CSCE 580 Artificial Intelligence The Resolution Refutation Proof Technique for First-Order Logic

1
CSCE 580Artificial IntelligenceThe Resolution
Refutation Proof Technique for First-Order Logic

• Fall 2009
• Marco Valtorta
• mgv_at_cse.sc.edu

2
Acknowledgment

• The slides are based on the textbook AIMA and
  other sources, including other fine textbooks and
  the accompanying slide sets
• The other textbooks I considered are
• David Poole, Alan Mackworth, and Randy Goebel.
  Computational Intelligence A Logical Approach.
  Oxford, 1998
• A second edition (by Poole and Mackworth) is
  under development. Dr. Poole allowed us to use a
  draft of it in this course
• Ivan Bratko. Prolog Programming for Artificial
  Intelligence, Third Edition. Addison-Wesley,
  2001
• The fourth edition is under development
• George F. Luger. Artificial Intelligence
  Structures and Strategies for Complex Problem
  Solving, Sixth Edition. Addison-Welsey, 2009

3
Outline

• Reducing first-order inference to propositional
  inference
• Unification
• Generalized Modus Ponens
• Forward chaining
• Backward chaining
• Resolution

4
Universal instantiation (UI)

• Every instantiation of a universally quantified
  sentence is entailed by it
• ?v aSubst(v/g, a)
• for any variable v and ground term g
• E.g., ?x King(x) ? Greedy(x) ? Evil(x) yields
• King(John) ? Greedy(John) ? Evil(John)
• King(Richard) ? Greedy(Richard) ? Evil(Richard)
• King(Father(John)) ? Greedy(Father(John)) ?
  Evil(Father(John))
• .
• .
• .

5
Existential instantiation (EI)

• For any sentence a, variable v, and constant
  symbol k that does not appear elsewhere in the
  knowledge base
• ?v a
• Subst(v/k, a)
• E.g., ?x Crown(x) ? OnHead(x,John) yields
• Crown(C1) ? OnHead(C1,John)
• provided C1 is a new constant symbol, called a
  Skolem constant
• Logical equivalence is not preserved, because
  skolemization adds new constants to formulas
  however, the new KB is satisfiable iff the old
  one is satisfiable (s-equivalence)
• In general, skolemization adds Skolem function,
  as in
• ?x ?y Is_Father(y,x), which skolemizes to ?x
  Is_Father(Father(x),x)

6
Reduction to propositional inference

• Suppose the KB contains just the following
• ?x King(x) ? Greedy(x) ? Evil(x)
• King(John)
• Greedy(John)
• Brother(Richard,John)
• Instantiating the universal sentence in all
  possible ways, we have
• King(John) ? Greedy(John) ? Evil(John)
• King(Richard) ? Greedy(Richard) ? Evil(Richard)
• King(John)
• Greedy(John)
• Brother(Richard,John)
• The new KB is propositionalized proposition
  symbols are
• King(John), Greedy(John), Evil(John),
  King(Richard), etc.

7
Reduction ctd.

• Every FOL KB can be propositionalized so as to
  preserve entailment
• (A ground sentence is entailed by new KB iff
  entailed by original KB)
• Idea propositionalize KB and query, apply
  resolution, return result
• Problem with function symbols, there are
  infinitely many ground terms,
• e.g., Father(Father(Father(John)))

8
Reduction ctd.

• Theorem Herbrand (1930). If a sentence a is
  entailed by an FOL KB, it is entailed by a finite
  subset of the propositionalized KB
• Note Herbrand showed that no new constants have
  to be introduced (so, in the example, the only
  constants needed are John and Richard), except
  for one in case the KB contains no constants, in
  which case one constant must be introduced
• Idea For n 0 to 8 do
• create a propositional KB by instantiating
  with depth-n terms
• see if a is entailed by this KB
• Problem works if a is entailed, may loop forever
  if a is not entailed
• Theorem Turing (1936), Church (1936) Entailment
  for FOL is semidecidable (algorithms exist that
  say yes to every entailed sentence, but no
  algorithm exists that also says no to every
  non-entailed sentence.)

