Resolution and Refutation Proofs

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Resolution and Refutation Proofs

• Introduction to Artificial Intelligence
• CS440/ECE448
• Lecture 13
• Homework due March 2

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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

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Generalized Modus Ponens (GMP)

• p1, p2,, pn, (p1?p2? ?pn ? q)
  where pi? 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
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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)

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Backward Chaining Example

1. Pig(y)? Slug(z) ? Faster (y, z)
2. Slimy(a)? Creeps(a) ? Slug(a)
3. Pig(Pat)
4. Slimy(Steve)
5. Creeps(Steve)

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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.

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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 ? ?

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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 ? pj1 ? ? pm ?q1 ? ?
  qk-1 ? qk1 ? ? qn))
• where ? Unify (pj, ?qk)

• Example
• KB ? Rich(x) ? Unhappy(x)
• Rich(Me)
• Substitution ? x/Me
• Conclusion Unhappy(Me)

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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 lets stick to CNF.

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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 ? Q by ?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

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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 ?.

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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 called H1.
• 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.

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Resolution proof

• p1 ? ? pj ? ? pm
• q1 ? ? qk ? ? qn
• Subst(?, (p1 ? ? pj-1 ? pj1 ? ? pm ?q1 ? ?
  qk-1 ? qk1 ? ? 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)

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Resolution Proof
? Rich(Me) ?PhD(x) ? HighlyQualified(x) PhD(x) ?
EarlySalary(x) ?HighlyQualified(x) ? Rich(x)
?EarlySalary(x) ? Rich(x)
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Resolution Proof
? Rich(Me) ?PhD(x) ? HighlyQualified(x) PhD(x) ?
EarlySalary(x) ?HighlyQualified(x) ? Rich(x)
?EarlySalary(x) ? Rich(x)
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Resolution Proof
? Rich(Me) ?PhD(x) ? HighlyQualified(x) PhD(x) ?
EarlySalary(x) ?HighlyQualified(x) ? Rich(x)
?EarlySalary(x) ? Rich(x)
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Resolution Proof
? Rich(Me) ?PhD(x) ? HighlyQualified(x) PhD(x) ?
EarlySalary(x) ?HighlyQualified(x) ? Rich(x)
?EarlySalary(x) ? Rich(x)
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True-or-False Question

• 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

• All birds fly.
• No bird swims
• Pete is a bird
• Does Pete Fly?

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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.
• Greens trick add ??, Ans (v1 ,, vn) instead.

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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.

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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.