Lecture 9: Resolution in First Order Logic

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Lecture 9 Resolution in First Order Logic
Theorem Proving in First Order Logic Unification R
esolution
Heshaam Faili hfaili_at_ece.ut.ac.ir University of
Tehran
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FOL Decidability (1)

• If a proof procedure is always guaranteed to find
  a proof of a particular sentence or of its
  negation, then the question of logical
  implication for the sentence is said to be
  decidable.
• To see if KB c, we generate all possible
  inferences from KB, stopping when we get f or c.

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FOL Decidability (2)

• In general, neither f nor c may be logically
  implied by KB.
• In this case, the proof procedure will never
  stop.
• The question of logical implication in this case
  is thus semidecidable
• If KB c or KB c , the proof procedure
  will eventually discover that. Otherwise, the
  procedure will run forever.

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FOL -- Inference Rules for quantifiers

• Universal and Existential Elimination
  
• Existential Introduction

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Resolution in FOL

• Mechanical proof procedure with a single rule
• Invented by J. A. Robinson in 1965
• Three main steps 1. Clause form
  transform all clauses to uniform format
  2. Unification find a variable assignment so
  that two clauses can be resolved 3.
  Resolution rule to obtain new conclusions

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1. Clause Form

• We want a uniform representation for our logical
  propositions, so that we can use a simple uniform
  proof procedure.
• Clausal form expresses all logical propositions
  using literals (atomic sentences or negated
  atomic sentences) and clauses (sets of literals
  representing their disjunction).
• Ex all the following in a single form a gt b
  a ? b b ? a (a ? b)

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• Clause Form Conversion

• Any set of sentences in predicate calculus can be
  converted to clausal form.
• Eight-step procedure for conversion
• 1. Replace implications
• 2. Distribute negations
• 3. Standardize variables
• 4. Replace existential
• 5. Remove universals
• 6. Distribute disjunctions
• 7. Replace operators
• 8. Rename variables

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1. Replace implications
Any sentence of the form j
gt y becomes j ? y at all
levels
2. Distribute negations
Repeatedly do the following y becomes
y (j ? y) becomes j ? y (j ? y)
becomes j ? y ?V j becomes
? V j ?V j becomes ?V j
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3. Standardize variables
Rename variables to ensure that each quantifier
has its own unique variable. Ex (?X
p(X)) ? (?X q(X)) becomes (?X p(X)) ? (?Y
q(Y))
4. Replace existential

• Consider (?Y ?X p(X,Y))
• The identity of X depends on the value of Y. We
  can replace X by a function of Y (?Y
  p(g(X),Y)) where g is a Skolem function

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Skolemization

• Replace each occurrence of an existentially
  quantified variable by a Skolem function.
• The functions arguments are the universally
  quantified variables that are bound by universal
  quantifiers whose scope includes the scope of the
  existential quantifier being eliminated.
• The Skolem functions must be new, i.e., not
  already present in any other sentences.
• If the existential quantifier is not within a
  universal quantifier, use a Skolem function of no
  arguments, i.e., a constant.

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Examples of Skolemization

• 1. ?X p(X) becomes p(g)
• 2. ?Y ?X p(X,Y) becomes ?Y p(g(Y),Y)
• 3. ?X ?Y p(X,Y) becomes ?Y p(g,Y)
• 4. ?Y ?X (((q(X) ? p(X,Y)) ? ?Z ?W(r(W)))
• becomes
• ?Y (((q(f(Y)) ? p(f(Y),Y)) ? ?Z (r(g(Y,Z)))
• where f and g are Skolem functions

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5. Remove universal quantifiers
Throw away "s and assume all variables are
universally quantified. "X "Y "Z q(h(Y)) ?
p(h(Y),X) ? r(X,Z) becomes q(h(Y)) ?
p(h(Y),X) ? r(X,Z)
6. Distribute disjunctions
Write the formula in conjunctive normal form (the
conjunction of sets of disjunctions) with
j ? (y ? q) becomes (j ? y) ? (j ? q)
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7. Replace operators

