Resolution method

1
Lesson 9

• Resolution method
• (continuing, examples)

2
Verifying an argument

• (Anybody) who knows Paul and Mary is sorry for
  Mary. ?x ( Z(x, P) ? Z(x, M) ? L(x, M) )
• Some are not sorry for Mary, though they know
  her. ?x ?L(x, M) ? Z(x, M)
• Somebody knows Mary, but does not know Paul. ?x
  Z(x, M) ? ?Z(x, P)
• ?x ?Z(x, P) ? ?Z(x, M) ? L(x, M) implication
  elimination (1. assumption)
• ?L(a, M) ? Z(a, M) Scolemising (2.
  assumption)
• ?y ?Z(y, M) ? Z(y, P) negation of conclusion
  (and renaming x)
• Clauses
• 1. ?Z(x, P) ? ?Z(x, M) ? L(x, M)
• 2. ?L(a, M) (2. assumption is a
  conjunction
• 3. Z(a, M) two clauses arise!)
• 4. ?Z(y, M) ? Z(y, P)
• 5. ?Z(a, P) ? ?Z(a, M) resolution 1., 2.,
  substitution a/x
• 6. ?Z(a, P) resolution 3., 5.
• 7. ?Z(a, M) resolution 4., 6., substitution
  a/y
• 8. resolution 3., 7.
• We obtained an empty clause, i.e., the negated
  conclusion contradicts the premises thus the
  non-negated original conclusion logically follows
  from the premises.
• The argument is valid.

3
Remark Compare the proof of this argument with
the semantic proof (lesson 6, slides 22, 23)

• We demonstrate the truth-conditions of the
  premises first the predicates Z and L are
  interpreted as the relations ZU and LU,
  respectively
• ZU , ?i1,m?, ? i1,k?, ?i2,m?,
  ?i2,k?,,??,m?,
• 1. premise 2. premise
• LU , ?i1,m?, ...., ?i2,m?,.........,
  ??,m?,
• and so on, by the method of an indirect proof

4
Indirect proof of the validity of an argument

• We make use of the following equivalence valid
  for closed formulas
• P1,...,Pn Z iff (P1 ?...? Pn) ? Z
• Moreover (P1 ?...? Pn) ? Z iff the negated
  formula is a contradiction (P1 ?...? Pn ? ?Z)
• Hence, the argument is valid iff the negated
  conclusion Z contradicts the conjunction of
  premises. Indeed, compare with the definition of
  the validity of an argument
• P1,...,Pn Z iff Z is true in every model
  of the set of premises P1,...,Pn iff
• ?Z is true in no model of the set of premises.

5
Proofs of the validity

• Prove, that the following statement
• If there is a philosopher who disagrees with all
  the philosophers, then he disagrees with himself
  as well
• is logically true.
• We analyse this sentence in following
  way(intended interpretation is over the set
  of individuals, P ? subset of philosophers, Q ?
  relation the couples of individuals where the
  first member disagrees with the second)
• We prove the logical validity of the formula
• ?x P(x) ? ?y (P(y) ? Q(x,y)) ? Q(x,x)
• First, negate the formula and transform it into
  clausul form
• ?x ?y P(x) ? ?P(y) ? Q(x,y) ? ?Q(x,x).
• The set of clauses1. P(x) 2. ?P(y) ? Q(x,y)3.
  ?Q(x,x) the substitution y/x is the most
  general unifier
• 4. Q(x,x) resolution 1. and 2.5.
  resolution 3. and 4.

6
Proofs of the validity

• Prove the logical validity of following
  statement
• There is somebody such that if he\she is a
  genius, then everybody is a genius.
• Formula ?x G(x) ? ?x G(x) (mind the
  brackets!)
• We prove, that the negated formula is a
  contradiction
• ?x G(x) ? ?x ?G(x), renaming variables gives
• ?x G(x) ? ?y ?G(y), step 6 ?x G(x) ? ?y
  ?G(y) and by Scolemizing we have
• ?x G(x) ? ?G(a).
• G(x)
• ?G(a)
• resolution 1. and 2., substitution a/x
• We obtained an empty clause, i.e., the negated
  formula is a contradiction so the original
  formula is a tautology.

