first order logic (1)

8/13/2019 First Order Logic (1)

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Inference in first-order logic I

CIS 391Introduction to Artificial Intelligence

AIMA Chapter 9.1-9.2 (through p. 278)

Chapter 9.5 (through p. 300)


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CIS 391 - Intro to AI 2

Outline

Reducing first-order inference to propositionalinference

Unification

Generalized Modus Ponens


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Universal instantiation (UI)

Every instantiation of a universally quantified sentence isentailed by it:

v Subst({v/g}, )

for any variable vand ground term g

E.g., x King(x) Greedy(x) Evil (x)yields any or allof:

King(John) Greedy(John) Evil(John)

King(Richard) Greedy(Richard)Evil(Richard)King(Father(John)) Greedy(Father(John))

Evil(Father(John))


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Existential instantiation (EI)

For any sentence , variable v, and constantsymbol k that does no t appear elsewherein theknowledge base:

v

Subst({v/k}, )

E.g., x Crown(x) OnHead(x,John)yields:Crown(C1) OnHead(C1,John)

provided C1is a new constant symbol, called aSkolem constant


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Existential Instantiation cont inued

UI can be applied several times to addnewsentences

the new KB is logically equivalent to the old

EI can be applied once to replacethe

existential sentence

the new KB is not equivalent to the old

but it is satisfiable iff the old KB wassatisfiable


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Reduction contd.

Claim: Every FOL KB can be propositionalized soas to preserve entailment

(A ground sentence is entailed by new KB iff entailed by

original KB)

Idea: propositionalize KB and query, applyresolution, return result

Problem: with function symbols, there are

infinitely many ground terms, e.g., Father(Father(Father(John)))


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Reduction contd.

Theorem: Herbrand (1930). If a sentence is entailed by anFOL KB, it is entailed by a f inite subset of thepropositionalized KB

Idea: For n= 0 to docreate a propositional KB by instantiating with depth-nterms

see if is entailed by this KB

Problem: works if is entailed, loops if is not entailed

Theorem: Turing (1936), Church (1936) Entailment for FOL is

semidecidable(algorithms exist that say yes to everyentailed sentence, but no algorithm exists that also says noto every nonentailed sentence.)


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Problems withpropositionalization

1. Propositionalization seems to generate lots of irrelevantsentences.

E.g., from:

x King(x) Greedy(x) Evil (x)King(John)

y Greedy(y)Brother(Richard,John)

it seems obvious that Evil(John),but propositionalization

produces lots of facts such as Greedy(Richard)that are irrelevant

2. With p k-ary predicates and nconstants, there are pnk

instantiations.

With function symbols, it gets much, much worse!


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Unification

We can get the inference Evil(John) immediately if we can find a

substitution such that King(x)and Greedy(x)match

King(John)and Greedy(y)

= {x/John,y/John}works

Unify(,) = if =

p q Knows(John,x) Knows(John,Jane)

Knows(John,x) Knows(y,OJ)

Knows(John,x) Knows(y,Mother(y))Knows(John,x) Knows(x,OJ)


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CIS 391 - Intro to AI11

Unification





Unify(,) = if =

p q Knows(John,x) Knows(John,Jane) {x/Jane}}

Knows(John,x) Knows(y,OJ)



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Unification





Unify(,) = if =


Knows(John,x) Knows(y,OJ) {x/OJ,y/John}}



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Unification





Unify(,) = if =



Knows(John,x) Knows(y,Mother(y)) {y/John,x/Mother(John)}}Knows(John,x) Knows(x,OJ) fai l


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Unification





Unify(,) = if =



Knows(John,x) Knows(y,Mother(y)) {y/John,x/Mother(John)}}Knows(John,x) Knows(x,OJ) fai l

Standardizing apart eliminates overlap of variables, e.g.,

Knows(z17,OJ)


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Unification

To unify Knows(John ,x) and Knows(y,z),= {y/John , x/z }or = {y/John , x/John , z/John }

The first unifier is more generalthan the second.

There is a single most general unifier(MGU) thatis unique up to renaming of variables.

MGU = { y/John, x/z }


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The unification algorithm


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The unification algorithm


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Generalized Modus Ponens (GMP)

p1

', p2

', , pn

', ( p1

p2

pn

q)q

p1'is King(John) p1 is King(x)

p2'is Greedy(y) p2is Greedy(x)is {x/John,y/John} qis Evil(x)q is Evi l(John)

GMP used with KB of defini te clauses (exactlyonepositive

literal)

All variables assumed universally quantified

wherepi'= pifor all i


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Soundness of GMP

Need to show that

p1', , pn', (p1 pnq) qprovided that pi '= pifor all i

Lemma: For any definite clause p, we have p pby UI

1. (p1 pnq) (p1 pnq)= (p1 pnq)2. p1', , pn' p1' pn' p1' pn'3. From 1 and 2, qfollows by ordinary Modus Ponens


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Resolution: brief summary

Full first-order version:

1 k, 1 n(1 i-1 i+1 k 1 j-1 j+1 n)where Unify(i, j) = . The two clauses are assumed to be standardized apart so that

they share no variables.