• Replace the conjunction
• S1 ? S2 ? . ? Sn
• with the set of clauses S1 , S2 , , Sn
• Convert each Si into a set of literals (that is,
  get rid of the disjunctions symbols ?, and
  write them in clause notation.
• Ex (p(X) ? q(Y)) ?(p(X) ? r(X,Y))
• becomes two clauses
• p(X), q(Y)
• p(X), r(X,Y)

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8. Rename variables

• Change variable names so that no variable
  symbol appears in more than one clause
• Ex p(X), q(Y) and p(X), r(X,Y)
• becomes p(X), q(Y) and p(Z), r(Z,W)
• we can do this because
• "X "Y(p(X) ? q(Y)) ?(p(X) ? r(X,Y))
• is equivalent to
• "X "Y(p(X) ? q(Y)) ? "Z "W(p(Z) ? r(Z,W))

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Example of entire conversion (1)

• "X (p(X) gt (("Y p(Y) gt (p(f(X,Y))) ?
• ("Y (q(X,Y) ? p(Y))))
• 1. "X (p(X) ? (("Y p(Y) ? (p(f(X,Y))) ?
• ("Y (q(X,Y) ? p(Y))))
• 2. "X (p(X) ? (("Y p(Y) ? (p(f(X,Y)) ?
• (?Y (q(X,Y) ? p(Y))))
• 3. "X (p(X) ? (("Y p(Y) ? (p(f(X,Y))) ?
• (?W (q(X,W) ? p(W))))

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Example of entire conversion (2)

• 4. "X (p(X) ? (("Y p(Y) ? (p(f(X,Y))) ?
• (q(X,g(X)) ? p(g(X)))))
• 5. p(X) ? ((p(Y) ? (p(f(X,Y))) ?
• (q(X,g(X)) ? p(g(X)))))
• 6. (p(X) ? p(Y)) ? (p(f(X,Y))) ?
• (p(X) ? (q(X,g(X))) ?
• (p(X) ? p(g(X)))))

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Example of entire conversion (3)

• 7. p(X), p(Y)), p(f(X,Y)),
• p(X) , q(X,g(X))),
• p(X), p(g(X))
• 8. p(X1), p(Y1)), p(f(X1,Y1)),
• p(X2) , q(X2,g(X2))),
• p(X3), p(g(X3))

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Unification -- Substitution

• Unification is the process of determining whether
  two expressions can be made identical by
  appropriate substitutions for their variables.
• The substitution applies to variables of both
  expressions and makes them syntactically
  equivalent. The result is a substitution
  instance
• Example of a unifying substitution
• S1 p(X,f(b),Z) S2 p(a,Y,W)
• U X/a Y/f(b), Z/W
• S1U S2U p(a,f(b),W) substitution
  instance

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Properties of the substitution

• 1. Each variable is associated with at most one
  expression
• 2. No variable with an associated expression
  occurs within any associated expression. That
  is, no left side appears on a right side
• U X/a, Y/f(X), Z/Y
• is not a legal substitution

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Unification of sentences

• A set of expressions S1,S2,.Sn
• is unifiable if there is a substitution U that
  makes them identical
• S1U S2 U Sn U

Ex U X/a, Y/b, Z/c unifies the expressions
p(a,Y,Z) and p(X,b,Z) and p(a,Y,c)
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Unifiers are not unique!

• Two expressions can have more that one unifier
  that makes them equivalent
• S1 p(X,Y,Y) and S2 p(a,Z,Z)
• are unified by
• U1X/a,Y/Z and U2X/a,Y/b, Z/b
• S1 U1 S2 U1 p(a,Z,Z) S1 U2
  S2 U2 p(a,b,b)

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Partial order between unifiers

• Some unifiers are more general than others
• U1 is more general than U2 if there
  exists a unifier U3 such that U1U3 U2
• Example
• U1 X/f(Y),Z/W,R/c is more general that
• U2 X/f(a),Z/b, R/c since U3 Y/a,W/b
• U1 U3 X/f(Y),Z/W,R/cY/a,W/b U2
• Unifiers form a partial order

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Most general unifier (MGU)

• For each set of sentences, there exists a most
  general unifier (mgu) that is unique up to
  variable renaming
• Ex for S1 p(X,Y,f(Z)) and S2 p(a,W,R)
  UX/a,Y/W,R/f(Z) is the mgu
• For the resolution procedure, we want to find the
  most general unifier of literals a constant,
  variable, or a functional object of length n.