9
Problems with propositionalization

• Propositionalization seems to generate lots of
  irrelevant sentences.
• E.g., from
• ?x King(x) ? Greedy(x) ? Evil(x)
• King(John)
• ?y Greedy(y)
• Brother(Richard,John)
• it seems obvious that Evil(John), but
  propositionalization produces lots of facts such
  as Greedy(Richard) that are irrelevant
• With p k-ary predicates and n constants, there
  are pnk instantiations.

10
Unification

• We can get the inference immediately if we can
  find a substitution ? such that King(x) and
  Greedy(x) match King(John) and Greedy(y)
• ? x/John,y/John works
• Unify(a,ß) ? if a? ß?
• p q ?
• Knows(John,x) Knows(John,Jane)
• Knows(John,x) Knows(y,OJ)
• Knows(John,x) Knows(y,Mother(y))
• Knows(John,x) Knows(x,OJ)
• Standardizing apart eliminates overlap of
  variables, e.g., Knows(z17,OJ)

11
Unification

• We can get the inference immediately if we can
  find a substitution ? such that King(x) and
  Greedy(x) match King(John) and Greedy(y)
• ? x/John,y/John works
• Unify(a,ß) ? if a? ß?
• p q ?
• Knows(John,x) Knows(John,Jane) x/Jane
• Knows(John,x) Knows(y,OJ)
• Knows(John,x) Knows(y,Mother(y))
• Knows(John,x) Knows(x,OJ)
• Standardizing apart eliminates overlap of
  variables, e.g., Knows(z17,OJ)

12
Unification

• We can get the inference immediately if we can
  find a substitution ? such that King(x) and
  Greedy(x) match King(John) and Greedy(y)
• ? x/John,y/John works
• Unify(a,ß) ? if a? ß?
• p q ?
• Knows(John,x) Knows(John,Jane) x/Jane
• Knows(John,x) Knows(y,OJ) x/OJ,y/John
• Knows(John,x) Knows(y,Mother(y))
• Knows(John,x) Knows(x,OJ)
• Standardizing apart eliminates overlap of
  variables, e.g., Knows(z17,OJ)

13
Unification

• We can get the inference immediately if we can
  find a substitution ? such that King(x) and
  Greedy(x) match King(John) and Greedy(y)
• ? x/John,y/John works
• Unify(a,ß) ? if a? ß?
• p q ?
• Knows(John,x) Knows(John,Jane) x/Jane
• Knows(John,x) Knows(y,OJ) x/OJ,y/John
• Knows(John,x) Knows(y,Mother(y)) y/John,x/Mothe
  r(John)
• Knows(John,x) Knows(x,OJ)
• Standardizing apart eliminates overlap of
  variables, e.g., Knows(z17,OJ)

14
Unification

• We can get the inference immediately if we can
  find a substitution ? such that King(x) and
  Greedy(x) match King(John) and Greedy(y)
• ? x/John,y/John works
• Unify(a,ß) ? if a? ß?
• p q ?
• Knows(John,x) Knows(John,Jane) x/Jane
• Knows(John,x) Knows(y,OJ) x/OJ,y/John
• Knows(John,x) Knows(y,Mother(y)) y/John,x/Mothe
  r(John)
• Knows(John,x) Knows(x,OJ) fail
• Standardizing apart eliminates overlap of
  variables, e.g., Knows(z17,OJ)

15
Unification

• To unify Knows(John,x) and Knows(y,z),
• ? y/John, x/z or ? y/John, x/John,
  z/John
• The first unifier is more general than the
  second.
• There is a single most general unifier (MGU) that
  is unique up to renaming of variables.
• MGU y/John, x/z

