m9 inference
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Inference in first-order logicChapter 9
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Outline
Reducing first-order inference topropositional inference
Unification Generalized Modus Ponens Forward chaining
Backward chaining Resolution
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Reduction to propositional
inferenceSuppose 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.
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Reduction contd.
Every FOL KB can be propositionalized so as topreserve entailment
(A ground sentence is entailed by new KB iff entailed byoriginal KB)
Idea: propositionalize KB and query, apply resolution,return result
Problem: with function symbols, there are infinitely many
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Reduction contd.Theorem: Herbrand (1930). If a sentence is entailed by an FOL KB, it
is entailed by a finite subset of the propositionalized KB
Idea: Forn = 0 to docreate a propositional KB by instantiating with depth-$n$ termssee if is entailed by this KB
Problem: works if is entailed, loops if is not entailed
Theorem: Turing (1936), Church (1936) Entailment for FOL issemidecidable (algorithms exist that say yes to every entailedsentence, but no algorithm exists that also says no to everynonentailed sentence.)
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Problems with propositionalization
Propositionalization seems to generate lots of irrelevant sentences.
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 produceslots of facts such as Greedy(Richard) that are irrelevant
Withp k-ary predicates and n constants, there are pnkinstantiations.
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Unification We can get the inference 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)
Standardizing apart eliminates overlap of variables, e.g.,Knows(z17,OJ)
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Unification We can get the inference 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) {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)
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Unification We can get the inference 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) {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)
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Unification We can get the inference 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) {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)
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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 general than 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) p2 is Greedy(x)
is {x/John,y/John} q is Evil(x)q is Evil(John)
GMP used with KB ofdefinite clauses (exactly one positive literal)
All variables assumed universally quantified
where pi' = pi for all i
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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., xOwns(Nono,x) Missile(x):
Owns(Nono,M1) andMissile(M1)
all of its missiles were sold to it by Colonel WestMissile(x) Owns(Nono,x) Sells(West,x,Nono)
Missiles are weapons:
Missile(x) Weapon(x)
An enemy of America cou
nts 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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Forward chaining algorithm
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Forward chaining proof
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Forward chaining proof
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Forward chaining proof
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Properties of forward chaining
Sound and complete for first-order definite clauses
Datalog = first-order definite clau
ses + no fu
nctions FC terminates for Datalog in finite number of iterations
May not terminate in general if is not entailed
This is unavoidable: entailment with definite clauses issemidecidable
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Efficiency of forward chaining
Incremental forward chaining: no need to match a rule oniteration kif a premise wasn't added on iteration k-1
match each rule whose premise contains a newly added positiveliteral
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
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Backward chaining algorithm
SUBST(COMPOSE(1, 2), p) = SUBST(2,SUBST(1, p))
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Backward chaining example
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Properties of backward chaining
Depth-first recursive proof search: space islinear in size of proof
Incomplete due to infinite loops fix by checking current goal against every goal on
stack
Inefficient due to repeated subgoals (bothsuccess and failure) fixusing caching of previous results (extra space)
Widelyu
sed forlogic programming
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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=[]
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Resolution: brief summary Full first-order version:
l1 lk, m1 mn
(l1 li-1 li+1 lk m1 mj-1 mj+1 mn)where Unify(li, mj) = .
The two clauses are assumed to be standardized apart so that theyshare no variables.
For ex
ample, Rich(x) Unhappy(x)Rich(Ken)
Unhappy(Ken)
with = {x/Ken}
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Conversion to CNF
Everyone who loves all animals is loved bysomeone:
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 xp, x p xp
x [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 a different one4.
x [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 enclosinguniversally quantified variables:
x [Animal(F(x)) Loves(x,F(x))] Loves(G(x),x)
5. Drop universal quantifiers:6.
[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)]
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Resolution proof: definite clauses