Logical Representations and Resolution

1
Logical Representations and Resolution
2
Boolean Logic

• Conjunctive normal form
• Resolution

3
Conjunctive Normal Form
A literal is a variable or a negated variable. A
clause is either a single literal or the
disjunction of two or more literals. P, P ? ?P,
and P ? ?Q ? R ? S are clauses. ?(R ? S )
and P ? ?Q are not clauses. A wff is in
conjunctive normal form iff it is either a single
clause or the conjunction of two or more
clauses. (P ? ?Q ? R ? S) ? (?P ? ?R) is in
cnf (P ? ?Q ? R ? S) ? (?P ? ?R) is not in cnf
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Conjunctive Normal Form Theorem

• Theorem Every wff can be converted to an
  equivalent wff in conjunctive normal form.
• Proof By construction conjunctiveBoolean(w
  wff)
• Eliminate ? and ? from w, using the fact that P ?
  Q is equivalent to ?P ? Q.
• Reduce the scope of each ? to a single term,
  using the facts
• Double negation ?(?P) P
• deMorgans laws
• ?(P ? Q) ? (?P ? ?Q)
• ?(P ? Q) ? (?P ? ?Q)
• Convert w to a conjunction of clauses using the
  fact that both ? and ? are associative and the
  fact that ? and ? distribute over each other.

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Conversion to Conjunctive Normal Form
Let w be the wff P ? ?(R ? ?Q). Then w can
be converted to conjunctive normal form as
follows Step 1 produces ?P ? ?(R ?
?Q). Step 2 produces ?P ? (?R ? Q). Step 3
produces (?P ? ?R) ? (?P ? Q).
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Resolution
Suppose we know
winter ? summer ?winter summer ?summer
nil winter ? summer ?winter ? cold summer
? cold
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Resolution

• From the premises (P ? Q) and (R ? ?Q),
• Conclude (P ? R).
• Resolution is sound If (P ? Q) and (R ? ?Q) are
  True, then
• If Q is True, R must be True.
• If ?Q True, P must be True.
• Since either Q or ?Q must be True, P ? R must be
  True.

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Resolution The Algorithm

• To prove ST from A
• Convert A to a list L of clauses in conjunctive
  normal form.
• Construct ?ST and add it to L.
• Resolve.

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Complementary Literals
A pair of complementary literals is a pair of
literals that are not mutually satisfiable. So
two literals are complementary iff one is
positive, one is negative, and they contain the
same propositional symbol. For example, Q and ?
Q. Choose two parent clauses that contain a pair
of complementary literals. For example C1
R1 ? R2 ? ? Rj ? Q C2 S1 ? S2 ? ? Sk ? ?Q.
Resolve C1 and C2 to derive R1 ? R2 ? ?
Rj ? S1 ? S2 ? ? Sk
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Resolution Generating nil
Consider ?Q Q nil The empty clause
is unsatisfiable since it contains no literals
that can be made True. So if it is ever
generated, the resolution procedure halts and
reports that, since adding ?ST to A has led to a
contradiction, ST is a theorem given A.
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Resolution The Algorithm

• resolve-Boolean(A, ST)
• Construct L, the list of of clauses from A.
• Negate ST, convert the result to conjunctive
  normal form, and add the resulting clauses to L.
• Until either nil is generated or no progress is
  being made do
• 3. 1 Choose two parent clauses.
• 3. 2 Resolve the parent clauses together.
• 3. 3 If the resolvent is not nil and is not in
  L, add it to L.
• If nil was generated, a contradiction has been
  found. Return success. ST must be true.
• If nil was not generated and there was nothing
  left to do, return failure.

