Resolution

1
Resolution
2
An example

• John is a lawyer.
• Lawyers are rich.
• Rich people have big houses.
• Big houses are a lot of work.
• We would like to conclude that Johns house is a
  lot of work.
• Natural languages are ambiguous so we can have
  different axiomatizations.

3
Axiomatization 1

• lawyer(john)
• ?x lawyer(x) ? rich(x)
• ?x rich(x) ? ?y house(x,y)
• ?x,y rich(x) ? house(x,y) ? big(y)
• ?x,y ( house(x,y) ? big(y) ? work(y) )
• 3 and 4, say that rich people do have at least
  one house and all their houses are big.
• Conclusion we want to show
• house(john, jhouse) ? work(john, jhouse)
• Or, do we want to conclude that John has at least
  one house that needs a lot of work? I.e.
• ?y house(john,y) ? work(y)

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Axiomatization 2

• lawyer(john)
• ?x lawyer(x) ? rich(x)
• ?x rich(x) ? ?y house(x,y) ? big(y)
• ?x,y ( house(x,y) ? big(y) ? work(y) )
• Now, 3 says that rich people have at least one
  house which is big.
• Conclusion we want to show
• house(john, jhouse) ? work(john, jhouse)
• Well, this is not anymore a conclusion we can or
  want to derive!
• So, now we want to conclude that John has at
  least one house that needs a lot of work? I.e.
• ?y house(john,y) ? work(y)

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Lets fix axiomatization 1

• lawyer(john)
• ?x lawyer(x) ? rich(x)
• ?x rich(x) ? ?y house(x,y)
• ?x,y rich(x) ? house(x,y) ? big(y)
• ?x,y ( house(x,y) ? big(y) ? work(y) )
• Conclusion we want to show John has at least
  one house that needs a lot of work. I.e.
• ?y house(john,y) ? work(y)
• Think about sentence 3. It says that every rich
  person x, has a house which house? The one that
  belongs to him, I.e. lets name it houseof(x).
  Now, 3 can be rewritten as
• ?x rich(x) ? house(x, houseof(x))
• This is called skolemization.

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Universal quantifiers

• Since universal quantifiers appear so often, it
  is common simply to omit them, using the
  convention that any variable not associated with
  a quantifier is implicitly universally
  quantified.
• So, the premises become
• lawyer(john)
• lawyer(x) ? rich(x)
• rich(x) ? house(x, houseof(x))
• rich(x) ? house(x,y) ? big(y)
• house(x,y) ? big(y) ? work(y)

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Clausal Form

• A literal is either an atomic sentence or a
  negation of an atomic sentence.
• A clausal sentence is either a literal or a
  disjunction of literals.
• A clause is a set of literals.
• p(a)
• Øp(a)
• p(a), q(b)
• The empty clause is unsatisfiable.
• Before applying resolution we want to convert the
  sentences into clausal form.

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Inseado

• Implications Out
• j1 Þ j2 Øj1 Ú j2
• j1 Ü j2 j1 Ú Øj2
• j1 Û j2 (Øj1 Ú j2 ) Ù (j1 Ú Øj2 )
• Negations In
• ØØj j
• Ø(j1 Ù j2 ) Øj1 Ú Øj2
• Ø(j1 Ú j2 ) Øj1 Ù Øj2
• Ø"n.j n.Øj
• Øn.j "n.Øj

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Inseado (continued)

• Standardize variables (I.e. renaming)
• "x.p(x) Ú "x.q(x) "x.p(x) Ú "y.q(y)
• The scope of quantifiers determines the scope of
  variables. So, in order to avoid name conflicts
  we rename the variables.
• Existentials Out
• x.p(x) p(a)
• "xy.(p(x) Ù z.q(x, y, z)) "xy.(p(x) Ù q(x,
  y, f (x, y)))
• This is called skolemization. Recall example with
  lawyers.
• When the existentially quantified variable is not
  in the scope of any universally quantified
  variable, we just replace it with an object
  constant (name doesnt matter). See first case.