16
The unification algorithm
17
The unification algorithm
Occurs check when unifying a variable x with a
term T, check whether x occurs in T. Note that
this should be done after the already found
substitutions are applied. For example, unifying
t(x,f(x)) and t(g(y),y) occur checks because
after x\g(y) is applied, one obtains
t(g(y),f(g(y))) and t(g(y),y). Unifying f(g(y))
with y leads to the attempted substitution
y\g(y), which triggers the occurs check, leading
to failure.
18
Generalized Modus Ponens (GMP)

• p1', p2', , pn', ( p1 ? p2 ? ? pn ?q)
• q?
• E.g.,
• p1' is King(John) p1 is King(x)
• p2' is Greedy(y) p2 is Greedy(x)
• ? is x/John,y/John q is Evil(x)
• q ? is Evil(John)
• GMP used with KB of definite clauses (exactly one
  positive literal)
• All variables assumed universally quantified

where pi'? pi ? for all i
19
Soundness of GMP

• Need to show that
• p1', , pn', (p1 ? ? pn ? q) q?
• provided that pi'? pi? for all I
• Lemma For any sentence p, we have p p? by UI
• (p1 ? ? pn ? q) (p1 ? ? pn ? q)? (p1? ?
  ? pn? ? q?)
• p1', , pn' p1' ? ? pn' p1'? ? ? pn'?
• From 1 and 2, q? follows by ordinary Modus Ponens

20
Example knowledge base

• The law says that it is a crime for an American
  to sell weapons to hostile nations. The country
  Nono, an enemy of America, has some missiles, and
  all of its missiles were sold to it by Colonel
  West, who is American.
• Prove that Col. West is a criminal

21
Example knowledge base ctd.

• ... it is a crime for an American to sell weapons
  to hostile nations
• American(x) ? Weapon(y) ? Sells(x,y,z) ?
  Hostile(z) ? Criminal(x)
• Nono has some missiles, i.e., ?x Owns(Nono,x) ?
  Missile(x)
• Owns(Nono,M1) and Missile(M1)
• all of its missiles were sold to it by Colonel
  West
• Missile(x) ? Owns(Nono,x) ? Sells(West,x,Nono)
• Missiles are weapons
• Missile(x) ? Weapon(x)
• An enemy of America counts as "hostile
• Enemy(x,America) ? Hostile(x)
• West, who is American
• American(West)
• The country Nono, an enemy of America
• Enemy(Nono,America)
• Note This definite clause KB has no functions.
  It is therefore a Datalog KB

22
Forward chaining algorithm
23
Forward chaining proof
24
Forward chaining proof
25
Forward chaining proof
26
Properties of forward chaining

• Sound and complete for first-order definite
  clauses
• Datalog first-order definite clauses no
  functions
• FC terminates for Datalog in finite number of
  iterations
• May not terminate in general if a is not entailed
• This is unavoidable entailment with definite
  clauses is semidecidable

27
Efficiency of forward chaining

• Incremental forward chaining no need to match a
  rule on iteration k if a premise wasn't added on
  iteration k-1
• ? match each rule whose premise contains a newly
  added positive literal
• The rete algorithm builds a dataflow network to
  allow for reuse of matchings it is used in
  production system languages such as OPS-5
• Matching itself can be expensive
• Database indexing allows O(1) retrieval of known
  facts
• e.g., query Missile(x) retrieves Missile(M1)
• Forward chaining is widely used in deductive
  databases

28
Hard matching example
Diff(wa,nt) ? Diff(wa,sa) ? Diff(nt,q) ?
Diff(nt,sa) ? Diff(q,nsw) ? Diff(q,sa) ?
Diff(nsw,v) ? Diff(nsw,sa) ? Diff(v,sa) ?
Colorable() Diff(Red,Blue) Diff (Red,Green)
Diff(Green,Red) Diff(Green,Blue) Diff(Blue,Red)
Diff(Blue,Green)

• Colorable() is inferred iff the CSP has a
  solution
• CSPs include 3SAT as a special case, hence
  matching is NP-hard

29
Backward chaining algorithm

• SUBST(COMPOSE(?1, ?2), p) SUBST(?2, SUBST(?1,
  p))

