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Propositional Logic

Propositional Logic. Russell and Norvig: Chapter 6 Chapter 7, Sections 7.1—7.4 Slides adapted from: robotics.stanford.edu/~latombe/cs121/2003/home.htm. sensors. environment. ?. agent. actuators. Knowledge base. Knowledge-Based Agent. Types of Knowledge.

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Propositional Logic

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  1. Propositional Logic Russell and Norvig: Chapter 6Chapter 7, Sections 7.1—7.4Slides adapted from: robotics.stanford.edu/~latombe/cs121/2003/home.htm

  2. sensors environment ? agent actuators Knowledge base Knowledge-Based Agent

  3. Types of Knowledge • Procedural, e.g.: functionsSuch knowledge can only be used in one way -- by executing it • Declarative, e.g.: constraintsIt can be used to perform many different sorts of inferences

  4. Logic Logic is a declarative language to: • Assert sentences representing facts that hold in a world W (these sentences are given the value true) • Deduce the true/false values to sentences representing other aspects of W

  5. entail Sentences Sentences represent represent Conceptualization World W Facts about W Facts about W hold hold Connection World-Representation

  6. Examples of Logics • Propositional calculusA  B  C • First-order predicate calculus( x)( y) Mother(y,x) • Logic of BeliefB(John,Father(Zeus,Cronus))

  7. Model • Assignment of a truth value – true or false – to every atomic sentence • Examples: • Let A, B, C, and D be the propositional symbols • m = {A=true, B=false, C=false, D=true} is a model • m’ = {A=true, B=false, C=false} is not a model • With n propositional symbols, one can define 2n models

  8. Model of a KB • Let KB be a set of sentences • A model m is a model of KB iff it is a model of all sentences in KB, that is, all sentences in KB are true in m • Given a vocabulary A, B, C and D, how many models for A^B -> C are there? • for A^B -> B?

  9. valid sentenceor tautology Satisfiability of a KB A KB is satisfiable iff it admits at least one model; otherwise it is unsatisfiable KB1 = {P, QR} is satisfiableKB2 = {PP} is satisfiable KB3 = {P, P} is unsatisfiable

  10. Logical Entailment • KB : set of sentences •  : arbitrary sentence • KB entails – written KB  – iff every model of KB is also a model of  • Alternatively, KB  iff • {KB,} is unsatisfiable • KB   is valid

  11. Inference Rule • An inference rule {, } consists of 2 sentence patterns  and  called the conditions and one sentence pattern  called the conclusion • If  and  match two sentences of KB then the corresponding  can be inferred according to the rule 

  12. Inference • I: Set of inference rules • KB: Set of sentences • Inference is the process of applying successive inference rules from I to KB, each rule adding its conclusion to KB

  13. {  , }   {, }  Example: Modus Ponens Battery-OK  Bulbs-OK  Headlights-Work Battery-OK  Starter-OK Empty-Gas-Tank  Engine-Starts Engine-Starts Flat-Tire  Car-OK Battery-OK  Bulbs-OK

  14. KB  iff KB   is valid  Connective symbol (implication) Logical entailment Inference 

  15. Soundness • An inference rule is sound if it generates only entailed sentences • All inference rules previously given are sound, e.g.:modus ponens: {   , }  • The following rule:{   , }  is unsound, which does not mean it is useless  

  16. Completeness • A set of inference rules is complete if every entailed sentences can be obtained by applying some finite succession of these rules • Modus ponens alone is not complete, e.g.:from A  B and B, we cannot get A

  17. Proof The proof of a sentence  from a set of sentences KB is the derivation of by applying a series of sound inference rules

  18. Proof The proof of a sentence  from a set of sentences KB is the derivation of by applying a series of sound inference rules Battery-OK  Bulbs-OK  Headlights-Work Battery-OK  Starter-OK Empty-Gas-Tank  Engine-Starts Engine-Starts Flat-Tire  Car-OK Headlights-Work Battery-OK Starter-OK Empty-Gas-Tank Car-OK Battery-OK  Starter-OK  (5+6) Battery-OK  Starter-OK Empty-Gas-Tank  (9+7) Engine-Starts  (2+10) Engine-Starts  Flat-Tire  (3+8) Flat-Tire  (11+12)