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Resolution An Example
Prove R given Given Axioms Clauses
P P (P ? Q) ? R ?P ? ?Q ? R (S
? T) ? Q ?S ? Q ?T ? Q T
T Add ? R ? R
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Resolution An Example
P ?P ? ?Q ? R ?S ? Q ?T ? Q T ? R
?P ? ?Q ? R ? R ?P ? ?Q P ?T ? Q
?Q ?T T nil
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Resolution Only Select One Pair to Resolve
P ? Q (1) ?P ? ?Q ? R (2) Prove R ?P ?
?Q ? R ?R ?P ? ?Q P ? Q ? nil
But is R entailed by the two facts we have been
given?
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Facts in English
(1) Marcus was a man. (2) Marcus was a
Pompeian. (3) All Pompeians were Romans. (4)
Caesar was a ruler. (5) All Romans were either
loyal to Caesar or hated him. (6) Everyone is
loyal to someone. (7) People only try to
assassinate rulers they are not loyal to. (8)
Marcus tried to assassinate Caesar.
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An Example - Facts in FOL
(1) Marcus was a man. man(Marcus) (2) Marcus was
a Pompeian. Pompeian(Marcus) (3) All Pompeians
were Romans. ?x Pompeian(x) ? Roman(x) (4)
Caesar was a ruler. ruler(Caesar) (5) All Romans
were either loyal to Caesar or hated him. ?x
Roman(x) ? loyalto(x, Caesar) ? hate(x,
Caesar) (6) Everyone is loyal to someone. ?x ?y
loyalto(x, y) (7) People only try to assassinate
rulers they are not loyal to. ?x ?y person(x) ?
ruler(y) ? tryassassinate(x, y) ? ?loyalto(x,
y) (8) Marcus tried to assassinate
Caesar. tryassassinate(Marcus, Caesar)
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Question Answering
(1) Marcus was a man. man(Marcus) (2) Marcus was
a Pompeian. Pompeian(Marcus) (3) All Pompeians
were Romans. ?x Pompeian(x) ? Roman(x) (4)
Caesar was a ruler. ruler(Caesar) (5) All Romans
were either loyal to Caesar or hated him. ?x
Roman(x) ? loyalto(x, Caesar) ? hate(x,
Caesar) (6) Everyone is loyal to someone. ?x ?y
loyalto(x, y) (7) People only try to assassinate
rulers they are not loyal to. ?x ?y person(x) ?
ruler(y) ? tryassassinate(x, y) ? ?loyalto(x,
y) (8) Marcus tried to assassinate
Caesar. tryassassinate(Marcus, Caesar) Was
Marcus loyal to Caesar?
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Reasoning Backward
(1) man(Marcus) (2) Pompeian(Marcus) (3) ?x
Pompeian(x) ? Roman(x) (4) ruler(Caesar) (5) ?x
Roman(x) ? loyalto(x, Caesar) ? hate(x,
Caesar) (6) ?x ?y loyalto(x, y) (7) ?x ?y
person(x) ? ruler(y) ? tryassassinate(x, y) ?