7

• Every member of the management is an obligation
  owner or a shareholder.
• No member of the management is an obligation
  owner and also a shareholder.
• Every obligation owner is a member of the
  management.
• 4. No obligation owner is a shareholder.
• ?x V(x) ? (O(x) ? A(x))
• ?x V(x) ? ?(O(x) ? A(x))
• ?x O(x) ? V(x)
• ?x O(x) ? ?A(x)
• Clauses 1. ?V(x) ? O(x) ? A(x) 1. premise
• 2. ?V(y) ? ?O(y) ? ?A(y) 2. premise
• 3. ?O(z) ? V(z) 3. premise
• 4. O(k) negated conclusion
• 5. A(k) (after Scolemization)
• 6. ?O(y) ? ?A(y) resolution 2., 3.,
  substitution z/y
• 7. ?A(k) resolution 4., 6., substitution
  y/k
• 8. resolution 5., 7.
• Note that we do not use the first clause in the
  proof sequence. So conclusion is a consequent of
  the second and the third premise (the first one
  is superfluous for infering the conclusion).

8

• Everybody who likes George will collaborate
  with Michael.
• Michael is a friend of nobody who is a friend of
  Luke.
• Peter will collaborate only with the friends of
  Charles.
• If Charles is a friend of Luke, then Peter does
  not like George.
• ?x L(x, G) ? C(x, M)
• ?x F(x, L) ? ?F(M, x)
• ?x C(P, x) ? F(x, Cr)
• F(Cr, L) ? ?L(P, G)
• Clauses
• 1. ?L(x, G) ? C(x, M) 1. premise
• 2. ?F(y, L) ? ?F(M, y) 2. premise
• 3. ?C(P, z) ? F(z, Cr) 3. premise
• 4. F(Cr, L) negated
• 5. L(P, G) conclusion
• 6. ?F(M, Cr) resolution 4., 2., substitution
  y/Cr
• 7. ?C(P, M) resolution 3., 6., substitution
  z/M
• 8. ?L(P, G) resolution 1., 7., substitution
  x/P

9
What are the logical consequences of premises?

• Variant of the preceding example. Premises do
  not contain existential statement.
• Everybody who likes George will collaborate
  with Michael.
• Michael is a friend of nobody who is a friend of
  Luke.
• Peter will collaborate only with the friends of
  Charles.
• Charles is a friend of Luke.
• ???
• Clauses
• 1. ?L(x, G) ? C(x, M) 1. premise
• 2. ?F(y, L) ? ?F(M, y) 2. premise
• 3. ?C(P, z) ? F(z, Cr) 3. premise
• 4. F(Cr, L) 4. premise
• 5. ?F(M, Cr) inference, resolution 2,4, y/Cr
• 6. ?L(P, G) ? F(M, Cr) inference, resolution
  1,3, x/P, z/M
• 7. ?L(P, G) inference, resolution 5,6
• 8. ?C(P, M) inference, resolution 3,5, z/M

10
Indirect proof of the validity of an argument

• All men like football and beer.
• Xaver likes only those who like football and
  beer.
• There is somebody who likes football but does not
  like beer.
• Anybody who is not a man is a woman. (we must
  not omit tacit premises)
• Some women are not liked by Xaver.
• ?x M(x) ? (R(x,f) ? R(x,p)) 1. premise
• ?x R(Xa,x) ? (R(x,f) ? R(x,p)) 2. premise
• ?x R(x,f) ? ?R(x,p) 3. premise
• ?x ?M(x) ? Z(x) 4. premise
• ? ?x Z(x) ? ?R(Xa,x) negated conclusion

11
Indirect proof of the argument

• Clauses
• 1. ?M(x) ? R(x,f) first
• 2. ?M(x) ? R(x,p) premise
• 3. ?R(Xa,y) ? R(y,f) second
• 4. ?R(Xa,y) ? R(y,p) premise
• 5. R(k,f) third premise
• 6. ?R(k,p) after Scolemization x/k
• 7. M(z) ? Z(z) 4. premise
• 8. ?Z(u) ? R(Xa,u) negated conclusion
• 9. ?R(Xa,k) resolution 4., 6. (y/k)
• 10. ?Z(k) resolution 8., 9. (u/k)
• 11. M(k) resolution 7., 10. (z/k)
• 12. R(k,p) resolution 2., 11. (x/k)
• 13. resolution 6., 12.
• We have found out again that the negated
  conclusion contradicts the premises, so that the
  argument is valid.
• Do you know which premises were not necessary for
  inferring the conclusion?

