Artificial Intelligence
CS 165A
Tuesday, November 6, 2007
 Inference in FOL (Ch 9)
1
George Boole (1815-1864)
British
• More than 100 years later, he didn’t know
about Leibniz, but proceeded to bring to life
part of Leibniz’ dream
• His insight: Logical relationships are
expressible as a kind of algebra
– Letters represent classes (rather than numbers)
– So logic can be viewed as a form of
mathematics
• Published The Laws of Thought
• He extended Aristotle's simple syllogisms to a broader
range of reasoning
– Syllogism: Premise_1, Premise_2  Conclusion
– His logic: Propositional logic
2
Gottlob Frege (1848-1925)
German
• He provided the first fully developed system of
logic that encompassed all of the deductive
reasoning in ordinary mathematics.
• He intended for logic to be the foundation of
mathematics – all of mathematics could be
based on, and derived from, logic
• In 1879 he published Begriffsschrift, subtitled
“A formula language, modeled upon that of
arithmetic, for pure thought”
– This can be considered the ancestor of all
current computer programming languages
– Made the distinction between syntax and
semantics critical
• He invented what we today call predicate calculus (or first-order logic)
3
Notes
• HW#3 due Thursday
• Midterm exam one week from today
– Short-answer conceptual questions and HW-like questions
– Topics: Reading, lectures, material through chapter 8
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What AI is, main issues and areas in AI
Problem solving and intelligent agents
Blind search, informed search, adversarial search
Logic and inference in propositional calculus
Basics of first-order logic
– Closed book
– You may bring one 8.5”x11” piece of paper with your notes

Some formulas, procedures will be given (posted in advance)
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Reminder
• Term
– Constant, variable, function( )
• Atomic sentence
– Predicate( ), term1 = term2
• Literal
– An atomic sentence or a negated atomic sentence
• Sentence
– Atomic sentence, sentences with quantifiers and/or connectives
5
Review
Simple example of inference in FOL
KB0
Bob is a buffalo
Buffalo(Bob)
Pat is a pig
Pig(Pat)
Buffaloes outrun pigs
Buffalo(x)  Pig(y)  Outrun(x,y)
Does Bob outrun Pat?
KB entails Outrun(Bob, Pat)?
S
KB0
KB0 |– Buffalo(Bob)  Pig(Pat)
(And-Introduction)
KB1
KB1 |– Buffalo(Bob)  Pig(Pat)  Outrun(Bob, Pat)
(Universal Instantiation) [coming soon]
KB2
KB2 |– Outrun(Bob, Pat)
(Modus Ponens)
KB3
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Using FOL to express knowledge
• One can express the knowledge of a particular domain in
first-order logic
• Example: The “kinship domain”
–
–
–
–
Objects: people
Properties: gender, family relationships
Unary predicates: Male, Female, MotherOf, FatherOf
Binary predicates: Parent, Sibling, Brother, Sister, Son,
Daughter, Father, Mother, Uncle, Aunt, Grandparent, Grandfather,
Grandmother, Husband, Wife, Spouse, Brother-in-law,
Stepmother, etc….
– Functions: MotherOf, FatherOf…
• Note: There is usually (always?) more than one way to
specify knowledge
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Kinship domain
• Write down what we know (what we want to be in the KB)
– One’s mother is one’s female parent
  m, c Mother(m, c)  Female(m)  Parent(m, c)
  m, c TheMotherOf(c) = m  Female(m)  Parent(m, c)
– One’s husband is one’s male spouse
  w, h Husband(h, w)  Male(h)  Spouse(h, w)
– One is either male or female
  x Male(x)  Female(x)
– Parent-child relationship
  p, c Parent(p, c)  Child(c, p)
– Grandparent-grandchild relationship
  g, c Grandparent(g, c)   p Parent(g, p)  Parent(p, c)
– Etc…
• Now we can reason about family relationships. (How?)
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Kinship domain (cont.)
Assertions (“Add this sentence to the KB”)
TELL( KB,  m, c Mother(c) = m  Female(m)  Parent(m, c) )
TELL( KB,  w, h Husband(h, w)  Male(h)  Spouse(h, w) )
TELL( KB,  x Male(x)  Female(x) )
TELL( KB, Female(Mary)  Parent(Mary, Frank)  Parent(Frank, Ann) )
– Note: TELL( KB, S1  S2 )  TELL( KB, S1) and TELL( KB, S2)
(because of and-elimination and and-introduction)
Queries (“Does the KB entail this sentence?”)
ASK( KB, Grandparent(Mary, Ann) )  True
ASK( KB,  x Child(x, Frank) )  True
– But a better answer would be  { x / Ann }
– This returns a substitution or binding
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Implementing ASK: Inference
• We want a sound and complete inference algorithm so that
we can produce (or confirm) entailed sentences from the
KB
KB

