Inference in first-order logic

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(without any variable)

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))

.

.

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Existential instantiation (EI)

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

v αSubst({v/k}, α)

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

Reduction to propositional inference Suppose the KB contains just the following:

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

Instantiating the universal sentence in all possible ways, we have: King(John) Greedy(John) Evil(John) King(Richard) Greedy(Richard) 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 no algorithm exists that also says no to every nonentailed sentence.

Problems with propositionalization

Propositionalization seems to generate lots of irrelevant sentences. Example

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

With p k-ary predicates and n constants, there are p·nk 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(John,z27) Knows(z17,OJ)

{x/Jane}

{x/OJ, y/John}

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

No substitution possible yet.

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

Unification finds substitutions that make different logical expressions look identical UNIFY takes two sentences and returns a unifier for them, if one

exists UNIFY(p,q) = where SUBST(,p) = SUBST (,q)

Basically, find a that makes the two clauses look alike

Unification

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

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

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

UNIFY(Knows(John,x), Knows(x,Elizabeth)) = fail Last example fails because x would have to be both John and Elizabeth We can avoid this problem by standardizing:

The two statements now read UNIFY(Knows(John,x), Knows(z,Elizabeth))

This is solvable: UNIFY(Knows(John,x), Knows(z,Elizabeth)) = {x/Elizabeth,z/John}

Unification

To unify Knows(John,x) and Knows(y,z) Can use 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 }

Unification

Unification algorithm:Recursively explore the two expressions side

by sideBuild up a unifier along the wayFail if two corresponding points do not match

The unification algorithm

The unification algorithm

Simple Example

Brother(x,John)Father(Henry,y)Mother(z,John)

Brother(Richard,x)Father(y,Richard)Mother(Eleanore,x)

Generalized Modus Ponens (GMP)

p1', p2', … , pn', ( p1 p2 … pn q) qθ

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

p2' is Greedy(y) p2 is 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

where pi'θ = pi θ for all i

Soundness of GMP Need to show that

p1', …, pn', (p1 … pn q) ╞ qθprovided that pi'θ = piθ for all I

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

¨ (p1 … pn q) ╞ (p1 … pn q)θ = (p1θ … pnθ qθ)¨ p1', \; …, \;pn' ╞ p1' … pn' ╞ p1'θ … pn'θ ¨ From 1 and 2, qθ follows by ordinary Modus Ponens

Storage and Retrieval

Use TELL and ASK to interact with Inference Engine Implemented with STORE and FETCH

STORE(s) stores sentence s FETCH(q) returns all unifiers that the query q unifies with

Example: q = Knows(John,x) KB is:

Knows(John,Jane), Knows(y,Bill), Knows(y,Mother(y)) Result is

1={x/Jane}, 2=, 3= {John/y,x/Mother(y)}

Storage and Retrieval

First approach: Create a long list of all propositions in Knowledge Base Attempt unification with all propositions in KB

Works, but is inefficient Need to restrict unification attempts to sentences that

have some chance of unifying Index facts in KB

Predicate Indexing Index predicates:

All “Knows” sentences in one bucket All “Loves” sentences in another

Use Subsumption Lattice (see below)

Storing and Retrieval Subsumption Lattice

Child is obtained from parent through a single substitution Lattice contains all possible queries that can be unified with it.

Works well for small lattices Predicate with n arguments has a 2n lattice

Structure of lattice depends on whether the base contains repeated variables

Knows(John,John)

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

Knows(x,y)

Forward Chaining

Forward Chaining Idea:

Start with atomic sentences in the KB Apply Modus Ponens

Add new atomic sentences until no further inferences can be made

Works well for a KB consisting of Situation Response clauses when processing newly arrived data

Forward Chaining

First Order Definite Clauses Disjunctions of literals of which exactly one is positive: Example:

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

First Order Definite Clauses can include variables Variables are assumed to be universally quantified

Greedy(y) means y Greedy(y) Not every KB can be converted into first definite

clauses

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) 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)

Forward chaining algorithm

Forward chaining proof

Forward chaining proof

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 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

Diff(wa,nt) Diff(wa,sa) Diff(nt,q) Diff(nt,sa) Diff(q,nsw) Diff(q,sa) Diff(nsw,v) Diff(nsw,sa) Diff(v,sa) Colorable()

Diff(Red,Blue) Diff (Red,Green) Diff(Green,Red) Diff(Green,Blue) Diff(Blue,Red) Diff(Blue,Green)

