cs335 introduction to aifid.cl/courses/ai/firstorderlogic/slides_firstorderlogic.pdffirst order...

CS335 Introduction to AI

Francisco Iacobelli

July 20, 2015

First Order LogicCan intelligent systems deduce stuff?

One Sherlock Holmes

Another Sherlock

BBC One’s Sherlock (2010-) with Benedict Cumberbach and Martin Freeman.

First Order LogicRepresentation Of Language

I Whorf (1956) suggest that communities determine lang.categories

I Wanner (1975) subject remember the content of what theyread better than the actual words.

I Mitchell et al. (2008) could predict –with above chanceaccuracy– the areas of the brain that would activate withcertain words(fMRI)

First Order LogicFormal/Natural Languages

I Objects (cat, dog, house, John, etc.)I Relations (has color, bigger than, comes between, etc.)I Facts: (One value for a given input: has father, has head,

can swim)

Facts have a truth value. true or false

Formal LanguagesOntological and Epistemological Commitments

Language Ontological Commitment1 Epistemological Commitment2

Propositional Logic facts true/false/unknownFirst-Order Logic facts, objects, relations true/false/unknownTemporal Logic facts,objects, relations, time true/false/unknownProbability Theory facts degree of belief ∈ [0, 1]Fuzzy Logic facts with degree of truth known interval value

1What exists in the world2Agent’s beliefs about facts

RelationshipsModels for first order logic

94 Chapter 8. First-Order Logic

R J$

left leg

on headbrother

brother

person personking

crown

left leg

Figure 8.2 FILES: figures/fol-model.eps (Tue Nov 3 16:22:52 2009). A model containing fiveobjects, two binary relations, three unary relations (indicated by labels on the objects), and one unaryfunction, left-leg.

Relations are just tuples:brotherhood = {< Richard , John >,< John,Richard >};on − head : {< crown, John >}

First-Order-LogicSyntax

Sentence → AtomicSentence|ComplexSentenceAtomicSentence → Predicate|Predicate(Term, . . .)|Term = Term

ComplexSentence → (Sentence)|[Sentence]| ¬ Sentence| Sentence ∧ Sentence| Sentence ∨ Sentence| Sentence ⇒ Sentence| Sentence ⇔ Sentence| Quantifier Variable, . . . Sentence

Term → Function(Term, . . .)| Constant| Variable

Quantifier → ∀|∃Constant → A|X1|John| . . .Variable → a|x |s| . . .

Predicate → True|False|After |Loves|Raining| . . .Function → Mother |Left leg| . . .

Operator Precedence3 : ¬,∧,∨,⇒,↔3Otherwise the grammar is ambiguous

First Order LogicMore on Syntax

I Three kinds of symbolsI Constant: objectsI Predicate: relationsI Function: funcions (i.e. can return values other than truth

vals.)I Predicate and Function have arity.I Symbols have an interpretationI Terms: LeftLeg(John)I Atomic Sentences state facts: Brother(Richard , John)I Complex Sentence: Brother(R, J) ∧ Brother(J,R) or¬King(Richard)⇒ King(John)

I Universal Quantifiers: ∀King(x)⇒ Person(x)I Existential Quantifiers: ∃Crown(x) ∧OnHead(x , John)

First Order LogicTry this

What is the interpretation for:

I King(Richard) ∨ King(John)I ¬Brother(LeftLeg(Richard), John)I ∀x∀yBrother(x , y)⇒ Sibling(x , y)I In(Paris,France) ∧ In(Marseilles,France)I ∀c Country(c) ∧ Border(c,Ecuador)⇒

In(c,SouthAmerica)I ∃Country(c) ∧ Border(c,Spain) ∧ Border(c, Italy)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

First Order LogicMore Facts

• Richard has only two brothers, John and Geoffrey:

Brother(John,Richard) ∧ Brother(Geoffrey ,Richard) ∧ John 6= Geoffrey∧∀xBrother(x ,Richard)⇒ (x = John ∨ x = Geoffrey)

• No Region in South America borders any region in Europe

∀c,d In(c,SouthAmerica) ∧ In(d ,Europe)⇒ ¬Border(c,d)

• No two adjacent countries have the same map color

∀x , y Country(x) ∧ Country(y) ∧ Border(x , y)⇒¬(Color(x) = Color(y)) ∧ ¬(x = y)

Assertions and Queries in FOLASK and TELL

I TELL(KB,King(John))I TELL(KB, ∀x King(x)⇒ Person(x))

I ASK (KB,King(John)) return TrueI ASK (KB,∃x Person(x)) return True

I ASKVARS(KB,Person(x)) yields {x/John, x/Richard}, abinding list

First Order LogicKinship

“The son of my father is my brother”; “One’s grandmother is themother of one’s parent”; etc.