KB
i

• The resolution rule, along with a complete search
algorithm, provides a complete inference algorithm to
confirm or refute a sentence  in propositional logic
(Sec. 7.5)
– Based on proof by contradiction (refutation)
• Refutation: To prove that the KB entails P, assume P and
show a contradiction:
(KB  P  False)  (KB  P)
Prove this!
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Inference in First-Order Logic
• Inference rules for propositional logic:
– Modus ponens, and-elimination, and-introduction, or-introduction,
resolution, etc.
– These are valid for FOL also
• But since these don’t deal with quantifiers and variables,
we need new rules, especially those that allow for
substitution (binding) of variables to objects
– These are called lifted inference rules
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Substitution and variable binding
• Notation for substitution:
– SUBST ( Binding list, Sentence )
 Binding list: { var / ground term, var / ground term, … }
 “ground term” = term with no variables
– SUBST( {var/gterm}, Func (var) ) = Func (gterm)
 SUBST (, p)
– Examples:
 SUBST ( {x/Mary}, FatherOf (x) ) = FatherOf (Mary)
 SUBST ( {x/Joe, y/Lisa}, Siblings (x,y) ) = Siblings (Joe, Lisa)
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Three new inference rules using SUBST(, p)
• Universal Instantiation
v 
SUBST ({ v / g },  )
g – ground term
• Existential Instantiation
v 
SUBST ({ v / k },  )
k – constant that does not appear
elsewhere in the knowledge base
• Existential Introduction

v SUBST ({ g / v },  )
v – variable not in 
g – ground term in 
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To Add to These Rules
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Universal Instantiation – examples
v 
SUBST ({ v / g },  )
g – ground term
• x Sleepy(x)
– SUBST({x/Joe}, )
 Sleepy(Joe)
• x Mother(x)  Female(x)
– SUBST({x/Mary}, )
 Mother(Mary)  Female(Mary)
– SUBST({x/Dad}, )
 Mother(Dad)  Female(Dad)
• x, y Buffalo(x)  Pig(y)  Outrun(x,y)
– SUBST({x/Bob}, )

y Buffalo(Bob)  Pig(y)  Outrun(Bob,y)
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Existential Instantiation – examples
v 
SUBST ({ v / k },  )
k – constant that does not appear
elsewhere in the knowledge base
• x BestAction(x)
– SUBST({x/B_A}, )
 BestAction(B_A)
– “B_A” is a constant; it is not in our universe of actions
• y Likes(y, Broccoli)
– SUBST({y/Truman}, )
 x Likes(Truman, Broccoli)
– “Truman” is a constant; it is not in our universe of people
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Existential Introduction – examples

v SUBST ({ g / v },  )
v – variable not in 
g – ground term in 
• Likes(Jim, Broccoli)
– SUBST({Jim/x}, )
 x Likes(x, Broccoli)
• x Likes(x, Broccoli)  Healthy(x)
– SUBST({Broccoli/y}, )
 y x Likes(x, y)  Healthy(x)
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What’s our goal here?
• Formulate a search process:
– Initial state
 KB
– Operators
 Inference rules
– Goal test
 KB contains S
• What is a node?
– KB + new sentences (generated by applying the inference rules)
– In other words, the new state of the KB
• What kind of search to use?
– I.e., which node to expand next?
• How to apply inference rules?   
– Need to match the premise pattern 
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Applying inference rules
• FOL query
Path = proof
Path shows reasoning
(here’s why I think you have
cancer)
– ASK(KB, Outruns(Bob, Pat))
– ASK(KB,  x BestAction(x, t))
– ASK(KB,  x Equals(x, AuntOf(HusbandOf(Mary))))
• Proof – From axioms (KB) to hypothesis (S)
– Does the KB entail a sentence S – if so, what steps were taken?
• We need to apply inference rules and appropriate
substitutions to reach the desired sentence
– Answer: Yes, SUBST({x/Grab}, Action(x, t))
– Answer: Yes, the KB entails S (and here’s how…)
• Reasoning systems may require the solution path, not just
the final answer
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Concerns
• Is the inference procedure sound and complete?
• Is it efficient?
– We need to be concerned about the branching factor
– Generalized modus ponens (GMP) is one way of reducing
branching factor
• How do we know which operators (inference rules) are
valid at any particular time?
– Inference rule: Pig(x)  Cuddly(x)
– KB: Pig(Bob)
Pig(x) and Pig(Bob) are unified by the substitution  = {x/Bob}
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Problem of branching factor
•
10 inference rules
– Branching factor can be huge!
•
Common inference pattern:
– And-Introduction
– Universal Instantiation
– Modus ponens
•
Buffaloes outrun pigs example
Let’s introduce a new inference rule which reduces the
branching factor by combining these three steps into one
– Generalized Modus Ponens
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Generalized Modus Ponens
• For atomic sentences pi , pi' , and q , where there is a
substitution  such that SUBST(, pi') = SUBST(, pi) for all
i, then
Specific facts
General rule



