First-Order Logic (FOL) aka. predicate calculus

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First-Order Logic (FOL)aka. predicate calculus

• Tuomas Sandholm
• Carnegie Mellon University
• Computer Science Department

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First-Order Logic (FOL) Syntax

• User defines these primitives
• Constant symbols (i.e., the "individuals" in the
  world) E.g., Mary, 3
• Function symbols (mapping individuals to
  individuals) E.g., father-of(Mary) John,
  color-of(Sky) Blue
• Predicate symbols (mapping from individuals to
  truth values) E.g., greater(5,3), green(Grass),
  color(Grass, Green)

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First-Order Logic (FOL) Syntax

• FOL supplies these primitives
• Variable symbols. E.g., x,y
• Connectives. Same as in PL not (), and (), or
  (v), implies (gt), if and only if (ltgt)
• Quantifiers Universal (A) and Existential (E)

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Quantifiers

• Universal quantification corresponds to
  conjunction ("and") in that (Ax)P(x) means that P
  holds for all values of x in the domain
  associated with that variable.
• E.g., (Ax) dolphin(x) gt mammal(x)
• Existential quantification corresponds to
  disjunction ("or") in that (Ex)P(x) means that P
  holds for some value of x in the domain
  associated with that variable.
• E.g., (Ex) mammal(x) lays-eggs(x)
• Universal quantifiers are usually used with
  "implies" to form "if-then rules."
• E.g., (Ax) cs15-381-student(x) gt smart(x) means
  "All cs15-381 students are smart."
• You rarely use universal quantification to make
  blanket statements about every individual in the
  world (Ax)cs15-381-student(x) smart(x) meaning
  that everyone in the world is a cs15-381 student
  and is smart.

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Quantifiers

• Existential quantifiers are usually used with
  "and" to specify a list of properties or facts
  about an individual.
• E.g., (Ex) cs15-381-student(x) smart(x) means
  "there is a cs15-381 student who is smart."
• A common mistake is to represent this English
  sentence as the FOL sentence (Ex)
  cs15-381-student(x) gt smart(x)
• Switching the order of universal quantifiers does
  not change the meaning (Ax)(Ay)P(x,y) is
  logically equivalent to (Ay)(Ax)P(x,y).
  Similarly, you can switch the order of
  existential quantifiers.
• Switching the order of universals and
  existentials does change meaning
• Everyone likes someone (Ax)(Ey)likes(x,y)
• Someone is liked by everyone (Ey)(Ax)likes(x,y)

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First-Order Logic (FOL) Syntax

• Sentences are built up of terms and atoms
• A term (denoting a real-world object) is a
  constant symbol, a variable symbol, or a function
  e.g. left-leg-of ( ). For example, x and f(x1,
  ..., xn) are terms, where each xi is a term.
• An atom (which has value true or false) is either
  an n-place predicate of n terms, or, if P and Q
  are atoms, then P, P V Q, P Q, P gt Q, P ltgt Q
  are atoms
• A sentence is an atom, or, if P is a sentence and
  x is a variable, then (Ax)P and (Ex)P are
  sentences
• A well-formed formula (wff) is a sentence
  containing no "free" variables. I.e., all
  variables are "bound" by universal or existential
  quantifiers.
• E.g., (Ax)P(x,y) has x bound as a universally
  quantified variable, but y is free.

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Translating English to FOL

• Every gardener likes the sun.(Ax) gardener(x) gt
  likes(x,Sun)
• You can fool some of the people all of the
  time.(Ex)(At) (person(x) time(t)) gt
  can-fool(x,t)
• You can fool all of the people some of the
  time.(Ax)(Et) (person(x) time(t) gt
  can-fool(x,t)
• All purple mushrooms are poisonous.(Ax)
  (mushroom(x) purple(x)) gt poisonous(x)

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Translating English to FOL

• No purple mushroom is poisonous.(Ex) purple(x)
  mushroom(x) poisonous(x) or,
  equivalently,(Ax) (mushroom(x) purple(x)) gt
  poisonous(x)
• There are exactly two purple mushrooms.(Ex)(Ey)
  mushroom(x) purple(x) mushroom(y) purple(y)
  (xy) (Az) (mushroom(z) purple(z)) gt
  ((xz) v (yz))
• Deb is not tall.tall(Deb)
• X is above Y if X is on directly on top of Y or
  else there is a pile of one or more other objects
  directly on top of one another starting with X
  and ending with Y.(Ax)(Ay) above(x,y) ltgt
  (on(x,y) v (Ez) (on(x,z) above(z,y)))