  19. Inference Problem • Given: • KB: a set of sentence • : a sentence • Answer: • KB  ?

  20. Deduction vs. Satisfiability Test KB  iff {KB,} is unsatisfiable • Hence: • Deciding whether a set of sentences entails another sentence, or not • Testing whether a set of sentences is satisfiable, or not • are closely related problems

  21. Complementary Literals • A literal is a either an atomic sentence or the negated atomic sentence, e.g.: P or P • Two literals are complementary if one is the negation of the other, e.g.: P and P

  22. Unit Resolution Rule • Given two sentences:L1 …  Lp and Mwhere Li,…, Lp and M are all literals, and M and Li are complementary literals • Infer:L1 … Li-1Li+1 …  Lp

  23. Engine-Starts  Car-OK Examples From:Engine-Starts  Car-OK Engine-Starts Infer:Car-OK Modus ponens From:Engine-Starts  Car-OK Car-OK Infer:Engine-Starts Modus tolens

  24. Shortcoming of Unit Resolution From: • Engine-Starts  Flat-Tire Car-OK • Engine-Starts Empty-Gas-Tank we can infer nothing!

  25. Full Resolution Rule • Given two sentences:L1 …  Lp and M1 …  Mqwhere L1,…, Lp, M1,…, Mq are all literals, and Li and Mj are complementary literals • Infer:L1…Li-1Li+1…LkM1…Mj-1Mj+1…Mkin which only one copy of each literal is retained (factoring)

  26. Example From: Engine-Starts  Flat-Tire Car-OK Engine-Starts Empty-Gas-Tank Infer: Empty-Gas-Tank  Flat-Tire Car-OK

  27. Example From: P  Q ( P  Q) Q  R ( Q  R) Infer: P  R ( P  R)

  28. Not All Inferences are Useful! From: Engine-Starts  Flat-Tire Car-OK Engine-Starts  Flat-Tire Infer: Flat-Tire Flat-Tire Car-OK

  29. Not All Inferences are Useful! From: Engine-Starts  Flat-Tire Car-OK Engine-Starts  Flat-Tire Infer: Flat-Tire Flat-Tire Car-OK tautology

  30. Not All Inferences are Useful! From: Engine-Starts  Flat-Tire Car-OK Engine-Starts  Flat-Tire Infer: Flat-Tire Flat-Tire Car-OK  True tautology

  31. Full Resolution Rule • Given two clauses:L1 …  Lp and M1 …  Mq • Infer the clause:L1…Li-1Li+1…LkM1…Mj-1Mj+1…Mk

  32. Sentence  Clause Form Example: (A B)  (C  D) 1. Eliminate (A B)  (C  D)2. Reduce scope of  (A  B)  (C  D)3. Distribute  over (A  (C  D))  (B  (C  D)) (A  C)  (A  D)  (B  C)  (B  D) Set of clauses: {A  C , A  D , B  C , B  D}

  33. Resolution Refutation Algorithm RESOLUTION-REFUTATION(KB,a) clauses set of clauses obtained from KB and a new  {} Repeat: For each C, C’ in clauses dores  RESOLVE(C,C’) If res contains the empty clause then return yes new  new U resIf newclauses then return no clauses  clauses U new

  34. Example Battery-OK Bulbs-OK  Headlights-Work Battery-OK Starter-OK  Empty-Gas-Tank  Engine-Starts Engine-Starts  Flat-Tire  Car-OK Headlights-Work Battery-OK Starter-OK Empty-Gas-Tank Car-OK Flat-Tire

  35. Summary • Propositional Logic • Model of a KB • Logical entailment • Inference rules • Resolution rule • Clause form of a set of sentences • Resolution refutation algorithm

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