30
Backward chaining example
31
Backward chaining example
32
Backward chaining example
33
Backward chaining example
34
Backward chaining example
35
Backward chaining example
36
Backward chaining example
37
Backward chaining example
38
Properties of backward chaining

• Depth-first recursive proof search space is
  linear in size of proof
• Incomplete due to infinite loops
• ? fix by checking current goal against every goal
  on stack
• Inefficient due to repeated subgoals (both
  success and failure)
• ? fix using caching of previous results (extra
  space)
• Widely used for logic programming

39
Logic programming Prolog

• Algorithm Logic Control
• Basis backward chaining with Horn clauses
  bells whistles
• Widely used in Europe, Japan (basis of 5th
  Generation project)
• Compilation techniques ? 60 million LIPS
• Program set of clauses head - literal1,
  literaln.
• criminal(X) - american(X), weapon(Y),
  sells(X,Y,Z), hostile(Z).
• Depth-first, left-to-right backward chaining
• Built-in predicates for arithmetic etc., e.g., X
  is YZ3
• Built-in predicates that have side effects (e.g.,
  input and output predicates, assert/retract
  predicates)
• Closed-world assumption ("negation as failure")
• e.g., given alive(X) - not dead(X).
• alive(joe) succeeds if dead(joe) fails

40
Prolog

• Appending two lists to produce a third
• append(,Y,Y).
• append(XL,Y,XZ) - append(L,Y,Z).
• query ?-append(A,B,1,2)
• answers A B1,2
• A1 B2
• A1,2 B

41
Backward chaining is incomplete

• Goal trees are built by backward chaining
• C is true, as shown in the following (natural
  deduction) proof by cases (split on D)
• It is impossible to prove C by backward chaining
• Two additions make backward chaining complete
• addition of contrapositives (ex C -gt D)
• ancestor checking

42
PTTP A Prolog Technology Theorem Prover

• Prolog is not a full theorem prover for three
  main reasons
• It uses an unsound unification algorithm without
  the occurs check
• Its inference system is complete for Horn
  clauses, but not for more general formulas, as
  shown in the previous slide
• Its unbounded depth-first search strategy is
  incomplete. Also, it cannot display the proofs it
  finds
• The Prolog Technology Theorem Prover (PTTP)
  overcomes these limitations by
• transforming clauses so that head literals have
  no repeated variables and unification without the
  occurs check is valid remaining unification is
  done using complete unification with the occurs
  check in the body
• adding contrapositives of clauses (so that any
  literal, not just a distinguished head literal,
  can be resolved on) and the model- elimination
  procedure reduction rule that matches goals with
  the negations of their ancestor goals
• using a sequence of bounded depth-first searches
  to prove a theorem
• retaining information on what formulas are used
  for each inference so that the proof can be
  printed
• Proof of soundness and completeness follows from
  the soundness and completeness of resolution
  refutation

43
Resolution brief summary

• Full first-order version
• l1 ? ? lk, m1 ? ? mn
• (l1 ? ? li-1 ? li1 ? ? lk ? m1 ? ?
  mj-1 ? mj1 ? ? mn)?
• where Unify(li, ?mj) ?.
• The two clauses are assumed to be standardized
  apart so that they share no variables.
• For example,
• ?Rich(x) ? Unhappy(x)
• Rich(Ken)
• Unhappy(Ken)
• with ? x/Ken
• Apply resolution steps to CNF(KB ? ?a) complete
  for FOL
• Full disclosure For completeness, one needs
  either
• general (non-binary) resolution, in which subsets
  of literals that are unifiable are resolved, or
• Factoring, which replaces unifiable atoms
  literals in a clause with a single atom