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Basic functions for MGU procedure

• Constant(exp) returns true if exp is a
  constant Ex a, f(g(a,b,c)) are constants
• Variable(exp) returns true if exp is a simple
  variable
• Length(exp) returns the number of items in a
  function Ex f(a,g(Y)) is an object of
  length 2.
• MguVar(Var, exp) returns
• false if Var is included in exp
• Var/exp otherwise

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Recursive procedure for MGU
function Mgu(exp1,exp2) returns unifier if
exp1 exp2 return if Variable(exp1) then
return MguVar(exp1,exp2) if Variable(exp2)
then return MguVar(exp2,exp1) if
Constant(exp1) or Constant(exp1) return false
if not(Length(exp1) Length(exp2) return false
unifier for i 1 to Length(exp1) do
s Mgu(Part(exp1,i),Part(exp2,i)) if s
is false return false unifier
Compose(unifier,s) exp1 Substitute(exp1,unifie
r) exp2 Substitute(exp2,unifier) return
unifier end
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Example of MGU trace (1)

• Find the mgu unifier for
• p(f(X,g(a,Y)), g(a,Y)) and p(f(X,Z),Z).
• 1. Mgu is called on p and p, returns
• 2. Mgu is called recursively on f(X, g(a,Y))
  and f(X,Z)
• 3. Mgu is called on f and f, returns
• 4. Mgu is called on X and X, returns
• 5. Mgu is called on g(a,Y) and Z since Z is a
  variable, it returns (via Mguvar) Z/g(a,Y)
  after checking that Z is not in g(a,Y).

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Example of MGU trace (2)

• 6. Z/g(a,Y) is composed with the previous
  (empty) substitution
• 7. The entire substitution is applied to both
  expressions, yielding f(X, g(a,Y))
• 8. Since i 3, Mgu returns Z/g(a,Y), which is
  then applied to the top level expressions. All
  other checks show equality. The result is
• p(f(X,g(a,Y)),g(a,Y))

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Performance Issues

• Predicate Indexing Storing the answers of some
  queries
• Knows(John, x) store all x (in a hash table)
  which John knows
• Using subsumption lattice to be more efficient

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Subsumption Lattice
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3. Resolution

• Given a clause containing the literal j and
  another clause the literal j , we can infer the
  clause consisting of all the literals of both
  clauses without j and j.

Ex 1. p, q KB 2. q, r KB 3. p, r 1,2
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Why does this work?

• Consider the two clauses
• winter, summer
• winter, cold
• At any point, winter is either true or false. If
  it is true, then cold must be true to guarantee
  the truth of clause 2. If winter is false, then
  summer must be true to guarantee the truth of
  clause 1. So regardless of winter's truth value,
  either cold or summer must be true, i.e., we can
  conclude
• cold, summer

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Other Resolution Examples
1. p, q KB 2. p, q KB 3. q 1,2
merge the q's
1. p, q KB 2. p KB 3. q 1,2
much like Modus Ponens
We can derive the empty clause, showing that the
KB has a contradiction

• 1. p KB
• 2. p KB
• 3. 1,2

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Resolution Rule

• For clauses containing variables, we can resolve
  j in one clause with y in another clause, as
  long as j and y have a mgu U. The resulting
  clause is the union of the original 2 clauses,
  with j and y removed, and with U applied to the
  remaining literals.
• r,jand y,d
• rU,dU
• where j U y U

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Examples of Resolutions

• Ex1 1. p(X), q(X,Y) KB
• 2. p(a), r(b,Z) KB
• 3. q(a,Y), r(b,Z) 1,2

• Ex2 Two clauses may resolve in more than one
  way since f and y may be chosen in different
  ways
• 1. p(X,X), q(X), r(X) KB
• 2. p(a,Z), q(b) KB
• 3. q(a), r(a),q(b) 1,2
• 4. p(b,b), r(b), p(a,Z) 1,2

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Resolution Deduction

• A resolution deduction of a clause j from a data
  base KB is a sequence of clauses in which
• 1. j is an element of the sequence
• 2. Each element is either a member of KB or the
  result of applying the resolution principle to
  clauses earlier in the sequence.

