (x P)  (x Q) becomes (x P)  (y Q) • Move quantifiers left in order, e.g., x P  y Q becomes x y P  Q • Eliminate  by Skolemization (next slide) • Drop universal quantifiers • Distribute  over  , e.g., (P  Q)  R becomes (P  R)  (Q R) • Flatten nested conjunctions & disjunctions, e.g. (P  Q)  R  P  Q  R

  11. Skolemization (Thoralf Skolem 1920) • The process of removing existential quantifiers by elimination. • Simple case: No universal quantifiers.  Existential Elimination Rule • For example: x Rich(x) becomes Rich(G1) where G1 is a new ``Skolem constant'‘. • More tricky when  is inside .

  12. Skolemization – continued • More tricky when  is inside  E.g., ``Everyone has a heart'' x Person(x)   y Heart(y)  Has(x,y) • Incorrect: x Person(x)  Heart(H1)  Has(x,H1) This means everyone has the same heart calledH1. • Problem is that for each person, we need another “heart” – i.e., consider the “heart” to be a function of the person. • Correct:  Person(x)  Heart(H(x))  Has(x,H(x)) where H is a new symbol (``Skolem function'') • Skolem function arguments: all enclosing universally quantified variables.

  13. Resolution proof p1  …  pj …  pm q1  …  qk …  qn Subst(, (p1  …  pj-1  pj+1 …  pm q1  …  qk-1  qk+1 …  qn)) • To prove : • Negate . • Convert to CNF. • Add to CNF KB. • Infer contradiction using the resolution rule (a contradiction is detected when resolution derives the empty clause). • E.g., to prove Rich(Me),add Rich(Me) to the CNF KB, then: PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

  14. Resolution Proof  Rich(Me) PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

  15. Resolution Proof  Rich(Me) PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

  16. Resolution Proof  Rich(Me) PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

  17. Resolution Proof  Rich(Me) PhD(x)  HighlyQualified(x) PhD(x)  EarlySalary(x) HighlyQualified(x)  Rich(x) EarlySalary(x)  Rich(x)

  18. All birds fly. No bird swims Pete is a bird Does Pete Fly? The Knowledge base 1. Bird(x) Flies(x) 2. Bird(y)Swims(y) Bird(Pete) The query  Flies ( Pete) Applying Resolution 5. Swims(Pete) 2,3  Bird(Pete) 1,4 { } 3,6 True-or-False Question

  19. Fill-in-the-blanks Question • Given a database KB and a sentence  with free variables v1, …, vn what are the bindings that make  true? • Prove that given KB,  v1, …, vn  • i.e., add  to the database, derive a contradiction and find what is the subsitution leading to it. • Green’s trick: add {, Ans (v1 ,…, vn)} instead.

  20. Decidability How hard is it to determine if KB entails ? • Propositional logic (zeroth order) is decidable: • Can determine whether or not KB entails  in finite time. • Second order logic is undecidable: • Cannot determine whether KB entails  in finite time. • First order logic is semi-decidable: • If KB entails  or , then a proof will be found in finite time. • But, if KB neither entails  or , then proof process may never terminate.

  21. Resolution properties • Resolution is sound. • Resolution refutation is complete. (See Section 9.5 for proof.) • Resolution (and any proof procedure) is at best semi-decidable. • Note: checking consistency of a database is also semidecidable.

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