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Inseado (concluded)

• Alls Out
• "x.( p(x) Ù q(x, y, f (x, y))) p(x) Ù q(x,
  y, f (x, y))
• Distribution
• j1 Ú (j2 Ù j3) (j1 Ú j2) Ù (j1 Ú j3)
• (j1 Ù j2) Ú j3 (j1 Ú j3) Ù (j2 Ú j3)
• Operators out
• j1 Ù... Ùjn j1
• ...
• jn
• j1 Ú... Újn j1,..., jn

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Example
?x (rich(x) ? ?y house(x,y)) I ?x (?rich(x) ?
?y house(x,y)) N ?x (?rich(x) ? ?y
house(x,y)) S ?x (?rich(x) ? ?y house(x,y)) E ?x
(?rich(x) ? house(x,houseof(x)) A ?rich(x) ?
house(x,houseof(x)) D ?rich(x) ?
house(x,houseof(x)) O ?rich(x),
house(x,houseof(x)) ?x,y (rich(x) ? house(x,y)
? big(y)) I ?x,y (?(rich(x) ? house(x,y)) ?
big(y)) N ?x,y (?rich(x) ? ?house(x,y) ?
big(y)) S ?x,y (?rich(x) ? ?house(x,y) ?
big(y)) E ?x,y (?rich(x) ? ?house(x,y) ?
big(y)) A ?rich(x) ? ?house(x,y) ?
big(y) D ?rich(x) ? ?house(x,y) ?
big(y) O ?rich(x), ?house(x,y), big(y)

• lawyer(john)
• ?x (lawyer(x) ? rich(x))
• ?x (rich(x) ? ?y house(x,y))
• ?x,y (rich(x) ? house(x,y) ? big(y))
• ?x,y ( house(x,y) ? big(y) ? work(y) )
• ?x (lawyer(x) ? rich(x))
• I ?x (?lawyer(x) ? rich(x))
• N ?x (?lawyer(x) ? rich(x))
• S ?x (?lawyer(x) ? rich(x))
• E ?x (?lawyer(x) ? rich(x))
• A ?lawyer(x) ? rich(x)
• D ?lawyer(x) ? rich(x)
• O ?lawyer(x), rich(x)

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Example (continued)

• ?x,y ( house(x,y) ? big(y) ? work(y) )
• I ?x,y (?(house(x,y) ? big(y)) ? work(y) )
• N ?x,y (?house(x,y) ? ?big(y) ? work(y) )
• S ?x,y (?house(x,y) ? ?big(y) ? work(y) )
• E ?x,y (?house(x,y) ? ?big(y) ? work(y) )
• A ?house(x,y) ? ?big(y) ? work(y)
• D ?house(x,y) ? ?big(y) ? work(y)
• O ?house(x,y), ?big(y), work(y)
• Negated conclusion
• ??y (house(john,y) ? work(y))
• I ??y (house(john,y) ? work(y))
• N ?y (?house(john,y) ? ?work(y))
• S ?y (?house(john,y) ? ?work(y))
• E ?y (?house(john,y) ? ?work(y))
• A ?house(john,y) ? ?work(y)
• D ?house(john,y) ? ?work(y)
• O ?house(john,y), ?work(y)

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Example (concluded)

• Here are all the clauses we got through INSEADO
  from the premises and the negated conclusion.
• lawyer(john)
• ?lawyer(x1), rich(x1)
• ?rich(x2), house(x2,houseof(x2))
• ?rich(x3), ?house(x3,y1), big(y1)
• ?house(x4,y2), ?big(y2), work(y2)
• ?house(john,y3), ?work(y3)
• Note We rename the variables, in order to not
  have variable name clashes between clauses.

