Inference in first-order logic

Chapter 9

Outline

Reducing first-order inference to propositional inference

Unification

Generalized Modus Ponens

Forward chaining

Backward chaining

Resolution

Universal instantiation (UI)

Every instantiation of a universally quantified sentence is entailed by it:

"v α
Subst({v/g}, α)

for any variable v and ground term g

E.g., "x King(x) Ù Greedy(x) Þ Evil(x) yields:

King(John) Ù Greedy(John) Þ Evil(John)

King(Richard) Ù Greedy(Richard) Þ Evil(Richard)

King(Father(John)) Ù Greedy(Father(John)) Þ Evil(Father(John))

.

.

.

Existential instantiation (EI)

For any sentence α, variable v, and constant symbol k that does not appear elsewhere in the knowledge base:

$v α

Subst({v/k}, α)

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

Crown(C₁) Ù OnHead(C₁,John)

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

Reduction to propositional inference

Suppose the KB contains just the following:

"x King(x) Ù Greedy(x) Þ Evil(x)

King(John)

Greedy(John)

Brother(Richard,John)

Instantiating the universal sentence in all possible ways, we have:

King(John) Ù Greedy(John) Þ Evil(John)

King(Richard) Ù Greedy(Richard) Þ Evil(Richard)

King(John)

Greedy(John)

Brother(Richard,John)

The new KB is propositionalized: proposition symbols are

King(John), Greedy(John), Evil(John), King(Richard), etc.

Reduction contd.

Every FOL KB can be propositionalized so as to preserve entailment

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

Idea: propositionalize KB and query, apply resolution, return result

Problem: with function symbols, there are infinitely many ground terms,

e.g., Father(Father(Father(John)))

Reduction contd.

Theorem: Herbrand (1930). If a sentence α is entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB

Idea: For n = 0 to ∞ do

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 is
semidecidable (algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every non-entailed sentence.)

Problems with propositionalization

Propositionalization seems to generate lots of irrelevant sentences.

E.g., from:

"x King(x) Ù Greedy(x) Þ 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

With p k-ary predicates and n constants, there are p·n^k instantiations.

Unification

We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

θ = {x/John,y/John} works

Unify(α,β) = θ if αθ = βθ

p q θ

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)

Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)

Unification

We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

θ = {x/John,y/John} works

Unify(α,β) = θ if αθ = βθ

p q θ

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

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

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

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

Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)

Unification

We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

θ = {x/John,y/John} works

Unify(α,β) = θ if αθ = βθ

p q θ

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

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

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

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

Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)

Unification

We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

θ = {x/John,y/John} works

Unify(α,β) = θ if αθ = βθ

p q θ

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

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

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

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

Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)

Unification

We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

θ = {x/John,y/John} works

Unify(α,β) = θ if αθ = βθ

p q θ

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

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

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

Knows(John,x) Knows(x,OJ) {fail}

Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)

Unification

To unify Knows(John,x) and Knows(y,z),

θ = {y/John, x/z } or θ = {y/John, x/John, z/John}

The first unifier is more general than the second.

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

MGU = { y/John, x/z }

The unification algorithm

Generalized Modus Ponens (GMP)

p₁', p₂', … , p_n', ( p₁ Ù p₂ Ù … Ù p_n Þq)

qθ

p₁' is King(John) p₁ is King(x)

p₂' is Greedy(y) p₂is Greedy(x)

θ is {x/John,y/John} q is Evil(x)

q θ is Evil(John)

GMP used with KB of definite clauses (exactly one positive literal)

All variables assumed universally quantified

Soundness of GMP

Need to show that

p₁', …, p_n', (p₁ Ù … Ù p_n Þ q) ╞ qθ

provided that p_i'θ = p_iθ for all I

Lemma: For any sentence p, we have p ╞ pθ by UI

(p₁ Ù … Ù p_n Þ q) ╞ (p₁ Ù … Ù p_n Þ q)θ = (p₁θ Ù … Ù p_nθ Þ qθ)

p₁', …, p_n' ╞ p₁' Ù … Ù p_n' ╞ p₁'θ Ù … Ù p_n'θ

From 1 and 2, qθ follows by ordinary Modus Ponens

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 base contd.