I Domain: People.I Unary predicates: Male,FemaleI Relations:

Parent ,Sibling,Brother ,Sister ,Child ,Daughter ,Son,Spouse,Wife,Husband ,Grandparent ,Grandchild ,Cousin,Aunt ,Uncle

I Functions: Mother ,Father

First Order LogicKinship

“One’s mother is one’s female parent”∀m, c Mother(c) = m↔ Female(m) ∧ Parent(m, c)

“A sibling is another child of one’s parents”∀x , y Sibling(x , y)↔ x 6= y ∧ ∃p Parent(p, x) ∧ Parent(p, y)

“Wendy is female”

Female(wendy)

These are axioms

First Order LogicKinship

“One’s mother is one’s female parent”∀m, c Mother(c) = m↔ Female(m) ∧ Parent(m, c)

“A sibling is another child of one’s parents”∀x , y Sibling(x , y)↔ x 6= y ∧ ∃p Parent(p, x) ∧ Parent(p, y)

“Wendy is female”

Female(wendy)

These are axioms

First Order LogicKinship

“One’s mother is one’s female parent”∀m, c Mother(c) = m↔ Female(m) ∧ Parent(m, c)

“A sibling is another child of one’s parents”∀x , y Sibling(x , y)↔ x 6= y ∧ ∃p Parent(p, x) ∧ Parent(p, y)

“Wendy is female”

Female(wendy)

These are axioms

First Order LogicKinship

“One’s mother is one’s female parent”∀m, c Mother(c) = m↔ Female(m) ∧ Parent(m, c)

“A sibling is another child of one’s parents”∀x , y Sibling(x , y)↔ x 6= y ∧ ∃p Parent(p, x) ∧ Parent(p, y)

“Wendy is female”

Female(wendy)

These are axioms

First Order LogicKinship

“One’s mother is one’s female parent”∀m, c Mother(c) = m↔ Female(m) ∧ Parent(m, c)

“A sibling is another child of one’s parents”∀x , y Sibling(x , y)↔ x 6= y ∧ ∃p Parent(p, x) ∧ Parent(p, y)

“Wendy is female”

Female(wendy)

These are axioms

First Order LogicWumpus: Time domain included

Can be represented more concisely

I at time step3:Percept([Stench,Breeze,Glitter ,None,None],3)

I at time step6:Percept([None,Breeze,None,None,Scream],6)

I Actions can be:Turn(Right),Turn(Left),Forward ,Shoot ,Grab,Climb

And we can ASK the best action at time step 5ASKVARS(∃a BestAction(a,5))

FOL: WumpusEncoding complex rules

Can encode:I Raw percepts:∀t , s,g,m, c Percept([s,b,Glitter ,m, c], t)⇒ Glitter(t)

I Reflex actions: ∀t Glitter(t)⇒ BestAction(Grab, t)

Instead of encoding stuff like:

Adjacent(Square1,2,Square1,1)Adjacent(Square3,4,Square4,4)

Encode:∀x , y ,a,b Adjacent([x , y ], [a,b])⇔

(x = a∧(y = b−1∨y = b+1))∨(y = b∧(x = a−1∨x = a+1))

First Order LogicLegos

I Define pieces: Long(p),Short(p),etc.I and restrictions: ∀p Long(p)↔ ¬(Long(p) ∧ Short(p))I Define rules: ∀x , y , z CanConnectWithOverlap(x , y , z)↔

x 6= y ∧ Piece(x) ∧ Piece(y) ∧ Number(z) ∧ Value(z) ≤Overlap(x , y)...

I ∀x , y Short(x) ∧ Short(y) ∧Overlap(x , y) < 1⇒WeakLink(x , y)

PrologExample

Alain Colmerauer (1972)

Download SWI Prolog

And a quick tutorial

Creating a Knowledge Base

I Identify the TaskI Assemble the relevant knowledgeI Decide on a vocabulary of predicates, functions and

constantsI Encode general knowledge about the domain (rules)I Encode a description of the problemI Pose queries to the inference procedure and get answersI Debug the KB

Inferencein First Order Logic

Given ∀x King(x) ∧Greedy(x)⇒ Evil(x)

One can infer

I King(John) ∧Greedy(John)⇒ Evil(John)I King(Richard) ∧Greedy(Richard)⇒ Evil(Richard)I King(Father(John)) ∧Greedy(Father(John))⇒

Evil(Father(John))

Inferencein First Order Logic

I Universal Instantiation (in a ∀ rule, substitute all symbols)I Existential Instantiation (in a ∃ rule, substitute one symbol,

use the rule and discard)I Skolem Constants is a new variable that represents a new

inference.