?loyalto(x, y) (8) tryassassinate(Marcus, Caesar)
?loyalto(Marcus, Caesar) (Marcus/x)
(Caesar/y) person(Marcus) ruler(Caesar)
tryassassinate(Marcus, Caesar)
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Reasoning Backward
(1) man(Marcus) (2) Pompeian(Marcus) (3) ?x
Pompeian(x) ? Roman(x) (4) ruler(Caesar) (5) ?x
Roman(x) ? loyalto(x, Caesar) ? hate(x,
Caesar) (6) ?x ?y loyalto(x, y) (7) ?x ?y
person(x) ? ruler(y) ? tryassassinate(x, y) ?
?loyalto(x, y) (8) tryassassinate(Marcus,
Caesar) (9) ?x man(x) ? person(x)
?loyalto(Marcus, Caesar) (Marcus/x)
(Caesar/y) person(Marcus) ruler(Caesar)
tryassassinate(Marcus, Caesar)
(Marcus/x) man(Marcus)
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Functions and Predicates
(1) Marcus was a man. man(Marcus) (2) Marcus was
a Pompeian. Pompeian(Marcus) (3) Marcus was born
in 40 A.D. born(Marcus, 40) (4) All men are
mortal. ?x man(x) ? mortal(x) (5, 6)
AllPompeians died when the volcano erupted in 79
A.D. erupted(volcano, 79) ? ?x Pompeian(x) ?
died(x,79) (7) No mortal lives longer than 150
years. ?x ?t1 ?t2 mortal(x) ? born(x, t1) ?
gt(t2-t1, 150) ? dead(x,t2) (8) It is now
2004. now 2004 Is Marcus alive?
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Functions and Predicates
(1) man(Marcus) (2) Pompeian(Marcus) (3)
born(Marcus, 40) (4) ?x man(x) ? mortal(x) (5)
erupted(volcano, 79) (6) ?x Pompeian(x) ?
died(x,79) (7) ?x ?t1 ?t2 mortal(x) ? born(x,
t1) ? gt(t2-t1, 150) ? dead(x,t2) (8) now 2007
? alive(Marcus, now)
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Functions and Predicates - Filling in the Blanks
(1) man(Marcus) (2) Pompeian(Marcus) (3)
born(Marcus, 40) (4) ?x man(x) ? mortal(x) (5)
erupted(volcano, 79) (6) ?x Pompeian(x) ?
died(x,79) (7) ?x ?t2 ?t2 mortal(x) ? born(x,
t1) ? gt(t2-t1, 150) ? dead(x,t2) (8) now
2007 (9a) ?x ?t alive(x, t) ? ?dead(x, t) (9b) ?x
?t ?dead(x,t) ? alive(x, t) (10) ?x ?t2 ?t2
died(x, t1) ? gt(t2, t1) ? dead(x, t2)
? alive(Marcus, now)
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Showing that Marcus is Not Alive
? alive(Marcus, now) (9a) (Marcus/x)
(now/t) dead(Marcus, now)
(10)(Marcus/x)(now/t2)
(7) (Marcus/x)(now/t2)
died(Marcus, t1) gt(now, t1)
mortal(Marcus) born(Marcus, t1)
gt(now-t1,150) (5) (Marcus/x)
(4) (Marcus/x) (3) (40/t1)
subst (79/t1) Pompeian(Marcus) gt(now,
79) man(Marcus) born(Marcus, 40)
gt(now-40,150) (2)
subst (1)
subst T gt(2004,
79) T T
gt(2007-40,150)
eval eval T