p1 , p2 ,  , pn , ( p1  p2    pn  q )
SUBST ( , q )
Instantiated conclusion
• GMP features:
– Efficient – it combines several inferences into one
– Sensible – substitutions are purposeful (unification)
– Efficient – when sentences are in a certain canonical form (Horn
clauses), this is the only inference rule necessary (!!!)
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Generalized Modus Ponens Example
• Go from
– Buffalo(Bob), Pig(Pat), (Buffalo(x)  Pig(y)  Outrun(x,y))
to
– Outrun(Bob, Pat)
• More generally, go from
– p1', p2', (p1  p2  q)
to
– SUBST(, q)
where SUBST(, pi') = SUBST(, pi)
• Reminder: What is  ?
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Unification
• Unification takes two atomic sentences p and q and returns a
substitution that would make p and q look the same (or else it fails)
– UNIFY(p, q) =  where SUBST(, p) = SUBST(, q)
–  is the unifier of the sentences p and q
• Examples
– p = Knows(John, x)
q = Knows(John, Jane)
  = {x/Jane}
– p = Knows(John, x)
q = Knows(y, Fred)
  = {x/Fred, y/John}
– p = Knows(John, x)
q = Knows(y, MotherOf(y))
  = {x/MotherOf(John), y/John}
– p = Knows(John, x)
q = Knows(x, Mary)
  = fail
– p = Knows(John, x1)
q = Knows(x2, Mary)
  = {x1/Mary, x2/John}
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From GMP to GR
• GMP requires sentences in Horn clause
• Prolog is based on this: Horn clause form and GMP inference
– Horn clause: an implication of positive literals
• It is “reasonably powerful,” but
– Cannot handle disjunctions or negations
– Many legal sentences cannot be expressed in a Horn clause
– I.e., this form is incomplete
• It turns out we can do better by putting the sentences of the KB into a
different format and using just one inference rule, which is both
complete and sound!
– KB Format: Conjunctive normal form (CNF)
– Inference Rule: Generalized (first-order) resolution
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Note: Completeness
• An inference procedure i is complete iff
– KB i  whenever KB

– “The procedure finds the needle in the haystack when there is one”
• An inference procedure using GMP is incomplete
– There are sentences entailed by the KB that the procedure cannot
infer
• A complete inference procedure exists: Generalized
Resolution
– Generalized resolution will produce  if KB entails 
– However, what if  is not entailed by the KB? Will resolution tell
us this?
 No, there exists no certain procedure to show this in general
– Related to the famous “Halting problem”
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Remember: p  q  p  q
Resolution
p  q , q  r
pr
Let’s rewrite it:
Review
Simple resolution
p  q, q  r
p  r
or:
a  b, b  c
ac
So resolution is really the “chaining” of implications….
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Generalized (first-order) resolution
• Generalized resolution is the basis for a complete inference
procedure
– Can derive new implications (P  Q)
 GMP can only derive atomic conclusions (Q)
– Any sentence can be put into its normal form
• Just like “regular” resolution except
– Uses UNIFY/SUBST rather than =
– Deals with more than 2 disjuncts
• GR is complete, but semidecidable
– Not always able to show that a sentence is not entailed by the KB
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Reminder: Atomic sentences vs. Literals
• Atomic sentences
– Predicate and its arguments (terms)
 P(x), Q(y), Siblings(Bill, z), Equal(x, y)
• Literals
– Atomic sentences and negated atomic sentences
 P(x), Q(y), Siblings(Bill, z), Equal(x, y)
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Two versions of Generalized Resolution
For literals pi and qi , where UNIFY(pj , qk) = 
p1   p j   pm
Disjunctions
q1   qk   qn
SUBST ( , p1    pm  q1    qn ) except p j and qk
For atomic sentences pi, qi, ri, si , where UNIFY(pj , qk) = 
p1   p j   pn1  r1   rn2
s1   sn3  q1   qk   qn4
Implications
SUBST ( , p1   pn1  s1   sn3  r1   rn2  q1   qn4 ) except p j and qk
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