Inferencein First Order Logic

Suppose KB:I ∀x King(x) ∧Greedy(x)⇒ Evil(x)I King(John)I Greedy(John)I Brother(Richard , John)

Apply UI using {x/John}and{x/Richard}I King(John) ∧Greedy(John)⇒ Evil(John)I King(Richard) ∧Greedy(Richard)⇒ Evil(Richard)

And discard the Universally quantified sentence. We can getthe KB to be propositions.

Inferencein First Order Logic

Suppose KB:I ∀x King(x) ∧Greedy(x)⇒ Evil(x)I King(John)I ∀y Greedy(y)

Apply UI using {x/John}and{y/John}

InferenceGeneralized Modus Ponens

for atomic sentences pi ,p′

i and q, where there is a substitutionθ such that SUBST (θ,p

′

i ) = SUBST (θ,pi), for all i

p′

1,p′

2, . . . ,p′n, (p1 ∧ p2 ∧ . . . ∧ pn ⇒ q)SUBST (θ,q)

p′

1 = King(John) p1 = King(x)p

′

2 = Greedy(y) p2 = Greedy(x)θ = {x/John, y/John} q = Evil(x)SUBST (θ,q) .

InferenceUnification

UNIFY (p,q) = θ Where SUBST (θ,p) = SUBST (θ,q)

For example:I We ask ASKVARS(Knows(John, x)) (Whom does John

know?)I UNIFY (Knows(John, x),Knows(John, Jane)) = {x/Jane}I UNIFY (Knows(John, x),Knows(y ,Bill)) ={y/John, x/Bill}

I UNIFY (Knows(John, x),Knows(y ,Mother(y))) ={y/John, x/Mother(John)}

Algorithm in the book (goes variable by variable recursively unifying)

InferencePutting it all together

“The Law says that it is a crime for an American to sellweapons to hostile nations. The country Nono, and enemy ofAmerica, has some missiles , and all of its missiles were sold toit by Colonel West, who is American”

Prove that Colonel Wes is a Criminal

InferenceThe KB

“The Law says that it is a crime for an American to sellweapons to hostile nations. The country Nono, and enemy ofAmerica, has some missiles , and all of its missiles were sold toit by Colonel West, who is American”

I R1:American(x) ∧ Weapon(y) ∧ Sells(x , y .z) ∧ Hostile(z) ⇒ Criminal(x)

I R2: Owns(Nono,M1) Nono has some missilesI R3: Missile(M1)

I R4: Missile(x) ⇒ Weapon(x) A missile is a weaponI R5: Missile(x) ∧ Owns(Nono, x) ⇒ Sells(West , x ,Nono) All missiles

sold by westI R6: Enemy(x ,America) ⇒ Hostile(x) Enemies of America are hostileI R7: American(West) West is americanI R8: Enemy(Nono,America)

InferenceGraph

102 Chapter 9. Inference in First-Order Logic

Hostile(Nono)

Enemy(Nono,America)Owns(Nono,M1)Missile(M1)American(West)

Weapon(M1)

Criminal(West)

Sells(West,M1,Nono)

Figure 9.4 FILES: figures/crime-fc.eps (Tue Nov 3 16:22:35 2009). The proof tree generated byforward chaining on the crime example. The initial facts appear at the bottom level, facts inferred onthe first iteration in the middle level, and facts inferred on the second iteration at the top level.

Forward Chaining ASKIterations

Iteration 1:I R5 satisfied with {x/M1} and R9: Sells(West ,M1,Nono) is

addedI R4 satisfied with {x/M1} and R10: Weapon(M1) is addedI R6 satisfied with {x/Nono} and R11: Hostile(Nono) is

added

Iteration 2:I R1 is satisfied with {x/West , y/M1, z/Nono} and

Criminal(West) is added.

Forward Chaining ASKIterations

Iteration 1:I R5 satisfied with {x/M1} and R9: Sells(West ,M1,Nono) is

addedI R4 satisfied with {x/M1} and R10: Weapon(M1) is addedI R6 satisfied with {x/Nono} and R11: Hostile(Nono) is

added

Iteration 2:I R1 is satisfied with {x/West , y/M1, z/Nono} and

Criminal(West) is added.

Inference in First Order LogicDiscussion

I Once we have facts that evaluate to T or FI We can apply Forward Chaining, Backwards Chaining and

ResolutionI The key is to understand UnificationI Very similar to Logical agents.