T
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A Harder One
Given ?x Roman(x) ? know(x, Marcus)
? hate(x, Caesar) ? (?y (?z hate(y, z)) ?
thinkcrazy(x, y)) Roman(Isaac) ?hate(Isaac,
Caesar) hate(Paulus, Marcus) ?thinkcrazy(Isaac,
Paulus) Prove ?know(Isaac, Marcus)
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Prenex Normal Form
A sentence in first-order logic is in prenex
normal form iff it is of the form ltquantifier
listgt ltmatrixgt, where ltquantifier listgt is a
list of quantified variables and ltmatrixgt is
quantifier-free. ?x (?y ((P(x) ? Q(y)) ? ?z
(R(x, y, z))) is not in pnf. ?x ?y ?z (P(x) ?
Q(y)) ? R(x, y, z) is in pnf. Its matrix is
(P(x) ? Q(y)) ? R(x, y, z).
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Clause Form

• A sentence is in clause form iff
• It has been converted to prenex normal form,
• Its quantifier list contains only universal
  quantifiers,
• Its quantifier list is no longer explicitly
  represented,
• It is in conjunctive normal form, and
• There are no variable names that appear in more
  than one clause.
• Begin with ?x (P(x) ? Q(x)) ? ?y (S(y))
• In prenex normal form ?x ?y (P(x) ? Q(x)) ? S(y)
• In clause form (?P(x) ? Q(x)) ? S(y)

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Clause Form Simplifies the Process
?x Roman(x) ? know(x, Marcus) ? hate(x,
Caesar) ? (?y (?z hate(y, z)) ? thinkcrazy(x,
y)) becomes ?Roman(x) ? ?know(x, Marcus) ?
hate(x, Caesar) ? ?hate(y, z) ? thinkcrazy(x,
y)
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Conversion to Clause Form - Step 1
1. Eliminate ?, using the fact that a ? b
is equivalent to ?a ? b
?x Roman(x) ? know(x, Marcus) ? hate(x,
Caesar) ? (?y (?z hate(y, z)) ? thinkcrazy(x,
y)) ?x ? Roman(x) ? know(x, Marcus)
? hate(x, Caesar) ? (?y ?(?z hate(y, z)) ?
thinkcrazy(x, y))
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Conversion to Clause Form - Step 2