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Inference Rules for FOL

• Inference rules for PL apply to FOL as well. For
  example, Modus Ponens, And-Introduction,
  And-Elimination, etc.
• New sound inference rules for use with
  quantifiers
• Universal EliminationIf (Ax)P(x) is true, then
  P(c) is true, where c is a constant in the domain
  of x. For example, from (Ax)eats(Ziggy, x) we can
  infer eats(Ziggy, IceCream).
• The variable symbol can be replaced by any ground
  term, i.e., any constant symbol or function
  symbol applied to ground terms only.
• Existential IntroductionIf P(c) is true, then
  (Ex)P(x) is inferred.
• For example, from eats(Ziggy, IceCream) we can
  infer (Ex)eats(Ziggy, x).
• All instances of the given constant symbol are
  replaced by the new variable symbol. Note that
  the variable symbol cannot already exist anywhere
  in the expression.
• Existential EliminationFrom (Ex)P(x) infer P(c).
  
• For example, from (Ex)eats(Ziggy, x) infer
  eats(Ziggy, Cheese).
• Note that the variable is replaced by a brand new
  constant that does not occur in this or any other
  sentence in the Knowledge Base. In other words,
  we don't want to accidentally draw other
  inferences about it by introducing the constant.
  All we know is there must be some constant that
  makes this true, so we can introduce a brand new
  one to stand in for that (unknown) constant.

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Inference Rules for FOL

• Paramodulation
• Given two sentences (P1 v ... v PN) and (ts v Q1
  v ... v QM) where each Pi and Qi is a literal and
  Pj contains a term t, derive new sentence (P1 v
  ... v Pj-1 v Pjs v Pj1 v ... v PN v Q1 v ... v
  QM) where Pjs means a single occurrence of the
  term t is replaced by the term s in Pj
• Example From P(a) and ab derive P(b)

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Inference Rules for FOL

• Generalized Modus Ponens (GMP)
• Combines And-Introduction, Universal-Elimination,
  and Modus Ponens
• E.g. from P(c), Q(c), and (Ax)(P(x) Q(x)) gt
  R(x), derive R(c)
• In general, given atomic sentences P1, P2, ...,
  PN, and implication sentence (Q1 Q2 ... QN)
  gt R, where Q1, ..., QN and R are atomic
  sentences, and subst(Theta, Pi) subst(Theta,
  Qi) for i1,...,N, derive new sentence
  subst(Theta, R)
• subst(Theta, alpha) denotes the result of
  applying a set of substitutions defined by Theta
  to the sentence alpha
• A substitution list Theta v1/t1, v2/t2, ...,
  vn/tn means to replace all occurrences of
  variable symbol vi by term ti.
• Substitutions are made in left-to-right order in
  the list.
• E.g. subst(x/IceCream, y/Ziggy, eats(y,x))
  eats(Ziggy, IceCream)

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Automated Inference in FOL

• Automated inference in FOL is harder than in PL
  because variables can take on potentially an
  infinite number of possible values from their
  domain. Hence there are potentially an infinite
  number of ways to apply the Universal-Elimination
  rule of inference
• Goedel's Completeness Theorem says that FOL
  entailment is semidecidable. That is, if a
  sentence is true given a set of axioms, there is
  a procedure that will determine this. However, if
  the sentence is false, then there is no guarantee
  that a procedure will ever determine this. In
  other words, the procedure may never halt in this
  case.
• Goedel's Incompleteness Theorem says that in a
  slightly extended language (that enables
  mathematical induction), there are entailed
  sentences that cannot be proved
• The truth table method of inference is not
  complete for FOL because the truth table size may
  be infinite
• Natural deduction is complete for FOL but is not
  practical for automated inference because the
  "branching factor" in a search is too large,
  caused by the fact that we would have to
  potentially try every inference rule in every
  possible way using the set of known sentences