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Resolution

• Propositional resolution was
• j1,..., j,..., jm
• y1,..., Øj,...,yn
• j1,..., jm,y1,..., yn
• Similarly, relational resolution is
• j1,..., j,..., jm
• y1,..., Øy,..., yn
• j1,..., jm,y1,..., yns
• where s mgu(j,y)
• mgu(j,y) Most General Unifier

Example. p(a, y),r(y) Øp(x,b) r(y) xa,
yb r(b) We are unifying (or making the same)
p(a, y) with p(x,b). So, what we should do?
Make xa and yb. The set xa, yb, which is a
mapping from variables to constants is called a
unifier.
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Unification

• Unification is the process of determining whether
  two expressions can be made identical by the
  appropriate substitutions for their variables.
• A substitution is a finite mapping of variables
  to terms. We write substitutions as sets of
  replacement rules, e.g.
• x a,y f (b),z v
• In each rule, the variable to which the arrow is
  pointing to is to be replaced by the term from
  which the arrow is pointing.
• In this case, x is to be replaced by a,
• y is to be replaced by f(b), and
• z is to be replaced by v.
• The variables being replaced constitute the
  domain of the substitution, and the terms
  replacing them constitute the range. In the
  example above
• domain is x, y, z, and range is a, b, v.
• The variables in the domain are called bound. The
  terms in the range are called variable bindings.

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• A substitution is pure when all replacement terms
  in the range are free of the variables in the
  domain of the substitution.
• Otherwise, the substitution is impure. The
  substitution shown above is pure whereas the one
  shown below is impure.
• x a,y f (b),z x
• The result of applying a substitution s to an
  expression j is the expression js obtained from
  the original expression by replacing every
  occurrence of every variable in the domain of the
  substitution by the term with which it is
  associated.
• Examples
• q(x, y) x a, y f (b), z v q(a, f (b))
• q(x, x) x a, y f (b), z v q(a,a)
• q(x,w) x a, y f (b), z v q(a,w)
• q(z,v) x a, y f (b), z v q(v,v)

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• If no bound variable of a pure substitution ?
  occurs in another pure substitution ?, then we
  can compose ? with ?, obtaining so a new
  substitution ?? which is also pure.
• We do the composition by applying ? at the
  bindings of ?, and then adding remaining bindings
  from ?.
• E.g. wg(x,y) xa, yb, zc wg(a,b), xa,
  yb, zc
• A set of expressions ?1, , ?n is unifiable if
  and only if there is a substitution ? that makes
  the expressions identical I.e.
• ?1? ?n?
• ? is said to be a unifer for the set.
• E.g. xa, yb, zc unifies the expressions
  p(a,y,z) and p(x,b,z)
• p(a,y,z) xa, yb, zc p(a,b,c) P(x,b,z)
  xa, yb, zc
• The above is not the only unifier.
• We can say for z, zw or zc.
• Well, the zw is more general than zc,
• because w is variable, which can bounded later to
  any other variable or constant, while c as
  constant cant be bounded to anything at all.

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• Formally We say that a substitution ? is more
  general than a unifier ? if there exists a
  substitution ? such that ?? ?.
• A most general unifier (mgu) ? has the property
  that any other unifier ? can be obtained from ?
  by some substitution
• ?? ?.
• The mgu is unique up to variable renaming.
• Example. p(x,y) and p(a,v) are unifiable because
  they have a unifier, e.g.
• xa,yb,vb
• The results of applying this substitution to the
  two expressions are
• p(x, y) x a, y b,v b p(a,b)
• p(a,v) x a,y b,v b p(a,b)
• We could have substituted
• c or d or f(c) or f(w) for y and v.
• In fact, we can unify the expressions without
  changing v at all by simply replacing y by v.
• So, the substitution xa,yv is more general
  than xa,yf(c),vf(c) since there is a
  substitution vf(c) that, when applied to the
  former, gives the latter
• x a,y v v f (c) x a, y f (c),v f (c)