Backward Chaining

Improves on Forward Chaining by not making irrelevant conclusions Alternatives to backward chaining:

restrict forward chaining to a relevant set of forward rules Rewrite rules so that only relevant variable bindings are

made: Use a magic set Example:

Rewrite rule: Magic(x)American(x) Weapon(x) Sells(x,y,z) Hostile(z)Criminal(x)

Add fact: Magic(West)

Backward Chaining

Idea:Given a query, find all substitutions that

satisfy the query.Algorithm:

Work on lists of goals, starting with original query Algo finds every clause in the KB that unifies with

the positive literal (head) and adds remainder (body) to list of goals

Backward chaining algorithm

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

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Backward chaining example

Properties of backward chaining

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

Incomplete due to infinite loops fix by checking current goal against every goal on stack

Inefficient due to repeated subgoals (both success and failure) 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) Program = set of clauses:

head :- literal1, … literaln. 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=[]

Prolog Has problems with repeated states and infinite

paths Example: Path finding in graphs

path(X,Z) :- link(X,Z) path(X,Z) :- path(X,Y),link(Y,Z)

A B Cpath(a,c)

link(a,c)fail link(Y,c)path(a,Y)

{ Y/b}

link(a,b){Y/b }

Prolog Has problems with repeated states and infinite

paths Example: Path finding in graphs

path(X,Z) :- path(X,Y),link(Y,Z) path(X,Z) :- link(X,Z)

A B C

path(a,c)

path(a,Y)

fail

link(Y,b)

path(a,Y’) link(Y’,Y)

path(a,Y) link(Y,b)

Resolution

Resolution for propositional logic is a complete inference procedure Existence of complete proof procedures in Mathematics would

entail: All conjectures can be established mechanically All mathematics can be established as the logical consequence of a set

of fundamental axioms Gődel 1930: Completeness Theorem for first order logic

Any entailed sentence has a finite proof No algorithm given until J.A. Robinson’s resolution algorithm in 1965

Gődel 1931: Incompleteness Theorem: Any logical system with induction is necessarily incomplete There are sentences that are entailed, but not proof can be given

Resolution

First order logic requires sentences in CNF Conjunctive Normal Form: Each clause is a

disjunction of literals, but literals can contain variables, which are assumed to be universally quantified

Example: Convertx American(x) Weapon(y) Sells(x,y,z) Hostile(z)

Criminal(x)

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

Conversion to CNF

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

Conversion to CNF

3. Standardize variables: each quantifier should use a different onex [y Animal(y) Loves(x,y)] [z Loves(z,x)]

4. 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)) 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)) Loves(G(x),x)] [Loves(x,F(x)) Loves(G(x),x)]

Resolution Inference Rule

Full first-order version:l1 ··· lk, m1 ··· mn

Subst(θ ,l1 ··· li-1 li+1 ··· lk m1 ··· mj-1 mj+1 ··· mn)θwhere Unify(li, mj) = θ.

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 θ = {x/Ken}

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

Resolution

Show KB ⊢ α by showing that KB α is unsatisfyable

Resolution Example

Everyone who loves all animals is loved by someone

Anyone who kills an animal is loved by no one. Jack loves all animals. Either Jack or Curiosity killed the cat, who is

named Tuna All dogs kill a cats Rintintin is a dog Question: Did Curiosity kill the cat?

Resolution Example

Everyone who loves all animals is loved by someone. Formulate in FOL

x [y [Animal(y) Loves(x,y)]] [z Loves(z,x)] Remove Implications

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

Move negation inward x [y Animal(y) Loves(x,y)] [z Loves(z,x)] x [y Animal(y) Loves(x,y)] [z Loves(z,x)]

Skolemize x [ Animal(F(x)) Loves(x,F(x))] [Loves(G(x),x)]

N.B.: Argument of Skolemization function are all universally qualified variables Drop universal quantifier

[ Animal(F(x)) Loves(x,F(x))] [Loves(G(x),x)] Use distributive law (and get two clauses)

Animal(F(x)) Loves(G(x),x); Loves(x,F(x)) Loves(G(x),x)

Resolution Example

Anyone who kills an animal is loved by no one. Transfer to FOL

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

Remove Implications x [y (Animal(y) Kills(x,y)] (z Loves(z,x)

Move negations inwards x [ y Animal(y) Kills(x,y)] (z Loves(z,x))

Remove quantifiers Animal(y) Kills(x,y) Loves(z,x)

Resolution Example

Jack loves all animals.FOL form

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

Remove implications x [Animal(x) Loves(Jack, x)]