For example,

Rich(x) Unhappy(x)Rich(Ken)

Unhappy(Ken)with = {x/Ken}

Apply resolution steps to CNF(KB ); complete for FOL


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Conversion to CNF for FOL

Everyone who loves all animals is loved by someone:x [ y An imal(y) Lo ves(x,y)] [y Loves(y,x)]

1. Eliminate biconditionals and implicationsx [y Animal(y) Loves(x,y)] [y Loves(y,x)]

2. Move inwards: x p x p, x p x px [y (Animal(y) Loves(x,y))] [y Loves(y,x)]

x [y Animal(y) Loves(x,y)] [y Loves(y,x)]

x [y Animal(y) Loves(x,y)] [y Loves(y,x)]


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Conversion to CNF contd.

3. Standardize variables: each quantifier should use adifferent onex [y An imal(y) Loves(x,y)] [z Loves(z,x)]

4. Skolemize: a more general form of existentialinstantiation.

Each existential variable is replaced by a Skolem functionof theenclosing universally quantified variables:x [Animal(F(x)) Loves(x,F(x))] Loves(G(x),x)

5. Drop universal quantifiers:[Animal(F(x)) Loves(x,F(x))] Loves(G(x),x)

6. Distribute over :[Animal(F(x)) Lov es(G(x),x)] [Loves(x,F(x)) Loves(G(x),x)]


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Example knowledge base

The law says that it is a crime for an American tosell weapons to hostile nations. The country

Nono, an enemy of America, has some missiles,

and all of its missiles were sold to it by Colonel

West, who is American.

Prove that Col. West is a criminal


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Example knowledge base contd.

... it is a crime for an American to sell weapons to hostile nations:

American(x) Weapon(y) Sells(x,y,z) Hostile(z) Criminal(x)

Nono has some missiles, i.e., x Owns(Nono,x) Missile(x):Owns(Nono,M1) and Missile(M1)

all of its missiles were sold to it by Colonel West

Missile(x) Owns(Nono,x) Sells(West,x,Nono)

Missiles are weapons:Missile(x) Weapon(x)

An enemy of America counts as "hostile:Enemy(x,America) Hostile(x)

West, who is American American(West)

The country Nono, an enemy of America Enemy(Nono,America)


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Resolution proof: definite clauses


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CIS 391 - Intro to AI

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Example proof: Did Curiosity kill the cat?

Jack owns a dog. Every dog owner is an animal lover. Noanimal lover kills an animal. Either Jack or Curiosity killedthe cat, who is named Tuna. Did Curiosity kill the cat?

The axioms can be represented as follows:

A. (x) Dog(x) ^ Owns(Jack,x)

B. (x) ((y) Dog(y) ^ Owns(x, y)) => AnimalLover(x)

C. (x) AnimalLover(x) => (y) Animal(y) => ~Kills(x,y)

D. Kills(Jack,Tuna) Kills(Curiosity,Tuna)

E. Cat(Tuna)

F. (x) Cat(x) => Animal(x)

(This and next 2 sl ides adapted from

CMSC421Fall 2006, Univ . of Maryland)


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Example: Did Curiosity kill the cat?1. Dog(spike)

2. Owns(Jack,spike)

3. ~Dog(y) v ~Owns(x, y) v AnimalLover(x)

4. ~AnimalLover(x1) v ~Animal(y1) v ~Kills(x1,y1)

5. Kills(Jack,Tuna) v Kills(Curiosity,Tuna)

6. Cat(Tuna)

7. ~Cat(x2) v Animal(x2)


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1. Dog(spike)

2. Owns(Jack,spike)

3. ~Dog(y) v ~Owns(x, y) v AnimalLover(x)

4. ~AnimalLover(x1) v ~Animal(y1) v ~Kills(x1,y1)

5. Kills(Jack,Tuna) v Kills(Curiosity,Tuna)

6. Cat(Tuna)

7. ~Cat(x2) v Animal(x2)

8. ~Kills(Curiosity,Tuna) negated goal

9. Kills(Jack,Tuna) 5,8

10. ~AnimalLover(Jack) V ~Animal(Tuna) 9,4 x1/Jack,y1/Tuna

11. ~Dog(y) v ~Owns(Jack,y) V ~Animal(Tuna) 10,3 x/Jack

12. ~Owns(Jack,spike) v ~Animal(Tuna) 11,1

13. ~Animal(Tuna) 12,2

14. ~Cat(Tuna) 13,7 x2/Tuna

15. False 14,6

Example: Did Curiosity kill the cat?

first order logic (1)

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