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Recursive procedure for the mgu of two
expressions(starting with a relation or function)
Mgu (x,y) if (xy) return if
(Variable(x)) return MguVar(x,y) if
(Variable(y)) return MguVar(y,x) if
(Constant(x) Constant(y)) exit(Non
unifiable) if (Length(x) ! Length(y))
exit(Non unifiable) i 0 g while
(TRUE) if (iLength(x)) return g s Mgu
(Part(x,i), Part(y,i)) g Compose(g,s) x
Substitute(x,g) y Substitute(y,g) i i
1 MguVar (x,y) if (x occurs in y)
exit(Non unifiable) return x?y Well,
this is for the computerIn doing examples by
hand, we can easily find the mgu by looking.
Variable(x) returns TRUE is x is
variable. Length(x) for x e.g. xF(A,G(y))
returns the number of arguments, I.e. 2.
Part(i) returns the i-th part. E.g.
Part(F(A,G(y)), 0) F Part(F(A,G(y)), 1)
A Part(F(A,G(y)), 2) G(y) Compose(g,s) is
self-explanatory. Substitute(x,g) is
self-explanatory.
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Unification example

• E1 p(x, g(y,y), x) with E2 p(g(a,z), v, v)
• pp -- part 0
• Now try to unify x with g(a,z) -- part 1
• x?g(a,z)
• Now apply this substitution to both E1 and E2
• E1 p(g(a,z), g(y,y), g(a,z)) E2 p(g(a,z),
  v, v)
• Now try to unify g(y,y) with v -- part 2
• v?g(y,y)
• Now apply this substitution to both E1 and E2
• E1 p(g(a,z), g(y,y), g(a,z)) E2 p(g(a,z),
  g(y,y), g(y,y))
• Now try to unify g(a,z) with g(y,y) -- part
  3
• Recursively, we find
• y?a and z?a.
• Finally, x?g(a,a), y?a, z?a, v?g(a,a)
• E1 p(g(a,a), g(a,a), g(a,a)) E2 p(g(a,a),
  g(a,a), g(a,a))

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• Example.
• p(a,b) ?
• ?p(x,y), q(x,y) ?
• ?q(z,w), r(z) ?
• ?r(v) ?
• q(a,b) 1,2
• ?q(v,w) 3,4
• 5,6

• When the conclusion is derived we have a
  contradiction. It is useful in proving things by
  refutation, I.e. assuming that what we want to
  prove is not true
• There are other possible conclusions that can be
  derived from the same premises. In the following
  graph there are the possible conclusions.

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1. p(a,b) ?
2. ?p(x,y), q(x,y) ?
3. ?q(z,w), r(z) ?
4. ?r(v) ?
5. q(a,b) 1,2
6. ?p(z,w), r(z) 2,3
7. ?q(v,w) 3,4
8. r(a) 3,5
9. r(a) 1,6
10. ?p(v,w) 4,6
11. ?p(v,w) 2,7
12. 1,10

• We keep two pointers (the slow and the fast)
• On each step we compare the sentences under the
  pointers. If we can resolve, we add the new
  derived sentence at the end of the list.
• At the start of the inference we initialize slow
  and fast at the top of the list.
• As long as the two pointers point to different
  positions, we leave the slow where it is and
  advance the fast.
• When they meet, we move the fast at the top of
  the list and we move the slow one position down
  the list.

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Lawyers example

• lawyer(john)
• ?lawyer(x1), rich(x1)
• ?rich(x2), house(x2,houseof(x2))
• ?rich(x3), ?house(x3,y1), big(y1)
• ?house(x4,y2), ?big(y2), work(y2)
• ?house(john,y3), ?work(y3)
• rich(john) 1,2 mgu x1?john
• ?lawyer(x2), house(x2,houseof(x2)) 2,3 mgu
  x1?x2
• ?lawyer(x2), ?house(x3,y1), big(y1) 2,4
• ?rich(x3), big(houseof(x3)) 3,4
• ?rich(x2), ?big(houseof(x2)), work(houseof(x2))
  3,5
• 4,5
• 3,6
• 5,6
• you can continue with the two finger methodbut
  its too long.