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Incompleteness of the generalized modus ponens
inference rule
?x P(x) gt Q(x) ?x ?P(x) gt Q(x) ?x Q(x) gt
S(x) ?x R(x) gt S(x)
? Cannot be converted to Horn
Want to conclude S(A)
S(A) is true if Q(A) or R(A) is true, and one of
those must be true because either P(A) or
?P(A) Incomplete there are entailed sentences
that the procedure cannot infer.
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Generalized Modus Ponens in Horn FOL

• Generalized Modus Ponens (GMP) is complete for
  KBs containing only Horn clauses
• A Horn clause is a sentence of the form(Ax)
  (P1(x) P2(x) ... Pn(x)) gt Q(x)where there
  are 0 or more Pi's, and the Pi's and Q are
  positive (i.e., un-negated) literals
• Horn clauses represent a subset of the set of
  sentences representable in FOL. For example, P(a)
  v Q(a) is a sentence in FOL but is not a Horn
  clause.
• Natural deduction using GMP is complete for KBs
  containing only Horn clauses. Proofs start with
  the given axioms/premises in KB, deriving new
  sentences using GMP until the goal/query sentence
  is derived. This defines a forward chaining
  inference procedure because it moves "forward"
  from the KB to the goal.

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Example of forward chaining

• Example KB All cats like fish, cats eat
  everything they like, and Ziggy is a cat. In FOL,
  KB
• (Ax) cat(x) gt likes(x, Fish)
• (Ax)(Ay) (cat(x) likes(x,y)) gt eats(x,y)
• cat(Ziggy)
• Goal query Does Ziggy eat fish?
• Proof
• Use GMP with (1) and (3) to derive 4.
  likes(Ziggy, Fish)
• Use GMP with (3), (4) and (2) to derive
  eats(Ziggy, Fish)
• So, Yes, Ziggy eats fish.

Data-driven
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Backward chaining

• Backward-chaining deduction using GMP is complete
  for KBs containing only Horn clauses. Proofs
  start with the goal query, find implications that
  would allow you to prove it, and then prove each
  of the antecedents in the implication, continuing
  to work "backwards" until we get to the axioms,
  which we know are true.
• Example Does Ziggy eat fish?
• To prove eats(Ziggy, Fish), first see if this is
  known from one of the axioms directly. Here it is
  not known, so see if there is a Horn clause that
  has the consequent (i.e., right-hand side) of the
  implication matching the goal.
• Proof Goal Driven
• Goal matches RHS of Horn clause (2), so try and
  prove new sub-goals cat(Ziggy) and likes(Ziggy,
  Fish) that correspond to the LHS of (2)
• cat(Ziggy) matches axiom (3), so we've "solved"
  that sub-goal
• likes(Ziggy, Fish) matches the RHS of (1), so try
  and prove cat(Ziggy)
• cat(Ziggy) matches (as it did earlier) axiom (3),
  so we've solved this sub-goal
• There are no unsolved sub-goals, so we're done.
  Yes, Ziggy eats fish.

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Resolution Procedure (aka Resolution Refutation
Procedure)

• Resolution procedure is a sound and complete
  inference procedure for FOL (BFS is complete for
  Horn, DFS may go into infinite loops)
• Resolution procedure uses a single rule of
  inference the Resolution Rule (RR), which is a
  generalization of the same rule used in PL
• Resolution Rule for PLFrom sentence P1 v P2 v
  ... v Pn and sentence P1 v Q2 v ... v Qm derive
  resolvent sentence P2 v ... v Pn v Q2 v ... v Qm
  
• Examples
• From P and P v Q, derive Q (Modus Ponens)
• From (P v Q) and (Q v R), derive P v R
• From P and P, derive False
• From (P v Q) and (P v Q), derive True

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Resolution Procedure

• Resolution Rule for FOL
• Given sentence P1 v ... v Pn
• and sentence Q1 v ... v Qm
• where each Pi and Qi is a literal, i.e., a
  positive or negated predicate symbol with its
  terms,
• if Pj and Qk unify with substitution list Theta,
  
• then derive the resolvent sentencesubst(Theta,
  P1 v ... v Pj-1 v Pj1 v ... v Pn v Q1 v ... Qk-1
  v Qk1 v ... v Qm)
• Example
• From clause P(x, f(a)) v P(x, f(y)) v Q(y)
• and clause P(z, f(a)) v Q(z),
• derive resolvent clause P(z, f(y)) v Q(y) v Q(z)
  using Theta x/z