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

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

Nono … has some missiles, i.e., $x Owns(Nono,x) Ù Missile(x):

Owns(Nono,M₁) and Missile(M₁)

… all of its missiles were sold to it by Colonel West

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

Missiles are weapons:

Missile(x) Þ Weapon(x)

An enemy of America counts as "hostile“:

Enemy(x,America) Þ Hostile(x)

West, who is American …

American(West)

The country Nono, an enemy of America …

Enemy(Nono,America)

Forward chaining algorithm

Forward chaining proof

Properties of forward chaining

Sound and complete for first-order definite clauses

Datalog = first-order definite clauses + no functions

FC terminates for Datalog in finite number of iterations

May not terminate in general if α is not entailed

This is unavoidable: entailment with definite clauses is semidecidable

Efficiency of forward chaining

Incremental forward chaining: no need to match a rule on iteration k if a premise wasn't added on iteration k-1

Þ match each rule whose premise contains a newly added positive literal

Matching itself can be expensive:

Database indexing allows O(1) retrieval of known facts

e.g., query Missile(x) retrieves Missile(M₁)

Forward chaining is widely used in deductive databases

Hard matching example

Colorable() is inferred iff the CSP has a solution

CSPs include 3SAT as a special case, hence matching is NP-hard

Backward chaining algorithm

SUBST(COMPOSE(θ₁, θ₂), p) = SUBST(θ₂, SUBST(θ₁, p))

Backward chaining example

Properties of backward chaining

Depth-first recursive proof search: space is linear in size of proof

Incomplete due to infinite loops

Þ fix by checking current goal against every goal on stack

Inefficient due to repeated subgoals (both success and failure)

Þ fix using caching of previous results (extra space)

Widely used for logic programming

Logic programming: Prolog

Algorithm = Logic + Control

Basis: backward chaining with Horn clauses + bells & whistles

Widely used in Europe, Japan (basis of 5th Generation project)

Compilation techniques Þ 60 million LIPS

Program = set of clauses = head :- literal₁, … literal_n.

criminal(X) :- american(X), weapon(Y), sells(X,Y,Z), hostile(Z).

Depth-first, left-to-right backward chaining

Built-in predicates for arithmetic etc., e.g., X is Y*Z+3

Built-in predicates that have side effects (e.g., input and output

predicates, assert/retract predicates)

Closed-world assumption ("negation as failure")

e.g., given alive(X) :- not dead(X).

alive(joe) succeeds if dead(joe) fails

Prolog

Appending two lists to produce a third:

append([],Y,Y).

append([X|L],Y,[X|Z]) :- append(L,Y,Z).

query:   append(A,B,[1,2]) ?

answers: A=[]    B=[1,2]

         A=[1]   B=[2]

         A=[1,2] B=[]

Resolution: brief summary

Full first-order version:

l₁ Ú ··· Ú l_k, m₁ Ú ··· Ú m_n

(l₁ Ú ··· Ú l_i-1 Ú l_i+1Ú ··· Ú l_k Ú m₁ Ú ··· Ú m_j-1 Ú m_j+1 Ú ··· Ú m_n)θ

where Unify(l_i, Øm_j) = θ.

The two clauses are assumed to be standardized apart so that they share no variables.

For example,

ØRich(x) Ú Unhappy(x)

Rich(Ken)

Unhappy(Ken)

with θ = {x/Ken}

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

Conversion to CNF

Everyone who loves all animals is loved by someone:

"x ["y Animal(y) Þ Loves(x,y)] Þ [$y Loves(y,x)]

1. Eliminate biconditionals and implications

"x [Ø"y ØAnimal(y) Ú Loves(x,y)] Ú [$y Loves(y,x)]

2. Move Ø inwards: Ø"x p ≡ $x Øp, Ø $x p ≡ "x Øp

"x [$y Ø(ØAnimal(y) Ú Loves(x,y))] Ú [$y Loves(y,x)]

"x [$y ØØAnimal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)]

"x [$y Animal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)]

Conversion to CNF contd.

Standardize variables: each quantifier should use a different one

"x [$y Animal(y) Ù ØLoves(x,y)] Ú [$z Loves(z,x)]

Skolemize: a more general form of existential instantiation.

Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables:

"x [Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x)

Drop universal quantifiers:

[Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x)

Distribute Ú over Ù :

[Animal(F(x)) Ú Loves(G(x),x)] Ù [ØLoves(x,F(x)) Ú Loves(G(x),x)]

Resolution proof: definite clauses


	Reducing first-order inference to propositional inference
	Unification
	Generalized Modus Ponens
	Forward chaining
	Backward chaining
	Resolution


	Every instantiation of a universally quantified sentence is entailed by it:
	"v α Subst({v/g}, α)
	for any variable v and ground term g

	E.g., "x King(x) Ù Greedy(x) Þ Evil(x) yields:
		King(John) Ù Greedy(John) Þ Evil(John)
		King(Richard) Ù Greedy(Richard) Þ Evil(Richard)
		King(Father(John)) Ù Greedy(Father(John)) Þ Evil(Father(John))
		.
		.
		.


	For any sentence α, variable v, and constant symbol k that does not appear elsewhere in the knowledge base:
	$v α
	Subst({v/k}, α)

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

	Crown(C₁) Ù OnHead(C₁,John)

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


	Suppose the KB contains just the following:
		"x King(x) Ù Greedy(x) Þ Evil(x)
		King(John)
		Greedy(John)
		Brother(Richard,John)

	Instantiating the universal sentence in all possible ways, we have:
		King(John) Ù Greedy(John) Þ Evil(John)
		King(Richard) Ù Greedy(Richard) Þ Evil(Richard)
		King(John)
		Greedy(John)
		Brother(Richard,John)

	The new KB is propositionalized: proposition symbols are

		King(John), Greedy(John), Evil(John), King(Richard), etc.


	Every FOL KB can be propositionalized so as to preserve entailment

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

	Idea: propositionalize KB and query, apply resolution, return result

	Problem: with function symbols, there are infinitely many ground terms,
		e.g., Father(Father(Father(John)))


	Theorem: Herbrand (1930). If a sentence α is entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB

	Idea: For n = 0 to ∞ do
		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 is semidecidable (algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every non-entailed sentence.)


	Propositionalization seems to generate lots of irrelevant sentences.

	E.g., from:
		"x King(x) Ù Greedy(x) Þ 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

	With p k-ary predicates and n constants, there are p·n^k instantiations.


	We can get the inference immediately if we can find a substitution θ such that King(x) and Greedy(x) match King(John) and Greedy(y)

	θ = {x/John,y/John} works

	Unify(α,β) = θ if αθ = βθ
	p q θ
	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)

	Standardizing apart eliminates overlap of variables, e.g., Knows(z₁₇,OJ)


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

	The first unifier is more general than the second.

	There is a single most general unifier (MGU) that is unique up to renaming of variables.
		MGU = { y/John, x/z }


	p₁', p₂', … , p_n', ( p₁ Ù p₂ Ù … Ù p_n Þq)
	qθ
	p₁' is King(John) p₁ is King(x)
	p₂' is Greedy(y) p₂is Greedy(x)
	θ is {x/John,y/John} q is Evil(x)
	q θ is Evil(John)

	GMP used with KB of definite clauses (exactly one positive literal)

	All variables assumed universally quantified


	Need to show that
	p₁', …, p_n', (p₁ Ù … Ù p_n Þ q) ╞ qθ
	provided that p_i'θ = p_iθ for all I

	Lemma: For any sentence p, we have p ╞ pθ by UI

		(p₁ Ù … Ù p_n Þ q) ╞ (p₁ Ù … Ù p_n Þ q)θ = (p₁θ Ù … Ù p_nθ Þ qθ)
		p₁', …, p_n' ╞ p₁' Ù … Ù p_n' ╞ p₁'θ Ù … Ù p_n'θ
		From 1 and 2, qθ follows by ordinary Modus Ponens


	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


	... it is a crime for an American to sell weapons to hostile nations:
		American(x) Ù Weapon(y) Ù Sells(x,y,z) Ù Hostile(z) Þ Criminal(x)
	Nono … has some missiles, i.e., $x Owns(Nono,x) Ù Missile(x):
		Owns(Nono,M₁) and Missile(M₁)
	… all of its missiles were sold to it by Colonel West
		Missile(x) Ù Owns(Nono,x) Þ Sells(West,x,Nono)
	Missiles are weapons:
		Missile(x) Þ Weapon(x)
	An enemy of America counts as "hostile“:
		Enemy(x,America) Þ Hostile(x)
	West, who is American …
		American(West)
	The country Nono, an enemy of America …
		Enemy(Nono,America)