• 2. Reduce the scope of each ? to a single term,
  using
• ?(?p) p
• deMorgans laws
• ??x P(x) ? ?x ?P(x)
• ??x P(x) ? ?x ?P(x)

?x ? Roman(x) ? know(x, Marcus) ? hate(x,
Caesar) ? (?y ?(?z hate(y, z)) ? thinkcrazy(x,
y)) ?x ? Roman(x) ? ? know(x, Marcus)
? hate(x, Caesar) ? (?y ?z ? hate(y, z) ?
thinkcrazy(x, y))
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Conversion to Clause Form - Step 3
3. Standardize variables so that each quantifier
binds a unique variable.
?x P(x) ? ?x Q(x) ?x P(x) ? ?y Q(y)
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Conversion to Clause Form - Step 4
4. Move all quantifiers to the left without
changing their relative order.
?x ? Roman(x) ? ? know(x, Marcus) ? hate(x,
Caesar) ? (?y ?z ? hate(y, z) ? thinkcrazy(x,
y)) ?x ?y ?z ? Roman(x) ? ? know(x, Marcus)
? hate(x, Caesar) ? (? hate(y, z) ?
thinkcrazy(x, y))
At this point, we have prenex normal form.
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Conversion to Clause Form - Step 5
5. Eliminate existential quantifiers through the
use of Skolem functions and constants.
?x Roman(x) Roman(S1) ?x ?z father-of(x, z)
?x father-of(x, S2(x))

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Conversion to Clause Form - Step 6
6. Drop the prefix since all remaining
quantifiers are universal.
?x ?y ?z ? Roman(x) ? ? know(x, Marcus)
? hate(x, Caesar) ? (? hate(y, z) ?
thinkcrazy(x, y)) ? Roman(x) ? ? know(x,
Marcus) ? hate(x, Caesar) ? (? hate(y, z) ?
thinkcrazy(x, y))
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Conversion to Clause Form - Step 7

• 7. Convert the matrix into a conjunction of
  disjuncts by using
• Associative properties of ? and ?.

? Roman(x) ? ? know(x, Marcus) ? hate(x,
Caesar) ? (? hate(y, z) ? thinkcrazy(x, y)) ?
Roman(x) ? ? know(x, Marcus) ? hate(x, Caesar) ?
? hate(y, z) ? thinkcrazy(x, y)

• Distributivity of ? and ?.

(P(x) ? Q(x)) ? T(x) (P(x) ? T(x)) ? (Q(x)) ?
T(x))
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Conversion to Clause Form - Step 8
8. Create a separate clause for each conjunct.
(P(x) ? T(x)) ? (Q(x)) ? T(x)) (P(x) ? T(x))
(Q(x)) ? T(x))
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Conversion to Clause Form - Step 9
9. Standardize apart the variables.
(P(x) ? T(x)) (Q(x)) ? T(x)) (P(x) ? T(x))
(Q(y)) ? T(y))
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Resolution in FOL

• To find a contradiction, we must show that the
  KB, augmented with ?P, is unsatisfiable.
• Herbrands theorem tells us
• To show that a set of clauses S is unsatisfiable,
  it is necessary to consider only interpretations
  over a particular set called the Herbrand
  universe of S, which is the set of all ground
  terms constructable from the following
• The function symbols in S, if any.
• The constant symbols in S, if any. If none, then
  the constant symbol A.
• A set of clauses S is unsatisfiable iff a finite
  subset of ground instances (in which all bound
  variables have had a value substituted for them)
  of S is unsatisfiable.
• Resolution is an algorithm that finds
  contradictions without enumerating most of the
  Herbrand universe.