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Unification

• Unification is a "pattern matching" procedure
  that takes two atomic sentences, called literals,
  as input, and returns "failure" if they do not
  match and a substitution list, Theta, if they do
  match.
• That is, unify(p,q) Theta means subst(Theta, p)
  subst(Theta, q) for two atomic sentences p and
  q.
• Theta is called the most general unifier (mgu)
• All variables in the given two literals are
  implicitly universally quantified
• To make literals match, replace
  (universally-quantified) variables by terms

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Unification Algorithm

• procedure unify(p, q, theta)
• Scan p and q left-to-right and find the first
  corresponding terms where p and q "disagree"
  where p and q not equal
• If there is no disagreement, return theta
  success
• Let r and s be the terms in p and q,
  respectively, where disagreement first occurs
• If variable(r) then theta union(theta, r/s)
  unify(subst(theta, p), subst(theta, q), theta)
• else if variable(s) then theta union(theta,
  s/r) unify(subst(theta, p), subst(theta, q),
  theta)
• else return "failure" end

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Unification

• Examples

Literal 1 Literal 2 Literal 3
parents(x, father(x), mother(Bill)) parents(x, father(x), mother(Bill)) parents(x, father(x), mother(Jane)) parents(Bill, father(Bill), y) parents(Bill, father(y), z) parents(Bill, father(y), mother(y)) x/Bill, y/mother(Bill) x/Bill, y/Bill, z/mother(Bill) Failure
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Unification

• Unify is a linear time algorithm that returns the
  most general unifier (mgu), i.e., a shortest
  length substitution list that makes the two
  literals match.
• (In general, there is not a unique minimum length
  substitution list, but unify returns one of
  them.)
• A variable can never be replaced by a term
  containing that variable. For example, x/f(x) is
  illegal. This "occurs check" should be done in
  the above pseudo-code before making the recursive
  calls.

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Resolution Procedure

• Proof by contradiction method
• Given a consistent set of axioms KB and goal
  sentence Q, we want to show that KB Q. This
  means that every interpretation I that satisfies
  KB, satisfies Q. But we know that any
  interpretation I satisfies either Q or Q, but
  not both. Therefore if in fact KB Q, an
  interpretation that satisfies KB, satisfies Q and
  does not satisfy Q. Hence KB union Q is
  unsatisfiable, i.e., that it's false under all
  interpretations.
• In other words, (KB - Q) ltgt (KB Q - False)
• What's the gain? If KB union Q is unsatisfiable,
  then some finite subset is unsatisfiable
• Resolution procedure can be used to establish
  that a given sentence Q is entailed by KB
  however, it cannot, in general, be used to
  generate all logical consequences of a set
  sentences. Also, the resolution procedure cannot
  be used to prove that Q is not entailed by KB.
• Resolution procedure won't always give an answer
  since entailment is only semidecidable. And you
  can't just run two proofs in parallel, one trying
  to prove Q and the other trying to prove Q,
  since KB might not entail either one.

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Resolution Algorithm

• procedure resolution-refutation(KB, Q)
• KB is a set of consistent, true FOL sentences
  
• Q is a goal sentence that we want to derive
• return success if KB - Q, and failure
  otherwise
• KB union(KB, Q)
• while false not in KB do
• pick 2 sentences, S1 and S2, in KB that contain
  literals that unify (if none, return "failure)
• resolvent resolution-rule(S1, S2)
• KB union(KB, resolvent)
• return "success"

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Resolution example (using PL sentences)

• From Heads I win, tails you lose prove that I
  win
• First, define the axioms in KB
• "Heads I win, tails you lose."(Heads gt IWin)
  or, equivalently, (Heads v IWin)(Tails gt
  YouLose) or, equivalently, (Tails v YouLose)
• Add some general knowledge axioms about coins,
  winning, and losing(Heads v Tails)(YouLose gt
  IWin) or, equivalently, (YouLose v IWin)(IWin
  gt YouLose) or, equivalently, (IWin v YouLose)
• Goal IWin

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Resolution example (using PL sentences)
Sentence 1 Sentence 2 Resolvent
IWin Heads Tails YouLose IWin Heads v IWin Heads v Tails Tails v YouLose YouLose v Iwin IWin Heads Tails YouLose IWin False
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Problems yet to be addressed

• Resolution rule of inference is only applicable
  with sentences that are in the form P1 v P2 v ...
  v Pn, where each Pi is a negated or nonnegated
  predicate and contains functions, constants, and
  universally quantified variables, so can we
  convert every FOL sentence into this form?
• Resolution strategy
• How to pick which pair of sentences to resolve?
• How to pick which pair of literals, one from each
  sentence, to unify?