	Sound and complete for first-order definite clauses

	Datalog = first-order definite clauses + no functions
	FC terminates for Datalog in finite number of iterations

	May not terminate in general if α is not entailed

	This is unavoidable: entailment with definite clauses is semidecidable


	Incremental forward chaining: no need to match a rule on iteration k if a premise wasn't added on iteration k-1
		Þ match each rule whose premise contains a newly added positive literal

	Matching itself can be expensive:
	Database indexing allows O(1) retrieval of known facts
		e.g., query Missile(x) retrieves Missile(M₁)

	Forward chaining is widely used in deductive databases


	Colorable() is inferred iff the CSP has a solution
	CSPs include 3SAT as a special case, hence matching is NP-hard









	SUBST(COMPOSE(θ₁, θ₂), p) = SUBST(θ₂, SUBST(θ₁, p))


	Depth-first recursive proof search: space is linear in size of proof
	Incomplete due to infinite loops
		Þ fix by checking current goal against every goal on stack
	Inefficient due to repeated subgoals (both success and failure)
		Þ fix using caching of previous results (extra space)
	Widely used for logic programming


	Algorithm = Logic + Control

	Basis: backward chaining with Horn clauses + bells & whistles
	Widely used in Europe, Japan (basis of 5th Generation project)
	Compilation techniques Þ 60 million LIPS

	Program = set of clauses = head :- literal₁, … literal_n.
	criminal(X) :- american(X), weapon(Y), sells(X,Y,Z), hostile(Z).

	Depth-first, left-to-right backward chaining
	Built-in predicates for arithmetic etc., e.g., X is Y*Z+3
	Built-in predicates that have side effects (e.g., input and output
	predicates, assert/retract predicates)
	Closed-world assumption ("negation as failure")
		e.g., given alive(X) :- not dead(X).
		alive(joe) succeeds if dead(joe) fails


	Appending two lists to produce a third:
	append([],Y,Y).
	append([X\|L],Y,[X\|Z]) :- append(L,Y,Z).

	query: append(A,B,[1,2]) ?

	answers: A=[] B=[1,2]
	A=[1] B=[2]
	A=[1,2] B=[]


	Full first-order version:
	l₁ Ú ··· Ú l_k, m₁ Ú ··· Ú m_n
	(l₁ Ú ··· Ú l_i-1 Ú l_i+1Ú ··· Ú l_k Ú m₁ Ú ··· Ú m_j-1 Ú m_j+1 Ú ··· Ú m_n)θ
	where Unify(l_i, Øm_j) = θ.

	The two clauses are assumed to be standardized apart so that they share no variables.
	For example,
	ØRich(x) Ú Unhappy(x)
	Rich(Ken)
	Unhappy(Ken)
	with θ = {x/Ken}

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


	Everyone who loves all animals is loved by someone:
		"x ["y Animal(y) Þ Loves(x,y)] Þ [$y Loves(y,x)]

	1. Eliminate biconditionals and implications
		"x [Ø"y ØAnimal(y) Ú Loves(x,y)] Ú [$y Loves(y,x)]

	2. Move Ø inwards: Ø"x p ≡ $x Øp, Ø $x p ≡ "x Øp
		"x [$y Ø(ØAnimal(y) Ú Loves(x,y))] Ú [$y Loves(y,x)]
		"x [$y ØØAnimal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)]
		"x [$y Animal(y) Ù ØLoves(x,y)] Ú [$y Loves(y,x)]


	Standardize variables: each quantifier should use a different one
		"x [$y Animal(y) Ù ØLoves(x,y)] Ú [$z Loves(z,x)]

	Skolemize: a more general form of existential instantiation.
		Each existential variable is replaced by a Skolem function of the enclosing universally quantified variables:
		"x [Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x)

	Drop universal quantifiers:
		[Animal(F(x)) Ù ØLoves(x,F(x))] Ú Loves(G(x),x)

	Distribute Ú over Ù :
		[Animal(F(x)) Ú Loves(G(x),x)] Ù [ØLoves(x,F(x)) Ú Loves(G(x),x)]