12
Proof by the resolution method

• If a number is even then its second power is
  even .
• If a number is even then its fourth power is
  even .
• ?y P(y) ? P?f(y)?
• ?x P(x) ? P(f(f(x))) negate it
• ?x P(x) ? ?P(f(f(x))) Scolemise
• P(a) ? ?P(f(f(a))) write down the clauses
• ?P(y) ? P?f(y)? premise
• P(a) negated
• ?P(f(f(a))) conclusion
• P(f(a)) resolution 1-2, substitution y/a
• P(f(f(a))) resolution 1-4, substitution y/f(a)
• resolution 3-5, contradiction
• The argument is valid.

13
Mathematical induction

• P(a), ?y P(y) ? P?f(y)? ? ?x P(x)
• Is the derived formula a logicall consequence of
  the premises? In other words, is the
  mathematical-induction rule correct, i.e., does
  it preserve truth?
• Clauses
• P(a)
• ?P(y) ? P(f(y))
• P(f(a)) inference 2.1., subst. y/a
• P(f(f(a))) inference 2.3., subst. y/f(a)
• P(f(f(f(a)))) inference 2.4., subst. y/f(f(a))
• and so on, ad infinitum.
• We cannot prove it by the indirect proof as well,
  because the negated conclusion is (after
  scolemising) ?P(b) and the terms b, a, f(a),
  f(f(a)), are not unifiable. We never prove that
  ?x P(x) the case of the actual infinity. We can
  only reach the potential infinity potentially
  we can go ad infinitum but not really.

14
Verifying the consistence

• There is a barber who shaves just all those who
  do not shave themselves.
• Does this barber shave himself?
• ?x ?y ?H(y,y) ? H(x,y) ? H(x,x)
• In Lesson 7 we have seen that the premise is a
  contradiction so such a barber does not exist.
  We prove it by the resolution method. We add the
  negated conclusion to the premise and write down
  the clauses.
• ?x ?y ?H(y,y) ? H(x,y) ? H(x,x)
• H(y,y) ? H(a,y) ? H(a,a) substitution y/a
• ?H(y,y) ? ?H(a,y) ? ?H(a,a) substitution y/a
• ?H(x,x)
• resolution 1., 2., substitution y/a
• We dont need the third clause (negated
  conclusion) in order to infer the contradiction.
  So the premise is contradictory.
• Such a barber does not exist.

15
Note

• Robinsons algorithm of the general resolution
  and unification operates also on particular
  literals of a clause. For instance, from the
  clauses
• 1. P(x,y) ? ?Q(a,f(y)) ? P(x,a) ? ?Q(y,x) ?
  P(f(a),a) ? ?Q(a,f(a))
• 2. ?P(f(z),z) ? ?P(v,a) ? ?P(f(a),a)
• we derive the resolvent ?Q(a,f(a)),
• after the unification x/f(a), y/a, z/a,
  v/f(a),
• because the first clause is unified into
  P(f(a),a) ? ?Q(a,f(a)) and the second into
  ?P(f(a),a).

16
Note Pay attention to equivalences

• Formula A ? B is not contradictory, it is
  satisfiable
• Formula A ? ?A is contradictory
• Clauses ad a)
• ?A ? B
• A ? ?B
• now in each step we can generate the resolvent
  by the resolution of only one pair of literals,
  so no empty clause is derived
• 3. B ? ?B
• 4. A ? ?A
• Clauses ad b)
• ?A ? ?A ? ?A
• A ? A ? A
• 1. and 2. resolve into the empty clause

17
Checking the consistence

• Mr X changed his job and got a more qualified one
  (K).
• Mr X knows the wage-payment system very well (M).
• If Mr X changed his job for the more qualified
  one, then it is right to discuss his case (P).
• If is it right to discuss his case, then he
  should not be a member of the commission (C).
• If he knows the wage-payment system very well,
  then he should be a member of the commission.
• What are the logical consequences?
• We check the consistence of the set of statements
  first.
• K ? M ? (K ? P) ? (P ? ?C) ? (M ? C)
• K
• M
• ?K ? P
• ?P ? ?C
• ?M ? C
• P resolution 1, 3
• ?C resolution 4, 6
• C resolution 2, 5
• contradiction resolution 7, 8
• The given set of statements is contradictory so
  anything follows.

18
Checking the validness of a formula

• ? ?x P(x) ? ?y ?x (?Q(x,y) ? ?z R(a,x,y)) ?
  ?z (P(b) ? Q(f(z), z)) ? R(a, f(b), z)
• We transform the antecedent to the clause form
• By the method of Skolem algorithm
• We transform it to the prenex form at first and
  then we scolemise it.

