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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

Resolution and Refutation Proofs

Introduction to Artificial Intelligence

CS440/ECE448

Lecture 13

Homework due March 2

last lecture
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
generalized modus ponens gmp
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

forward chaining example
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)
backward chaining example
Pig(y) Slug(z)  Faster (y, z)

Slimy(a) Creeps(a)  Slug(a)

Pig(Pat)

Slimy(Steve)

Creeps(Steve)

Backward Chaining Example
refutation proofs
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.
resolution inference rule
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  

generalized resolution
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)

canonical form
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.
converting any fol to cnf
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
skolemization
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 .
skolemization continued
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.
resolution proof
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)

resolution proof1
Resolution Proof

 Rich(Me)

PhD(x)  HighlyQualified(x)

PhD(x)  EarlySalary(x)

HighlyQualified(x)  Rich(x)

EarlySalary(x)  Rich(x)

resolution proof2
Resolution Proof

 Rich(Me)

PhD(x)  HighlyQualified(x)

PhD(x)  EarlySalary(x)

HighlyQualified(x)  Rich(x)

EarlySalary(x)  Rich(x)

resolution proof3
Resolution Proof

 Rich(Me)

PhD(x)  HighlyQualified(x)

PhD(x)  EarlySalary(x)

HighlyQualified(x)  Rich(x)

EarlySalary(x)  Rich(x)

resolution proof4
Resolution Proof

 Rich(Me)

PhD(x)  HighlyQualified(x)

PhD(x)  EarlySalary(x)

HighlyQualified(x)  Rich(x)

EarlySalary(x)  Rich(x)

true or false question
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
fill in the blanks question
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.
decidability
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.
resolution properties
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.