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Lawyers example in short

1. lawyer(john)
2. ?lawyer(x1), rich(x1)
3. ?rich(x2), house(x2,houseof(x2))
4. ?rich(x3), ?house(x3,y1), big(y1)
5. ?house(x4,y2), ?big(y2), work(y2)
6. ?house(john,y3), ?work(y3)
7. rich(john) 1,2
8. ?rich(john), ?work(houseof(john)) 3,6
9. ?work(houseof(john)) 7,8
10. house(john,houseof(john)) 3,7
11. ?house(john,y1), big(y1) 4,7
12. big(houseof(john)) 10,11
13. ?house(x4, houseof(john)), work(houseof(john)) 5
    ,12
14. work(houseof(john)) 10,13
15. 9,14

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Curiosity and the cat

• 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.
• Did Curiosity kill the cat?
• ?x ?y animal(y) ? loves(x,y) ? ?y loves(y,x)
• ?x ?y animal(y) ? kills(x,y) ? ?z ?loves(z,x)
• ?y animal(y) ? loves(jack,y)
• kills(jack,tuna) ? kills(curiosity,tuna)
• ?x cat(x) ? animal(x)
• ?kills(curiosity,tuna) negated goal

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Curiosity Inseado (1)

• ?x ?y (animal(y) ? loves(x,y)) ? ?y
  loves(y,x)
• I ?x ??y (?animal(y) ? loves(x,y)) ? ?y
  loves(y,x)
• N ?x ?y ?(?animal(y) ? loves(x,y)) ? ?y
  loves(y,x)
• ?x ?y (animal(y) ? ?loves(x,y)) ? ?y
  loves(y,x)
• S ?x ?y (animal(y) ? ?loves(x,y)) ? ?z
  loves(z,x)
• E ?x animal(f(x)) ? ?loves(x,f(x)) ?
  loves(g(x),x)
• A animal(f(x)) ? ?loves(x,f(x)) ?
  loves(g(x),x)
• D animal(f(x)) ? loves(g(x),x) ?
  ?loves(x,f(x)) ? loves(g(x),x)
• O animal(f(x)), loves(g(x),x)
• ?loves(x,f(x)), loves(g(x),x)

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Curiosity Inseado (2)

• ?x ?y animal(y) ? kills(x,y) ? ?z
  ?loves(z,x)
• I ?x ??y animal(y) ? kills(x,y) ? ?z
  ?loves(z,x)
• N ?x ?y ?animal(y) ? ?kills(x,y) ? ?z
  ?loves(z,x)
• S ?x ?y ?animal(y) ? ?kills(x,y) ? ?z
  ?loves(z,x)
• E ?x ?y ?animal(y) ? ?kills(x,y) ? ?z
  ?loves(z,x)
• A ?animal(y) ? ?kills(x,y) ? ?loves(z,x)
• D ?animal(y) ? ?kills(x,y) ? ?loves(z,x)
• O ?animal(y), ?kills(x,y), ?loves(z,x)

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Curiosity Inseado (3-5)

• ?y animal(y) ? loves(jack,y)
• INSE ?y ?animal(y) ? loves(jack,y)
• AD ?animal(y) ? loves(jack,y)
• O ?animal(y), loves(jack,y)
• kills(jack,tuna) ? kills(curiosity,tuna)
• INSEADO kills(jack,tuna), kills(curiosity,tuna)
• ?x cat(x) ? animal(x)
• INSE ?x ?cat(x) ? animal(x)
• AD ?cat(x) ? animal(x)
• O ?cat(x), animal(x)

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Clausal Curiosity

• animal(f(x1)), loves(g(x1),x1)
• ?loves(x2,f(x2)), loves(g(x2),x2)
• ?animal(y1), ?kills(x3,y1), ?loves(z,x3)
• ?animal(y2), loves(jack,y2)
• kills(jack,tuna), kills(curiosity,tuna)
• ?cat(x4), animal(x4)
• ?kills(curiosity,tuna)
• Note We rename the variables, in order to not
  have variable name clashes between clauses.

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