19
Checking the validness of a formula vs step 6

• Antecedent transformation to the clause form,
  procedure a)
• ?x P(x) ? ?y ?x (?Q(x,y) ? ?z R(a,x,y))
• Steps 2., 5. Renaming variable x and eliminating
  ?z
• ?x P(x) ? ?y ?u (?Q(u,y) ? R(a,u,y))
• Step 3. Elimination of the implication
• ?x P(x) ? ?y ?u (Q(u,y) ? R(a,u,y))
• Step 6. Moving the quantifier ?x to the left
• ?x P(x) ? ?y ?u (Q(u,y) ? R(a,u,y))
• Step 7. Skolemization (i.e., substituting g(y)
  for the varible u the symbol f is used in the
  conclusion, so we have to use a new one)
• ?x P(x) ? ?y (Q(g(y), y) ? R(a, g(y), y))
• Step 8. moving the quantifier to the left
• SK1 ?x ?y P(x) ? (Q(g(y), y) ? R(a, g(y),
  y)).

20
Checking the validness of a formula vs step 6

• Antecedent transformation to the clause form,
• procedure b) without step 6
• ?x P(x) ? ?y ?x (?Q(x,y) ? ?z R(a,x,y))
• Steps 2,5 Renaming variable x and eliminating
  ?z
• ?x P(x) ? ?y ?u (?Q(u,y) ? R(a,u,y))
• Step 3. Elimination of the implication
• ?x P(x) ? ?y ?u (Q(u,y) ? R(a,u,y))
• Step 7. Skolemization (i.e., substituting g(x,y),
  for the varible u, because the variable u is at
  the scope of ?x)
• ?x P(x) ? ?y (Q(g(x,y), y) ? R(a, g(x,y), y))
• Step 8. removing the quantifier to the left
• SK2 ?x ?y P(x) ? (Q(g(x,y), y) ? R(a, g(x,y),
  y)).

21
Checking the validness of a formula vs step 6

• The negated conclusion (is in the clausal form
  already)
• ?z (?P(b) ? ?Q(f(z), z)) ? ?R(a, f(b), z).
• We attempt to prove a contradiction with SK1
  (clausal form of the antecedent)
• We write the clauses down, procedure a)
• P(x) premise
• Q(g(y), y) ? R(a, g(y), y) premise
• ?P(b) ? ?Q(f(z), z) negated conclusion
• ?R(a, f(b), z) negated conclusion
• ?Q(f(z), z) resolution 1., 3., substitution x /
  b
• ???
• We cannot derive any more resolvents, because
  terms g(y) a f(z), or g(y) and f(b) are not
  unifiable.
• The Formula is not a tautology.

22
Checking the validness of a formula vs step 6

• We are not able to prove the contradiction with
  SK2 (procedure b) with the negated conclusion
  either
• P(x)
• Q(g(x,y), y) ? R(a, g(x,y), y)
• ?P(b) ? ?Q(f(z), z)
• ?R(a, f(b), z)
• ?Q(f(z), z) resolution 1., 3., substitution x / b
• However, we cant procede further, because the
  terms g(x,y) and f(z), possibly g(x,y) and f(b),
  are not unifiable.
• Conclusion The Formula is not a tautology.

23
Exercise, a puzzle

• Tom, Peter and John are members of a sport club.
  Every member of the club is a skier or a climber.
  No climber likes raining. All skiers like snow.
  Peter does not like what Tom likes, and does like
  what Tom does not like. Tom likes snow and
  raining.
• Question Is there in the club a sportsman who
  is a climber but not a skier?
• Solution Knowledge base ( query 11)
• SC(t)
• SC(p)
• SC(j)
• ?x SC(x) ? (SKI(x) ? CLIMB(x))
• ?x CLIMB(x) ? ?LIKE(x,r)
• ?x SKI(x) ? LIKE(x,s)
• ?x LIKE(t,x) ? ?LIKE(p,x)
• ?x ?LIKE(t,x) ? LIKE(p,x)
• LIKE(t,s)
• LIKE(t,r)
• ? ?x SC(x) ? CLIMB(x) ? ?SKI(x)

