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First-Order Logic. Limitations of propositional logic. Suppose you want to say “All humans are mortal” In propositional logic, you would need ~6.7 billion statements Suppose you want to say “Some people can run a marathon” You would need a disjunction of ~6.7 billion statements.

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limitations of propositional logic
Limitations of propositional logic
  • Suppose you want to say “All humans are mortal”
    • In propositional logic, you would need ~6.7 billion statements
  • Suppose you want to say “Some people can run a marathon”
    • You would need a disjunction of ~6.7 billion statements
first order logic1
First-order logic
  • Propositional logic assumes the world consists of atomic facts
  • First-order logic assumes the world contains objects, relations, and functions
syntax of fol
Syntax of FOL
  • Constants:John, Sally, 2, ...
  • Variables:x, y, a, b,...
  • Predicates:Person(John), Siblings(John, Sally), IsOdd(2), ...
  • Functions:MotherOf(John), Sqrt(x), ...
  • Connectives:, , , , 
  • Equality:=
  • Quantifiers:, 
  • Term: Constant or Variableor Function(Term1, ... , Termn)
  • Atomic sentence:Predicate(Term1, ... , Termn)or Term1 = Term2
  • Complex sentence: made from atomic sentences using connectives and quantifiers
semantics of fol
Semantics of FOL
  • Sentences are true with respect to a model and an interpretation
  • Model contains objects (domainelements) and relations among them
  • Interpretation specifies referents for

constantsymbols→objects

predicatesymbols→relations

functionsymbols→ functional relations

  • An atomic sentence Predicate(Term1, ... , Termn) is true iffthe objects referred to by Term1, ... , Termnare in the relation referred to by predicate
universal quantification
Universal quantification
  • x P(x)
  • Example: “Everyone at UNC is smart”

x At(x,UNC)  Smart(x)

Why not x At(x,UNC)  Smart(x)?

  • Roughly speaking, equivalent to the conjunction of all possible instantiations of the variable:

At(John, UNC)  Smart(John)  ...

At(Richard, UNC)  Smart(Richard)  ...

  • x P(x)is true in a model m iffP(x)is true with x being each possible object in the model
existential quantification
Existential quantification
  • x P(x)
  • Example: “Someone at UNC is smart”

x At(x,UNC)  Smart(x)

Why not x At(x,UNC)  Smart(x)?

  • Roughly speaking, equivalent to the disjunction of all possible instantiations:

[At(John,UNC)  Smart(John)] 

[At(Richard,UNC)  Smart(Richard)]  …

  • x P(x) is true in a model m iffP(x)is true with x being some possible object in the model
properties of quantifiers
Properties of quantifiers
  • x y is the same as yx
  • x yis the same as yx
  • x yis not the same as yx

x y Loves(x,y)

“There is a person who loves everyone”

yx Loves(x,y)

“Everyone is loved by at least one person”

  • Quantifier duality: each quantifier can be expressed using the other with the help of negation

x Likes(x,IceCream) x Likes(x,IceCream)

x Likes(x,Broccoli) xLikes(x,Broccoli)

equality
Equality
  • Term1= Term2is true under a given model if and only if Term1and Term2refer to the same object
  • E.g., definition of Sibling in terms of Parent:

x,ySibling(x,y)  [(x = y)  m,f(m = f)  Parent(m,x)  Parent(f,x)  Parent(m,y)  Parent(f,y)]

using fol the kinship domain
Using FOL: The Kinship Domain
  • Brothers are siblings

x,y Brother(x,y) Sibling(x,y)

  • “Sibling” is symmetric

x,y Sibling(x,y)  Sibling(y,x)

  • One\'s mother is one\'s female parent

m,c(Mother(c) = m)  (Female(m)  Parent(m,c))

why first order
Why “First order”?
  • FOL permits quantification over variables
  • Higher order logics permit quantification over functions and predicates:

P,x [P(x)  P(x)]

x,y (x=y)  [P (P(x)P(y))]

inference in fol
Inference in FOL
  • All rules of inference for propositional logic apply to first-order logic
  • We just need to reduce FOL sentences to PL sentences by instantiating variables and removing quantifiers
reduction of fol to pl
Reduction of FOL to PL
  • Suppose the KB contains the following:

x King(x)  Greedy(x)  Evil(x)

King(John) Greedy(John) Brother(Richard,John)

  • How can we reduce this to PL?
  • Let’s instantiate the universal sentence in all possible ways:

King(John)  Greedy(John)  Evil(John)

King(Richard)  Greedy(Richard)  Evil(Richard)

King(John) Greedy(John) Brother(Richard,John)

  • The KB is propositionalized
    • Proposition symbols are King(John), Greedy(John), Evil(John), King(Richard), etc.
reduction of fol to pl1
Reduction of FOL to PL
  • What about existential quantification, e.g.,

x Crown(x) OnHead(x,John) ?