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Converting FOL sentences to clause form

• Every FOL sentence can be converted to a
  logically equivalent sentence that is in a
  "normal form" called clause form
• Steps to convert a sentence to clause form
• Eliminate all ltgt connectives by replacing each
  instance of the form (P ltgt Q) by expression ((P
  gt Q) (Q gt P))
• Eliminate all gt connectives by replacing each
  instance of the form (P gt Q) by (P v Q)
• Reduce the scope of each negation symbol to a
  single predicate by applying equivalences such as
  converting
• P to P
• (P v Q) to P Q
• (P Q) to P v Q
• (Ax)P to (Ex)P
• (Ex)P to (Ax)P

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Converting FOL sentences to clause form

• Standardize variables rename all variables so
  that each quantifier has its own unique variable
  name. For example, convert (Ax)P(x) to (Ay)P(y)
  if there is another place where variable x is
  already used.
• Eliminate existential quantification by
  introducing Skolem functions. For example,
  convert (Ex)P(x) to P(c) where c is a brand new
  constant symbol that is not used in any other
  sentence. c is called a Skolem constant. More
  generally, if the existential quantifier is
  within the scope of a universal quantified
  variable, then introduce a Skolem function that
  depends on the universally quantified variable.
• For example, (Ax)(Ey)P(x,y) is converted to
  (Ax)P(x, f(x)). f is called a Skolem function,
  and must be a brand new function name that does
  not occur in any other sentence in the entire KB.
  
• Example (Ax)(Ey)loves(x,y) is converted to
  (Ax)loves(x,f(x)) where in this case f(x)
  specifies the person that x loves.
• (If we knew that everyone loved their mother,
  then f could stand for the mother-of function.)

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Converting FOL sentences to clause form

• Remove universal quantification symbols by first
  moving them all to the left end and making the
  scope of each the entire sentence, and then just
  dropping the "prefix" part. E.g., convert
  (Ax)P(x) to P(x)
• Distribute "and" over "or" to get a conjunction
  of disjunctions called conjunctive normal form.
• convert (P Q) v R to (P v R) (Q v R)
• convert (P v Q) v R to (P v Q v R)
• Split each conjunct into a separate clause, which
  is just a disjunction ("or") of negated and
  nonnegated predicates, called literals
• Standardize variables apart again so that each
  clause contains variable names that do not occur
  in any other clause

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Converting FOL sentences to clause form

• Examples
• Convert the sentence
• (Ax)(P(x) gt ((Ay)(P(y) gt P(f(x,y)))
  (Ay)(Q(x,y) gt P(y))))
• Eliminate ltgtNothing to do here.
• Eliminate gt(Ax)(P(x) v ((Ay)(P(y) v
  P(f(x,y))) (Ay)(Q(x,y) v P(y))))
• Reduce scope of negation(Ax)(P(x) v ((Ay)(P(y)
  v P(f(x,y))) (Ey)(Q(x,y) P(y))))

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Converting FOL sentences to clause form

1. Standardize variables(Ax)(P(x) v ((Ay)(P(y) v
   P(f(x,y))) (Ez)(Q(x,z) P(z))))
2. Eliminate existential quantification(Ax)(P(x) v
   ((Ay)(P(y) v P(f(x,y))) (Q(x,g(x))
   P(g(x)))))
3. Drop universal quantification symbols(P(x) v
   ((P(y) v P(f(x,y))) (Q(x,g(x)) P(g(x)))))
4. Convert to conjunction of disjunctions(P(x) v
   P(y) v P(f(x,y))) (P(x) v Q(x,g(x))) (P(x)
   v P(g(x)))

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Converting FOL sentences to clause form

• Create separate clauses
• P(x) v P(y) v P(f(x,y))
• P(x) v Q(x,g(x))
• P(x) v P(g(x))
• Standardize variables
• P(x) v P(y) v P(f(x,y))
• P(z) v Q(z,g(z))
• P(w) v P(g(w))

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Example Hoofers Club example from Charles Dyer

• Problem Statement Tony, Shi-Kuo and Ellen
  belong to the Hoofers Club. Every member of the
  Hoofers Club is either a skier or a mountain
  climber or both. No mountain climber likes rain,
  and all skiers like snow. Ellen dislikes
  whatever Tony likes and likes whatever Tony
  dislikes. Tony likes rain and snow.
• Query Is there a member of the Hoofers Club who
  is a mountain climber but not a skier?