24
Exercise, a puzzle

• Knowledge base (in Clausal form negated query
  11)
• SC(t) sport-club(tom).
• SC(p) sport-club(peter).
• SC(j) sport-club(john).
• ?SC(y) ? SKI(y) ? CLIMB(y) each club member is
  a skier or a climber
• ?CLIMB(z) ? ?LIKE(z,r) each climber does not
  like raining
• ?SKI(v) ? LIKE(v,s) each skier does not like
  snowing
• ?LIKE(t,x1) ? ?LIKE(p,x1) Tom and Peter have
  opposite
• LIKE(t,x2) ? LIKE(p,x2) tastes
• LIKE(t,s) like(tom, snow).
• LIKE(t,r) like(tom, raining).
• ?SC(x) ? ?CLIMB(x) ? SKI(x) Query
• Proof by resolution that 1-11 is inconsistent. In
  other words, we are looking for an instantiation
  of the variable x, that leads to a contradiction.
• ?LIKE(p,s) res. 9, 7 by substituting s for x1
• ?SKI(p) res. 12, 6 by substituting p for v
• ?SC(p) ? CLIMB(p) res. 13, 4 by substituting p
  for y
• CLIMB(p) res. 14, 2
• Resolution 11 2 13 15 by substituting
  p for x.
• (Obviously, the solution is p Peter)

25
Checking the validness of an argument

• Situation A couple of friends met a priest A,
  who said The first man whom I confessed was a
  murder. After a while Mr B came, looked at the
  priest and said I was the first man whom the
  priest A confessed.
• Question Did the priest offend against the
  confessional secrecy?
• Reení ?x (x f(A)) ? V(x)
• B f(A)
• ???
• (Intended interpretation Let f be a function
  that assigns to the priest the only man confessed
  by the priest as the first one)
• Clauses 1. ?(x f(A)) ? M(x) 1. premise
• 2. B f(A) 2. premise
• 3. M(B) inference resolution (B is a
  murder) 1.,2., substitution x/B
• Solution The priest offended against the
  confessional secrecy.

26
Prolog foundations (the logic)

• Method of (pure) logic programming is a special
  case of the general resolution method. Compared
  to the general resolution method, it has the
  following restrictions
• it works just with Horns clauses (they have one
  positive literal at most),
• it uses the linear strategy of generating
  resolvents together with the process of
  backtracking

27
Prolog foundations (the logic)

• In logic programming we use the following
  terminology
• Notation P- Q1, Q2,..,Qn. (is equivalent to
  ?Q1 ? ?Q2 ? ? ?Qn ? P,
• or to (Q1 ? Q2 ? ? Qn) ? P ) conditional
  statement (rule)
• Notation P. unconditional statement (fact)
• Notation?- Q1, Q2,..,Qn.(is equivalent to ?Q1 ?
  ?Q2 ? ? ?Qn) targets /target clauses,
  queries/
• Notation, YES contradiction /an empty clause /
  

28
Prolog foundations (the logic)

• Logic program is a sequence of conditional
  statements (rules) and unconditional statements
  (facts). Target clause set queries that the
  program tries to reply.
• Note Commands are understood as being declared
  by stating the rules and facts. So logic programs
  are declarative (not imperative). We only specify
  what is to be done and do not determine in
  which way to do it.

29
Prolog foundations (the logic)

• Example
• All students are younger than Peters mother.
  Charles and Mary are students. Who is younger
  than Peters mother?
• Notation in PL1 ?x St(x) ? Ml?x,matka(p?),
  St(k), St(m) ?
• Program in Prolog (Notice variables - capitals,
  constants - small letters)
• younger(X, mother(peter))- student(X). rule
• student(charles). fact
• student(mary). fact
• ?- younger(Y, mother(peter)). Query, target
  goal
• The interpreter performs unification and
  resolution following the linear strategy driven
  by target
• Target ?- younger(Y, mother(peter)) is unificated
  with younger (X, mother(peter)), YX
• New target is generated ?- student(Y)
• This target is unificated with the 2. fact in the
  database success with Ycharles
• Itll figure out the reply YES, Y charles
• We can enter semicolon which means that we
  are asking Who next?. It envokes backtracking,
  i.e., the process of returning to the last target
  and attempting to execute it again ?-
  student(Y). Now the interpreter cannot use the
  2. clause again (it remembers the place that has
  been used already) but it can use 3. clause
• e) Itll figure out the reply YES, Y Mary
• NO

30
Prolog foundations (the logic)

• As stated above, logic programs are declarative
  (not imperative). We specify what is to be done
  and do not determine in which way it is to be
  done.
• The interpreter finds the way how to reply the
  queries, i.e., it deduces which are the logical
  consequences of the specified knowledge base and
  which values have to be substitutiated for
  variables by unification (answers).
• However, the restriction to Horn clauses can
  cause troubles. See the example of a puzzle
  (slides 19, 20)
• Try to formulate this puzzle in Prolog!
• Summary of the restrictions
• one positive literal at most (no disjunction in
  the consequent),
• we are not able to specify negative facts
  directly.
• Negation as failure of inferring!