44
Conversion to CNF

• Everyone who loves all animals is loved by
  someone
• ?x ?y Animal(y) ? Loves(x,y) ? ?y Loves(y,x)
• 1. Eliminate biconditionals and implications
• ?x ??y ?Animal(y) ? Loves(x,y) ? ?y
  Loves(y,x)
• 2. Move ? inwards ??x p ?x ?p, ? ?x p ?x ?p
• ?x ?y ?(?Animal(y) ? Loves(x,y)) ? ?y
  Loves(y,x)
• ?x ?y ??Animal(y) ? ?Loves(x,y) ? ?y
  Loves(y,x)
• ?x ?y Animal(y) ? ?Loves(x,y) ? ?y Loves(y,x)

45
Conversion to CNF contd.

• Standardize variables each quantifier should use
  a different one
• ?x ?y Animal(y) ? ?Loves(x,y) ? ?z Loves(z,x)
• Skolemize a more general form of existential
  instantiation.
• Each existential variable is replaced by a Skolem
  function of the enclosing universally quantified
  variables
• ?x Animal(F(x)) ? ?Loves(x,F(x)) ?
  Loves(G(x),x)
• Drop universal quantifiers
• Animal(F(x)) ? ?Loves(x,F(x)) ? Loves(G(x),x)
• Distribute ? over ?
• Animal(F(x)) ? Loves(G(x),x) ? ?Loves(x,F(x))
  ? Loves(G(x),x)

46
Completeness of Resolution

• Resolution is refutation-complete if S is an
  unsatisfiable set of clauses, then the
  application of a finite number of resolution
  steps to S will yield a contradiction

47
Resolution proof definite clauses

• This is an input resolution proof

48
Resolution proof general case

• Curiosity killed the cat pp.298-300 AIMA-2
• Not a unit resolution proof not an input
  resolution proof

49
Resolution Strategies

• Breadth-First Strategy (complete)
• Set-of-Support Strategy (complete)
• At least one parent of each resolvent is selected
  from among the clauses resulting from the
  negation of the goal or from their descendants
  (the set of support)
• Unit-Preference Strategy (complete)
• Unit Resolution (complete for Horn clauses
  forward chaining)
• Each resolvent has a parent that is a unit clause
• Linear-Input Form Strategy (not complete)
• Each resolvent has at least one parent belonging
  to the base set (i.e. the set of clauses given as
  input)
• Complete for Horn clauses
• Called input resolution in AIMA
• Ancestry-Filtered Form Strategy (complete)
• Each resolvent has a parent that is either in the
  base set or an ancestor of the other parent
• Called linear resolution in AIMA

50
Example KB Nilsson, 1980

• Whoever can read is literate
• (1) R(x) ? L(x)
• Dolphins are not literate
• (2) D(x) ? L(x)
• Some dolphins are intelligent
• (3a) D(A)
• (3b) I(A)
• Goal Some who are intelligent cannot read
• (4) I(z) ? R(z)

• Whoever can read is literate
• ?xR(x) ?L(x)
• Dolphins are not literate
• ?xD(x) ?L(x)
• Some dolphins are intelligent
• ?xD(x) ? I(x)
• Goal Some who are intelligent cannot read
• ?xI(x) ? R(x)

51
An example proof

• (5) R(A) resolvent of 3b and 4
• (6) L(A) resolvent of 5 and 1
• (7) D(A) resolvent of 6 and 2
• (8) NIL resolvent of 7 and 3a

52
Breadth-first strategy
53
Set-of-Support Strategy
54
Linear-Input Form Strategy
55
Ancestry-Filtered Form Strategy
56
Four more examples

• Prove by resolution the result of exercise 8.2
  AIMA-2
• The goal is a universal formula!
• Group theory axioms and some consequences of them
  Schoening, example on pp.94-95
• (a) every dragon is happy if all its children can
  fly (b) green dragons can fly (c) a dragon is
  green if it is a child of at least one green
  dragon show that all green dragons are happy
• (a) Every barber shaves all persons who do not
  shave themselves (b) no barber shaves any person
  who shaves himself or herself show that there
  are no barbers
• Shows the need for factoring or non-binary
  resolution (p.297 AIMA-2)