  • Let’s instantiate the sentence with a new constant that doesn’t appear anywhere in the KB:

Crown(C1) OnHead(C1,John)

substitution
Substitution
  • Substitution of variables by ground terms:

SUBST({v/g},P)

    • Result of SUBST({x/Harry, y/Sally}, Loves(x,y)):

Loves(Harry,Sally)

    • Result of SUBST({x/John}, King(x)  Greedy(x)  Evil(x)):

King(John)  Greedy(John)  Evil(John)

universal instantiation ui
Universal instantiation (UI)
  • A universally quantified sentence entails every instantiation of it:

v P(v)SUBST({v/g}, P(v))

for any variable v and ground term g

  • E.g., x King(x)  Greedy(x)  Evil(x) yields:

King(John)  Greedy(John)  Evil(John)

King(Richard)  Greedy(Richard)  Evil(Richard)

King(Father(John))  Greedy(Father(John))  Evil(Father(John))

existential instantiation ei
Existential instantiation (EI)
  • An existentially quantified sentence entails the instantiation of that sentence with a new constant:

v P(v)

SUBST({v/C}, P(v))

for any sentence P, variable v, and constant Cthat does not appear elsewhere in the knowledge base

  • E.g., x Crown(x) OnHead(x,John) yields:

Crown(C1) OnHead(C1,John)

provided C1 is a new constant symbol, called a Skolem constant

propositionalization
Propositionalization
  • Every FOL KB can be propositionalized so as to preserve entailment
    • A ground sentence is entailed by the new KB iffit is entailed by the original KB
  • Idea: propositionalize KB and query, apply resolution, return result
  • Problem: with function symbols, there are infinitely many ground terms
    • For example, Father(X) yields Father(John), Father(Father(John)), Father(Father(Father(John))), etc.
propositionalization1
Propositionalization
  • Theorem (Herbrand 1930):
    • If a sentence αis entailed by an FOL KB, it is entailed by a finite subset of the propositionalizedKB
  • Idea: For n = 0 to Infinitydo
    • Create a propositional KB by instantiating with depth-n terms
    • See if α is entailed by this KB
  • Problem: works if α is entailed, loops if α is not entailed
  • Theorem (Turing 1936, Church 1936):
    • Entailment for FOL issemidecidable: algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every nonentailedsentence
propositionalization2
Propositionalization
  • Example KB:

x King(x)  Greedy(x)  Evil(x) y Greedy(y)

King(John) Brother(Richard,John)

  • What will propositionalization produce?

King(John)  Greedy(John)  Evil(John)

King(Richard)  Greedy(Richard)  Evil(Richard)

Greedy(John) Greedy(Richard)

King(John) Brother(Richard,John)

  • But what if all we want is to prove Evil(John)?
generalized modus ponens gmp
Generalized Modus Ponens (GMP)

p1\', p2\', … , pn\', (p1 p2 …  pnq)

such that SUBST(θ, pi\')= SUBST(θ, pi) for all i

SUBST(θ,q)

  • Used with definite clauses (exactly one positive literal)
  • All variables assumed universally quantified
  • Example:

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

p2\' is Greedy(y) p2 is Greedy(x)

θis {x/John,y/John}

qis Evil(x)

SUBST(θ,q)is Evil(John)

unification
Unification

UNIFY(α,β) = θ means that SUBST(θ, α) = SUBST(θ, β)

p qθ

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

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

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

Knows(John,x) Knows(x,Mary) {x1/John, x2/Mary}

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

  • Standardizing apart eliminates overlap of variables
  • Most general unifier
inference with gmp
Inference with GMP

p1\', p2\', … , pn\', (p1 p2 …  pnq)

such that SUBST(θ, pi\')= SUBST(θ, pi) for all i

SUBST(θ,q)

  • Forward chaining
    • Like search: keep proving new things and adding them to the KB until we can prove q
  • Backward chaining
    • Find p1, …, pn such that knowing them would prove q
    • Recursively try to prove p1, …, pn
example knowledge base
Example knowledge base
  • The law says that it is a crime for an American to sell 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
example knowledge base1
Example knowledge base

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)

Nonohas some missiles

x Owns(Nono,x)  Missile(x)

Owns(Nono,M1) 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 is American

American(West)

The country Nono is an enemy of America

Enemy(Nono,America)

forward chaining proof
Forward chaining proof

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

forward chaining proof1
Forward chaining proof

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

forward chaining proof2
Forward chaining proof

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example1
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example2
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example3
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example4
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

backward chaining example5
Backward chaining example

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

Owns(Nono,M1)  Missile(M1)

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

Missile(x)  Weapon(x) Enemy(x,America)  Hostile(x)

American(West) Enemy(Nono,America)

resolution fol version
Resolution: FOL version

p1···pk, q1···qn

such that UNIFY(pi, qj) = θ

SUBST(θ, p1···pi-1pi+1 ···pkq1···qj-1qj+1···qn)

  • For example,

Rich(x)  Unhappy(x)

Rich(Ken)

Unhappy(Ken)

with θ = {x/Ken}

  • Apply resolution steps to CNF(KB α); complete for FOL
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