Remove quantifier Animal(x) Loves(Jack, x)

Resolution Example

Either Jack or Curiosity killed the cat, who is named Tuna.FOL form

Kills(Jack,Tuna) Kills(Curiosity,Tuna); Cat(Tuna)

Resolution Example

All cats are animalsFOL form

x [Cat(x) Animal(x)

Remove implications x [Cat(x) Animal(x)]

Remove quantifier Cat(x) Animal(x)

Resolution Example

All dogs kill a cats FOL form

x [Dog(x) y[Cat(y) Kills(x,y)]] Remove implications

x [Dog(x) y[Cat(y) Kills(x,y)]] Skolemize

x [Dog(x) [Cat(H(x)) Kills(x,H(x))]] Drop universal quantifiers

Dog(x) [Cat(H(x)) Kills(x,H(x))] Distribute (and obtain two clauses)

Dog(x) Cat(H(x); Dog(x) Kills(x,H(x))]

Resolution Example

Rintintin is a dogFOL form

Dog(Rintintin)

Resolution Example

Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin)

Resolution Example

Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin)

Question: Did Curiosity kill the cat?

Kills(Curiosity,Tuna)]

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna)]

Cat(Tuna) , Cat(x) Animal(x)

Unify(Cat(Tuna), Cat(x)) = { x/Tuna }

First line thus resolves to:

Animal(Tuna)

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna)

Kills(Jack,Tuna) Kills(Curiosity,Tuna), Kills(Curiosity,Tuna)

Resolves to:

Kills(Jack,Tuna)

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna)

Animal(y) Kills(x,y) Loves(z,x), Animal(Tuna)Unify(Animal(Tuna), Animal(y)) = {y/Tuna}Resolves to:

Kills(x,Tuna) Loves(z,x),

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x),

Loves(x,F(x)) Loves(G(x),x), Animal(z) Loves(Jack, z)

Unify( Loves(x,F(x)) , Loves(Jack, z)) = { x / Jack, z / F(x)}

Resolvent clause is obtained by substituting the unification ruleLoves(G(Jack),Jack) Animal(F(Jack))

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack))

Animal(F(x)) Loves(G(x),x), Loves(G(Jack),Jack) Animal(F(Jack))

Unify(Animal(F(x)) , Animal(F(Jack)))= { x / Jack}

Resolvent clause is obtained by substituting the unification ruleLoves(G(Jack),Jack)

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack)) Loves(G(Jack),Jack)

Kills(x,Tuna) Loves(z,x), Loves(G(Jack),Jack)

Unify( Loves(z,x), Loves(G(Jack),Jack) ) = { x / Jack, z / G(Jack)}

Resolvent clause is obtained by substituting the unification rule Loves(G(Jack),Jack)

Resolution Example Animal(F(x)) Loves(G(x),x) Loves(x,F(x)) Loves(G(x),x) Animal(y) Kills(x,y) Loves(z,x) Animal(x) Loves(Jack, x) Kills(Jack,Tuna) Kills(Curiosity,Tuna) Cat(Tuna) Cat(x) Animal(x) Dog(x) Cat(H(x) Dog(x) Kills(x,H(x))] Dog(Rintintin) Kills(Curiosity,Tuna) Animal(Tuna) Kills(Jack,Tuna) Kills(x,Tuna) Loves(z,x) Loves(G(Jack),Jack) Animal(F(Jack)) Loves(G(Jack),Jack) Loves(G(Jack),Jack)

Loves(G(Jack),Jack), Loves(G(Jack),Jack)

Resolvent clause is empty. Proof succeeded

Resolution Resolution is refutation-complete

If a set of sentences is unsatisfiable, then resolution will be able to produce a contradiction

Theorem provers Use control in order to be more efficient

Focus of most research effort Separate control from knowledge base

Example: Otter (Organized Technique for Theorem proving and Effective Research) A set of clauses known as the SoS - Set of Support

The important facts about a problem Search if focused on resolving SoS with another axiom

A set of usable axioms Background knowledge about problem field

Rewrites / demodulators Rules to transform expressions into a canonical form

Set of parameters or clauses that defines the control strategy to allow user to control search and filtering functions to eliminate useless subgoals

Theorem Prover Successes

First formal proof of Gődel’s Incompleteness Theorem (1986)

Robbins algebra (a simple set of axioms) is Boolean algebra (1996)

Software verification:Remote agent spacecraft control program

(2000)

Resolution proof: definite clauses