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Unification
In propositional logic, it is easy to identify
complementary literals such as P and ?P. But in
FOL, what should we do about ?x,y (hate(x,
y)) ?hate(Marcus, Caesar) Or
?hate(Marcus, Caesar) hate(Paulus,
Caesar) Or ?x,y (P(x, x)) ?P(Marcus,
Caesar)
39
Before We Start
Before we consider a set of formulas and begin to
reason with them, we need to standardize the
variables apart Consider ?x, y Lived-In(x,
y) ?x, y Mother-Of(x, y) Although we have used
x and y in both of these formulas, the variables
in these formulas are logically distinct since
they occur inside the scope of quantifiers that
are independent (i.e., neither lies inside the
scope of the other). So an equivalent way to
write these facts is ?x1, y1 Lived-In(x1,
y1) ?x2, y2 Mother-Of(x2, y2)
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Unification

• We need an algorithm that does two things given a
  sentence S and a term T
• Determine whether T matches some term T in S,
  and
• Return a substitution that can be applied to the
  remainder of S.
• Example
• S ?x,y In(x, y) ? Climate(y, moderate) ?
  Climate(x, moderate)
• T In(Italy, Europe)
• The process we will call unification will
  determine that T matches the first term of S and
  it will return the substitution Italy/x,
  Europe/y. When we apply that to the remainder of
  S, we have
• Climate(Europe, moderate) ? Climate(Italy,
  moderate)

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The Unification Algorithm
unify-for-resolution(lit1, lit2) If
either lit1 or lit2 is a variable or a constant
then Case (checking the conditions in
order and executing only the first one that
matches) lit1 and lit2 are identical
return nil. / Succeed with
no substitution. lit1 is a variable that
occurs in lit2 return Fail. / These two
cases implement the lit2 is a variable
that occurs in lit1 return Fail.
occur check. lit1 is a variable return
(lit2/lit1). lit2 is a variable return
(lit1/lit2). otherwise return Fail.
/ No match. If the
initial predicate or function symbols of lit1 and
lit2 are not the same, return Fail. If lit1 and
lit2 do not have the same number of arguments,
return Fail. substitution-list nil. For i
1 to the number of arguments of lit1 do Let
S be the result of unify-for-resolution on the
ith argument of lit1 and of lit2. If S
contains Fail, return Fail. If S is not
equal to nil then Apply S to the
remainder of both lit1 and lit2. Append
S to substitution-list. Return
substitution-list.
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The Occur Check
Suppose we are trying to unify P(x,
x) P(g(x), g(x)) What happens if we skip
the occur check? x and g(x) unify and yield the
substitution g(x)/x. Now we must apply that
substitution g(x) g(g(x)) ??? g(g(x)) g(g(g
(x))) But if wed standardized apart the
variables before we started, this wouldnt have
been a problem.
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The Occur Check is Necessary Even with
Standardizing Apart
Suppose we are trying to unify P(x, f(x),
x) P(f(a), a, a) x and f(a) unify and
yield the substitution f(a)/x. Applying it
f(f(a)), f(a) a, a Without the occur
check, f(f(a)) and a will unify, yielding the
substitution f(f(a))/a. To apply it, we must
replace all as in the remaining terms. But that
process never terminates.
44
Most General Unifiers (MGUs)
Suppose we are trying to prove ?x,y (A(x, y) ?
B(x, y)) Using the fact ?z (A(John,
z)) Unifying A(x, y) with A(John, z) yields
John/x, z/y. Applying it ?z (B(John,
z)) But we could also have matched with the
substitution John/x, John/y, and been left trying
to prove B(John, John), which would be
harder. The algorithm we have presented always
returns the most general unifier. The MGU is
unique up to variable name substitution.
45
Unification Examples
Substitution Paulus/x father-of(Marcus
)/x x/y Flavius/x, Marcus/y
Inputs 1 Roman(x), Roman(Paulus)
2 Roman(x), Ancient(Paulus)
3 Roman(father-of(Marcus)), Roman(x)
4 Roman(father-of(Marcus)), Roman(Flavius),
5 Roman(x), Roman(y), 6 Roman(father-of(x
)), Roman(x) 7 Likes(x, y), Likes(Flavius,
Marcus)
Result Succeed Fail Succeed Fail Succeed Fa
il (fails occur check) Succeed
46
FOL Resolution An Example
Prove Mortal(Marcus) given ?x (Man(x) ?
Mortal(x)) ?Man(x) ? Mortal(x) Man(Marcus) Man(
Marcus) Add ?Mortal(Marcus) ?Man(x) ?
Mortal(x) ?Mortal(Marcus)
Marcus/x Man(Marcus) ? Man(Marcus) nil
47
FOL Resolution the Algorithm
resolve-FOL(A, ST) Construct L, the list of
clauses from A. Rename all variables in ST so
that they do not conflict with any variables in
L. Negate ST, convert the result to clause form,
and add the resulting clauses to L. Until
either nil is generated or no progress is being
made do Choose from L two (parent) clauses
that contain a pair CL of complementary literals.
Resolve the parent clauses together to
produce a resolvent Initially, let the
resolvent be the disjunction of all of the
literals in both parent clauses except for the
two literals in CL. Apply to all of the
literals in the resolvent the substitution that
was constructed when the literals in CL were
unified. Rename all of the variables in the
resolvent so that they do not conflict with any
of the variables in L. If the resolvent is
not nil and is not already in L, add it to L. If
nil was generated, a contradiction has been
found. Return success. ST must be true. If nil
was not generated and there was nothing left to
do, return failure. ST may or may not be true.
But no proof of ST has been found.
48
Heuristics to Aid the Resolution Procedure