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Resolution procedure with nondeterministic choices
function Resolution(KB) returns answer
while not(Empty_Clause(KB)) do c1
Choose_Clause(KB) c2 Choose_Clause(KB)
res Choose_Resolvents(c1,c2) KB KB
U res end return true end
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Ex Resolution deductionof the empty clause

• 1. p KB
• 2. p, q KB
• 3. q, r KB
• 4. r KB
• 5. q 1, 2
• 6. q 3, 4
• 7. 5, 6

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All possible resolutions (to 3 levels)
A resolution trace is a linear form of the graph
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Resolution Refutation

• As for Propositional Logic, we will use
  refutation resolution as the single rule for
  proving sentences about a KB
• We will use it to prove the unsatisfiability of a
  set of clauses, i.e., they contain a
  contradiction if we can derive the empty clause.
  
• Resolution refutation to prove c, prove
  that KB U c 0 like a proof by
  contradiction
• Rule is refutation-complete.

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Answering questions with Refutation-Resolution

• True/False questions we want to know if a
  conclusion follows from KB
• Ex father(art, jon)
• father(bob, kim)
• father(X, Y) gt? parent(X, Y) Is art the
  parent of jon?
• Queries fill-in-the blank we want to know also
  what instantiation makes it true Who is
  the parent of jon?

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True/False by Refutation Resolution
1. father(art, jon) KB 2. father(bob,
kim) KB 3. father(X,Y), parent(X,Y) KB 4.
parent(art,jon) 5. parent(art,jon) 1, 3 6.
parent(bob, kim) 2, 3 7. father(art, jon) 3,
4 8. 4, 5 9. 1, 7
negated goal clause
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Fill-in-the-blank (Greens method)

• To obtain one instantiation that makes the
  conclusion true (if any), form a disjunction of
  the negation of the goal c and its answer
  literal a term of form
• ans(X1,X2,Xn)
• where the variables X1,X2,Xn are the free
  variables in c. Ex Add to the previous KB
  the clause parent(X, jon), ans(X)
• Resolution halts when it derives a clause
  containing only the ans literal

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Example of fill-the-blank derivation

• 1. father(art, jon) KB
• 2. father(bob, kim) KB
• 3. father(X,Y), parent(X,Y) KB
• 4. parent(Z,jon), ans(Z) c
• 5. parent(art, jon) 1, 3
• 6. parent(bob, kim) 2, 3
• 7. father(w,jon), ans(W) 3, 4
• 8. ans(art) 4, 5
• 9. ans(art) 1, 7

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Properties of Fill-in-the-blank

• The answer may not be unique several different
  answers may result from the proof, depending on
  the clauses that were chosen for resolution.
• We may also get an answer of the form ans(a),
  ans(b) where one of the answers is right, but
  the clause doesnt tell us which one.
• 1. father(art,jon), father(bob,jon) KB
  2. father(X,jon),ans(X) c 3.
  father(bob,jon),ans(art) 1, 2 4.
  ans(art),ans(bob) 2, 3

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Forward Chaining
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Example of Forward Chaining
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Completeness of FC

• k number of predicate argument
• p number of predicates
• n number of constant symbols
• pnk distinct ground facts
• Propositional logic with above facts

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Efficient FC

• Problems of simple FC
• Inner loop involve pattern matching
• Algorithm rechecks every rule in each loop
• Might generate many irrelevant facts
• Should be addressed these problems

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Pattern Matching and CSP

• Every finite-domain CSP can be expressed as a
  single definite clause together with some
  associated ground facts.

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Incremental FC

• Every new fact inferred on iteration t must be
  derived from at least one new fact inferred on
  iteration t - 1.
• Retain the partial matching of any rule inorder
  to use it next iterations

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Irrelevant facts

• 3 ways to avoid
• Use backward chaining
• Restrict subset of rules
• Use magic set , some information about the goal
  in order to ignore irrelevant facts

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Backward Chaining
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BC example
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Logic Programming

• Algorithm Logic Control
• Prolog Programming in Logic

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• ?