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Example Hoofers Club

• Translation into FOL Sentences
• Let S(x) mean x is a skier, M(x) mean x is a
  mountain climber, and L(x,y) mean x likes y,
  where the domain of the first variable is Hoofers
  Club members, and the domain of the second
  variable is snow and rain. We can now translate
  the above English sentences into the following
  FOL wffs
• (Ax) S(x) v M(x)
• (Ex) M(x) L(x, Rain)
• (Ax) S(x) gt L(x, Snow)
• (Ay) L(Ellen, y) ltgt L(Tony, y)
• L(Tony, Rain)
• L(Tony, Snow)
• Query (Ex) M(x) S(x)
• Negation of the Query (Ex) M(x) S(x)

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Example Hoofers Club

• Conversion to Clause Form
• S(x1) v M(x1)
• M(x2) v L(x2, Rain)
• S(x3) v L(x3, Snow)
• L(Tony, x4) v L(Ellen, x4)
• L(Tony, x5) v L(Ellen, x5)
• L(Tony, Rain)
• L(Tony, Snow)
• Negation of the Query M(x7) v S(x7)

Write these on blackboard
37
Example Hoofers Club

• Resolution Refutation Proof

Clause 1 Clause 2 Resolvent MGU (i.e., Theta)
8 9 10 11 1 3 4 7 9. S(x1) 10. L(x1, Snow) 11. L(Tony, Snow) 12. False x7/x1 x3/x1 x4/Snow, x1/Ellen
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Example Hoofers Club

• Answer Extraction

Clause 1 Clause 2 Resolvent MGU (i.e., Theta)
M(x7) v S(x7) v (M(x7) S(x7)) 9 10 11 1 3 4 7 9. S(x1) v (M(x1) S(x1)) 10. L(x1, Snow) v (M(x1) S(x1)) 11. L(Tony, Snow) v (M(Ellen) S(Ellen)) 12. M(Ellen) S(Ellen) x7/x1 x3/x1 x4/Snow, x1/Ellen

• Answer to the query Ellen!

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Resolution procedure as search

• Resolution procedure can be thought of as the
  bottom-up construction of a search tree, where
  the leaves are the clauses produced by KB and the
  negation of the goal. When a pair of clauses
  generates a new resolvent clause, add a new node
  to the tree with arcs directed from the resolvent
  to the two parent clauses. The resolution
  procedure succeeds when a node containing the
  False clause is produced, becoming the root node
  of the tree.
• A strategy is complete if its use guarantees that
  the empty clause (i.e., false) can be derived
  whenever it is entailed

40
Some strategies for controlling resolution's
search

• Breadth-First
• Level 0 clauses are those from the original
  axioms and the negation of the goal. Level k
  clauses are the resolvents computed from two
  clauses, one of which must be from level k-1 and
  the other from any earlier level.
• Compute all level 1 clauses possible, then all
  possible level 2 clauses, etc.
• Complete, but very inefficient.
• Set-of-Support
• At least one parent clause must be from the
  negation of the goal or one of the "descendents"
  of such a goal clause (i.e., derived from a goal
  clause)
• Complete (assuming all possible set-of-support
  clauses are derived)
• Unit Resolution
• At least one parent clause must be a "unit
  clause," (clause containing a single literal)
• Not complete in general, but complete for Horn
  clause KBs
• Input Resolution
• At least one parent from the original KB (axioms
  and the negation of the goal)
• Not complete in general, but complete for Horn
  clause KBs
• Linear Resolution
• P and Q can be resolved together if either P is
  in the original KB or if P is an ancestor of Q in
  the proof tree
• Complete

41
Subsumption

• Motivation Helps keep the KB small gt less
  search
• Never keep more specific (subsumed) sentences
  in the KB than more general ones
• E.g., if the KB has P(x), then eliminate
• P(c)
• P(a) v P(b)