• Only resolve pairs of clauses that contain
  complementary literals.
• P(x) Q(x)
• Eliminate certain clauses as soon as they are
  generated
• Tautologies
• Clauses that are subsumed by other clauses.
• Set of support strategy whenever possible
  resolve with a clause that is part of the
  statement we are trying to prove.
• Unit preference strategy whenever possible,
  resolve with clauses that have a single literal.

49
Subsumption
Clause A subsumes clause B iff B must be true in
any interpretation in which A is true. (A is
more general than B.) True P ? Q ? R ?
?T W P ? Q R ? ?T P Q P ? Q
P ? Q ? W
50
Subsumption in FOL
Clause A subsumes clause B iff B must be true in
any interpretation in which A is true. (A is
more general than B.) True ?x
(P(x)) P(Marcus) ?x(P(x))
51
Using Resolution
Painter(Leonardo) Composer(Palistrina) Country(Ita
ly) Lived-In(Leonardo, Italy) In(Italy,Europe) Cli
mate(Europe, moderate) Painter(Rubens) Lived-In(Ru
bens, Europe) ?x,y (In(x, y) ? Climate(y,
moderate) ? Climate(x, moderate)) Prove that
Leonardo was a painter who lived in a moderate
climate. Prove ?y (Painter(Leonardo)
? Lived-In(Leonardo,y) ? Climate(y,
moderate)) Negate it ?(?y
(Painter(Leonardo) ? Lived-In(Leonardo,y) ?
Climate(y, moderate)))
52
Converting to Clause From
Painter(Leonardo) Composer(Palistrina) Country(Ita
ly) Lived-In(Leonardo, Italy) In(Italy,Europe) Cli
mate(Europe, moderate) Painter(Rubens) Lived-In(Ru
bens, Europe) ?x,y (In(x, y) ? Climate(y,
moderate) ? Climate(x, moderate)) ? In(x, y) ? ?
Climate(y, moderate) ? Climate(x,
moderate) ?(?y (Painter(Leonardo) ?
Lived-In(Leonardo,y) ? Climate(y, moderate)))
?y (?(Painter(Leonardo) ? Lived-In(Leonardo,y) ?
Climate(y, moderate))) ?Painter(Leonardo) ? ?
Lived-In(Leonardo,y) ? ? Climate(y, moderate)
53
Using Resolution
1. Painter(Leonardo) 2. Composer(Palistrina) 3.
Country(Italy) 4. Lived-In(Leonardo, Italy) 5.
In(Italy,Europe) 6. Climate(Europe,
moderate) 7. Painter(Rubens) 8.
Lived-In(Rubens, Europe) 9. Painter(Sargent) 10.
? In(x, y) ? ? Climate(y, moderate) ?
Climate(x, moderate)
?Painter(Leonardo) ?? Lived-In(Leonardo,y)??
Climate(y, moderate)
Painter(Leonardo) Leonardo/x Lived-In(
Leonardo, Italy) ? Lived-In(Leonardo,y1) ? ?
Climate(y1, moderate) Italy/y1 ?
Climate(Italy, moderate)
54
Using Resolution
1. Painter(Leonardo) 2. Composer(Palistrina) 3.
Country(Italy) 4. Lived-In(Leonardo, Italy) 5.
In(Italy,Europe) 6. Climate(Europe,
moderate) 7. Painter(Rubens) 8.
Lived-In(Rubens, Europe) 9. Painter(Sargent) 10.
? In(x, y) ? ? Climate(y, moderate) ?
Climate(x, moderate)
?Painter(Leonardo) ? ? Lived-In(Leonardo,y) ? ?
Climate(y, moderate) Painter(Leonardo)
Leonardo/x Lived-In(Leonardo, Italy) ?
Lived-In(Leonardo,y1) ? ? Climate(y1,
moderate) Italy/y1 ?
In(x, y) ? ? Climate(y, moderate)?Climate(x,
moderate) ? Climate(Italy,
moderate) Italy/x In(Italy,Europe) ?
In(Italy, y2) ? ? Climate(y2, moderate)
Europe/y2 Climate(Europe, moderate) ?
Climate(Europe, moderate) nil
55
Answering Questions
A Question Name a painter who lived in a
moderate climate. ?Painter(x) ?
Lived-In(x,y) ? Climate(y, moderate) Construct
an existence statement to be proved ?x, y
(Painter(x) ? Lived-In(x,y) ? Climate(y,
moderate)) Negate it and resolve, tracking the
bindings ? (?x, y (Painter(x) ? Lived-In(x,y)
? Climate(y, moderate))) ?x,y (?(Painter(x)
? Lived-In(x,y) ? Climate(y, moderate)))
?Painter(x) ? ? Lived-In(x,y) ? ? Climate(y,
moderate)
56
Tracking the Bindings
Painter(x) ? ?Painter(x) ? ? Lived-In(x,y) ? ?
Climate(y, moderate) Painter(Leonardo)
Leonardo/x Painter(Leonardo) ? ?
Lived-In(Leonardo,y1) ? ? Climate(y1,
moderate) Lived-In(Leonardo, Italy)
Italy/y1
