AI Proofs: Understanding Substitutions, Unification, and Refutation, Study notes of Computer Science

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Resolution and Refutation ProofsResolution and Refutation Proofs

Introduction to Artificial Intelligence

CS440/ECE

Lecture 13

Homework due March 2

Last lectureLast lecture

•^

Substitutions and unification

•^

Generalized Modus Ponens

•^

Resolution definition

This lecture^ This lecture

•^

Refutation proofs

•^

True-or-false questions

•^ •

FillFill

- in-in

- the-the

- blanks questions-blanks questions

•^ •

Resolution propertiesResolution properties

Reading^ Reading

•^

Chapter 9

Forward Chaining ExampleForward Chaining Example

•^

White: Facts added in turn

-^

Yellow:

The result of implication of rules.

1. Buffalo(x)

Pig(y)

Faster(x, y)

2. Pig(y)

Slug(z)

Faster(y, z)

3. Faster(x, y)

Faster(y, z)

Faster(x, z)

4. Buffalo(Bob)

[ Unifies with 1-a ]

5. Pig(Pat)

[ Unifies with 1-b, GMP Fires ][ Unifies with 2-a ]

6. Faster(Bob,Pat)

[ Unifies with 3-a, 3-b ]

7. Slug(Steve)

[ Unifies with 2-b, GMP Fires ]

8. Faster(Pat, Steve)

[ Unifies with 3-b and with 6,GMP Fires ]

9. Faster (Bob, Steve) 10. …

Backward Chaining ExampleBackward 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)

Resolution Inference RuleResolution 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

,^

γ^

TRUE

Generalized ResolutionGeneralized Resolution

Generalized resolution rule for first order logic (with variables)If

p

can be unified withj

qk

, then we can apply the resolution rule: p^1

p

∨j

p

m

q^1

q

∨k

q

n

Subst(

θ, (p

p

j-

p

j+

p

m^

q^1

q

k-

q

k+

q

))n

where

θ

= Unify (

pj

,^ ¬

qk

-^

Example:^ KB:

Rich(x)

Unhappy(x)

Rich(Me)

Substitution:

θ^

x/Me

Conclusion:

Unhappy(Me)

Converting any FOL to CNFConverting 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:1. Replace

P

Q

by

P

Q

2. Move

inwards to literals,

e.g.,

x P

becomes

x^

P

3. Standardize variables, e.g.,

x P)

∃x Q)

becomes (

x P)

∃y Q)

4. Move quantifiers left in order, e.g.,

x P

y Q

becomes

x^

∃y P

Q

5. Eliminate

by Skolemization (next slide)

6. Drop universal quantifiers7. Distribute

over

, e.g.,

(P

Q)

R

becomes

(P

R)

(Q

R)

8. Flatten nested conjunctions & disjunctions, e.g.

(P

Q)

R

P

Q

R

SkolemizationSkolemization

(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

G

is a new ``Skolem constant'‘.

-^

More tricky when

is inside

Resolution proofResolution proof

p^1

∨ …

∨

p

∨j …

∨

p

m

q^1

∨ …

∨

q

∨k …

∨

q

n

Subst(

θ, (p

∨ 1 …

∨

p

j-

∨ p

j+

∨ …

∨

p

m^

∨q

∨ 1 …

∨

q

k-

∨ q

k+

∨ …

∨

q

))n

•^

To prove

α

–^

Negate

α

–^

Convert to CNF.

-^

Add to CNF KB.

-^

Infer contradiction using the resolution rule (a contradiction isdetected 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 ProofResolution Proof

¬

Rich(Me) ¬

PhD(x)

∨

HighlyQualified(x)

PhD(x)

∨

EarlySalary(x)

¬

HighlyQualified(x)

∨

Rich(x)

¬

EarlySalary(x)

∨

Rich(x)

Resolution ProofResolution Proof

¬

Rich(Me) ¬

PhD(x)

∨

HighlyQualified(x)

PhD(x)

∨

EarlySalary(x)

¬

HighlyQualified(x)

∨

Rich(x)

¬

EarlySalary(x)

∨

Rich(x)

Resolution ProofResolution Proof

¬

Rich(Me) ¬

PhD(x)

∨

HighlyQualified(x)

PhD(x)

∨

EarlySalary(x)

¬

HighlyQualified(x)

∨

Rich(x)

¬

EarlySalary(x)

∨

Rich(x)

FillFill

inin

thethe

blanks Questionblanks Question

•^

Given a database KB and a sentence

with free

variables v

, …, v 1

n^

what are the bindings that

make

true?

•^

Prove that given KB,

v

, …, v 1

n^

•^

i.e., add

to the database, derive a contradiction

and find what is the subsitution leading to it.

•^

Green’s trick: add {

, Ans (v

1

,…, v

)} instead.n

DecidabilityDecidability

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.