# QQ:
153688106  # Countably infinite ver terminate

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ANU COMP3620/6320
Inference in FOL
Chapter 9: “AI: A Modern Approach”
Lecturer: Scott Sanner
Outline
• Reducing first-order inference to
propositional inference
• Unification
• Generalized Modus Ponens
• Forward chaining
• Backward chaining
• Resolution
Reduction of FOL 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 (for finitely
bounded domains) 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
nonentailed sentence.)
Problems with propositionalization
• Propositionalization seems to generate lots of irrelevant sentences.
• E.g., from knowledge base (KB):
∀x King(x) ∧ Greedy(x) ⇒ Evil(x)
King(John)
∀y Greedy(y)
Brother(Richard,John )
Query: Evil(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.
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Towards first-order
inference…
First need to define some “tools”
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(z 17 ,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(z 17 ,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(z 17 ,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(z 17 ,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(z 17 ,OJ)
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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
The unification algorithm
Generalized Modus Ponens
(GMP)
p 1 ', p 2 ', … , p n ', ( p 1 ∧ p 2 ∧ … ∧ p n ⇒q)
p 1 ' is King(John) p 1 is King(x)
p 2 ' is Greedy(y) p 2  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 p i 'θ = p i θ for all i
Soundness of GMP
• Need to show that
p 1 ', …, p n ', (p 1 ∧ … ∧ p n ⇒ q) ╞ qθ
provided that p i 'θ = p i θ for all I
• Lemma: For any sentence p, we have p ╞ pθ by UI
1. (p 1 ∧ … ∧ p n ⇒ q) ╞ (p 1 ∧ … ∧ p n ⇒ q)θ = (p 1 θ ∧ … ∧ p n θ ⇒ qθ )
2. p 1 ', ; …, ;p n ' ╞ p 1 ' ∧ … ∧ p n ' ╞ p 1 'θ ∧ … ∧ p n 'θ
3. 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
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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., ∃x Owns(Nono,x) ∧ Missile(x  ):
Owns(Nono,M 1 ) and Missile(M 1 )
… 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
• May not terminate in general if α is not entailed
• This is unavoidable: entailment with definite clauses is
semidecidable
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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(M 1 )
Forward chaining is widely used in deductive databases
Backward chaining algorithm
SUBST(COMPOSE(θ 1 , θ 2 ), p) =
SUBST(θ 2 , SUBST(θ 1 , p))
Backward chaining example Backward chaining example
Backward chaining example Backward chaining example
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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
In fact, can build a logical
programming language from
backchaining (linear resolution)
Prolog!
(not tested)
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We want a sound and complete
FOL inference system
Generalization of Resolution from
Propositional to First-order Logic
• Resolution requires CNF form
– Unlike propositional case, have quantifiers
• Get rid of existentials with Skolemization
• Remaining variables are universal, just drop quantifiers
• Resolution Inference
– Requires unification (MGU)
– If a ground propositionalized refutation proof exists
• Lifted resolution will find it
Skolemization
• We need to…
– Remove all existental quantifiers while preserving satisfiability
– Note: this does not yield a logically equivalent KB!
• It only preserves (un)satisfiability!
• Do this via Skolemization
– A more general form of existential instantation
– Each existential variable is replaced by a Skolem function of the
enclosing universally quantified variables
• ∃x ∃y Loves(x, y) → Loves(c 1 , c 2 )
• ∃x ∀y Loves(x, y) → ∀y Loves(c 1 , y)
• ∀x ∃y Loves(x, y) → ∀x Loves(x, f(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 implications
∀x [¬∀y ¬Animal(y) ∨ Loves(x,y)] ∨ [∃y Loves(y,x  )]
• 2. Push ¬ in: ¬∀x p ≡ ∃x ¬p, ¬ ∃x p ≡ ∀x ¬  p
∀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)]
Conversion to CNF contd.
3. Standardize variables: each quantifier should use a
different one
∀x [∃y Animal(y) ∧ ¬Loves(x,y)] ∨ [∃z Loves(z,x)]
4. Skolemize :
∀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: brief summary
• Full first- order version:
l 1 ∨ ··· ∨ l k , m 1 ∨ ··· ∨ m n
(l 1 ∨ ··· ∨ l i-1 ∨ l i+1  ∨ ··· ∨ l k ∨ m 1 ∨ ··· ∨ m j-1 ∨ m j+1 ∨ ··· ∨ m n )θ
where Unify(l i , ¬m j) = θ.
• Clauses are assumed to be standardized apart – 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
Resolve multiple literals
or use factoring rule for
completeness
Need paramodulation
rule for completeness
with equality
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Resolution proof: definite clauses Completeness
• Basic motivation from Herbrand’s theorem
– (Roughly) If a refutation of a set of clauses exists,
⇒ ∃ refutation with finite ground clauses
• Resolution with most general unifier (MGU)
– Represents most general resolution
– Covers resolutions over infinite number of
(propositional) ground clauses
– The ground refutation (if it exists) must be subsumed
by this “lifted” resolution procedure
⇒ ⇒ Resolution is complete for FOL!
Godel’s Incompleteness Theorem
• Wait…
– Resolution rule + factoring + paramodulation is
complete for FOL with equality
• But Godel said FOL is incomplete
– With + and * functions
• Reason is that + and * must be axiomatized
– But number of axioms needed to find a refutation (if it
exists) is uncountably infinite… cannot enumerate.
• So FOL with equality and +, * is incomplete
More Explanation of Last Slide
• Issue comes down to systematic enumeration
– Do there exist positive primes m, n where m+n=k?
• Enumerate all in finite time, return true or false.
• Decidable! (like propositional logic inference)
– Same, but m-n=k?
• Countably infinite to enumerate but if solution, will terminate.
• Semidecidable! (like FOL)
– Find some positive real numbers p, q such that
SomeTestableProperty(p,q) holds?
• Uncountably infinite to enumerate (no systematic
enumeration method), so may never terminate even if true.
• Undecidable! (FOL with +,* axiomatized b/c required
axioms are uncountably infinite)
Implications of
Incompleteness
We’ll never be able to infer the truth of
all statements that arise from an
axiomatic system powerful enough to
represent basic mathematics (+,*).
But rarely does KR&R need to query
complex theorems of mathematics!
Important Part: What’s on the Exam?
• All propositional logic content.
– And basic properties of model-theoretic semantics for any logic
having such a semantics.
• Reading / writing / interpreting FOL.
• Knowledge of FOL inference (CNF conversion,
resolution), but won’t ask you to do a FOL proof on the
exam (simply because it takes too long).
– First-order backward chaining / forward chaining not on exam
• I won’t test Chapter 6, but Charles might… ask!