Painter(Leonardo) ? ? Climate(Italy,
moderate) ? In(x, y) ? ? Climate(y,
moderate)?Climate(x, moderate)
Italy/x In(Italy,Europe)
Painter(Leonardo) ? ? In(Italy, y2) ? ?
Climate(y2, moderate)
Europe/y2 Climate(Europe, moderate)
Painter(Leonardo) ? ? Climate(Europe,
moderate) Painter(Leonardo)
57
A Reminder About Standardizing Variables Apart
Prove ?father(Chris, Bill) given ?father(x, y)
? ?woman(x) (1) father(x, y)
??woman(x) ?mother(x, y) ? woman(x) (2) moth
er(x, y) ? woman(x) mother(Chris,
Mary) (3) Add father(Chris, Bill)
?father(x, y) ? ?woman(x) ?mother(x, y) ?
woman(x)
?father(x, y) ? ?mother(x,
y) mother(Chris, Mary)
Chris/x, Mary/y
?father(Chris, Mary)
58
Back to Marcus
(1) man(Marcus) (2) Pompeian(Marcus) (3) ?x
Pompeian(x) ? Roman(x) (4) ruler(Caesar) (5) ?x
Roman(x) ? loyalto(x, Caesar) ? hate(x,
Caesar) (6) ?x ?y loyalto(x, y) (7) ?x ?y
man(x) ? ruler(y) ? tryassassinate(x, y) ?
?loyalto(x, y) (8) tryassassinate(Marcus,
Caesar) (1) man(Marcus) (2) Pompeian(Marcus) (3)
?Pompeian(x1) ? Roman(x1) (4) ruler(Caesar) (5)
?Roman(x2) ? loyalto(x2, Caesar) ? hate(x2,
Caesar) (6) loyalto(x3, S1(x3)) (7) ?man(x4) ?
?ruler(y1) ? ?tryassassinate(x4, y1) ?
?loyalto(x4, y1) (8) tryassassinate(Marcus,
Caesar)
59
Proving Marcus Not Loyal to Caesar
loyalto(M, C) ? man(x4) ? ?ruler(y1) ?
?tryassassinate(x4, y1) ? ?loyalto(x4,
y1) (M/x4)(C/y1) ? man(M) ? ?ruler(C) ?
?tryassassinate(M,C) man(M)
?ruler(C) ? ?tryassassinate(M,C) ruler(C
) ?tryassassinate(M,C)
tryassassinate(M,C) nil
60
Does Isaac Know Marcus?
Given 1. ?x Roman(x) ? know(x, Marcus)
? hate(x, Caesar) ? (?y (?z hate(y, z)) ?
thinkcrazy(x, y)) ?Roman(x) ? ? know(x,
Marcus) ? hate(x, Caesar) ? ? hate(y, z) ?
thinkcrazy(x, y) 2. Roman(Isaac) 3. ?hate(Isaac,
Caesar) 4. hate(Paulus, Marcus) 5.
?thinkcrazy(Isaac, Paulus) Prove ?know(Isaac,
Marcus) know(Isaac, Marcus)
61
Does Marcus Hate Caesar?
Try to prove hate(M, C) ?Hate(M, C)
62
Try to Prove that Marcus Does Not Hate Caesar
Try to prove ?hate(M, C) Hate(M, C)
63
Try (Again) to Prove that Marcus Does Not Hate
Caesar
(1) man(Marcus) (2) Pompeian(Marcus) (3)
?Pompeian(x1) ? Roman(x1) (4) ruler(Caesar) (5)
?Roman(x2) ? loyalto(x2, Caesar) ? hate(x2,
Caesar) (6) loyalto(x3, S1(x3)) (7) ?man(x4) ?
?ruler(y1) ? ?tryassassinate(x4, y1) ?
?loyalto(x4, y1) (8) tryassassinate(Marcus,
Caesar) (9) ?persecute(x5, y2) ? hate(y2,
x5) persecute(x, y) ? hate(y, x) (10) ?hate(x6,
y3) ? persecute(y3, x6) hate(x, y) ?
persecute(y, x)
Now try to prove ?hate(M, C) Hate(M, C)
64
A Winnie Example
"And if anyone knows anything about anything,"
said Bear to himself, "It's Owl who knows
something about something," he said, "Or my
name is not Winnie-the-Pooh," he said. "Which it
is," he added. "So there you are."
65
When Does Resolution Work?

• Resolution is sound and refutation-complete.
• If there is a contradiction, resolution will
  eventually find it if resolvents are chosen
  systematically.
• But
• It may take exponential time.
• Satisfiability is semi-decidable if a set of
  clauses is satisfiable, resolution will never
  find a conflict and it may not notice that it is
  failing to do so.

66
Satisfiability is Semi Decidable
Suppose we want to know whether P(x) is true.
There are three possibilities (1) P(x) is
entailed by KB and thus ?P(x) ? KB is
unsatisfiable. (2) ?P(x) is entailed by KB and
thus P(x) ? KB is unsatisfiable. (3)
67
Horn Clauses
A Horn Clause is a clause in which at most one
literal is positive. A(x) ? B(x) ? C(x)
becomes ?A(x) ? ? B(x) ? C(x) a Horn
clause Deciding entailment with Horn clauses can
be done in time that is linear in the size of the
KB by using either backward or forward chaining.