AR: clausal logic

1
AR clausal logic

• The step to Resolution

2
A deeper study
3
Representation-powerof Horn clauses

• Most predicate logic formulae can easily be
  rewritten in Horn clauses.

• Examples

?x cat(x) ? dog(x) ? pet(x)
?x poodle(x) ? dog(x) ? small(x)

• BUT

?x human(x) ? male(x) ? female(x)
?x dog(x) ? abnormal(x) ? has_4_legs(x)
4
Clausal form

• Horn clauses are a special case with m 1
• We assume that S T ? F consists only of
  clausal formulae.
• Goal prove that S is inconsistent.

5
Example Moores problem

• Given 3 blocks

• Prove that there is a blue block next to a red
  block!

6
Moores problem (2)

• PROBLEM Modus ponens is not suitable for
  blue(Block2) ? red(Block2)

7
Introducing negation in bodies is equivalent

• Example

8
Disjunction versus negation in general
?x1 ?xk A1 ? A2 ? Am ? B1 ? B2 ? Bn
9
The resolution principle

• Propositional case

• Correctness clear due to
• make from all other disjuncts negated body
  atoms,
• apply generalized modus ponens,
• move all these negated body atoms back as
  disjuncts in the head.

10
Resolution predicate logic

• where ? mgu(B,B).
• Correctness
• with a help of the correctness result for the
  ground case, applied to all instances of this
  rule

11
Also in other forms

• In conjunctive normal form
• with ? mgu(B,B).

A1 ? A2 ? ? Am ? B1 ? ? B ? ? Bn C1 ?
... ? B ? ? Ck ? D1 ? D2 ? ? Dl
(A1 ? A2 ? ? Am ? C1 ? .. .. ? Ck) ? ?
(B1 ? .. .. ? Bn ? D1 ? D2 ? ? Dl) ?
12
Once again Moores example
false ? next_to(b1,b2) ? blue(b1) ? red(b2)
13
Ph.D. example
14
Factoring in general
15
Why do we need factoring?

• Without factoring resolution is not complete !

• Example prove (p ? p) ? (q ? q) inconsistent

• Normalization

(p ? q) ? (p ? q) ? (p ? q) ? (p ? q)

• Clausal form

You can never get false ? !!!!
16
Reason?

• The length of a formula the number of atoms
  (false not included)

N M
N M - 2

• In the previous example all formulas had length 2

17
The resolution procedure
S initial theory (inconsistency to be
shown) Consistent false Inconsistent false
If false ? ? S Then Inconsistent true
If S contains no pair (F,G) resolvable and not
yet resolved
Then Consistent true
H factor( resolvent (F,G) )
S S ? H
End-while
18
Behavior under Horn clause resolution
Linear resolution !
false ?
19
Behavior under General resolution
General resolution !
20
Linear resolution

• The most important differences with Horn clauses

• we start with the goal
• we apply a Horn clause to compute a new goal
• etc.
• Clausal resolution is NOT linear
• Also factoring is sometimes needed

• Linear resolution (a proof is a linear sequence
  of resolution steps starting with a goal) is one
  of the most important strategies to make the
  resolution process efficient.

21
Non-determinism in the resolution procedure

• SELECT a pair (F,G) makes it a VERY
  non-deterministic procedure.

• The control problem for resolution is extremely
  difficult.

• A proof is no longer 1 (linear) branch in a tree,
  but a subgraph of all possible resolutions.

• Is it correct? Is it complete? ?

22
Correctness / Completeness ?

• Completeness There exists a complete strategy
  (standard example the Herbrand theorem prover).

• Correctness
• If the procedure returns Inconsistent

• Then false ? is added
• Then false ? is logically entailed by S
  (since the resolution step is correct).
• Thus, in all models of S false ? is also true
• Thus, S has no models

23
Correctness/Completeness 2 ?

• If the procedure returns Consistent

• Then ALL POSSIBLE resolution steps were done
  without discovering false ? .

• Now assume that the set was inconsistent.
• There exists a COMPLETE strategy that after
  some time derives false ?

• But it performs (a part of) the same resolution
  steps !!

24
AR for full predicate logic

• Normalization to clausal form

25
What else is needed for full predicate logic?

• Clausal logic is equivalent to full predicate
  logic

• every theory T in FOL (first order predicate
  logic) can automatically be converted in a
  clausal theory T, such that

T is inconsistent iff T is inconsistent
26
Propositional via conjunctive normal form

• Every formula is equivalent to a formula of the
  form

(A1 ? ... ? An) ? (B1 ? ? Bm) ? ? (C1 ? ?
Ck)

• where all Ai, Bi, , Ci are either atomic or
  atomic.

27
Predicate case main steps

• These 2 steps are interleaved.

28
Predicate case continued

• ?x rich(x) is replaced by rich(Sk) , with Sk
  being a new constant (skolem constant) that
  does not appear in the alphabet.
• More complex if ? appears nested inside ?

29
Predicate case continued

• Disjunctions

?x ?y ?z (p(x) ? q(y) ? r(y)) ?
(r(A) ? q(z)) ? s(x,y)

• Clausal form

30
Explicit Procedure
1. Eliminate ? en ? .
2. Move the negations inside (p) ? p, (p ?
q) ? p ? q, (analogously for ?) ?x ? ?x ,
?x ? ?x
3. Standardize variable names (make them
different).
8. Drop ?.
31
Marcus example

• Facts 1. , 2. , 4. and 8. were already o.k.
• ex. ruler(Caesar)

• 3. ?x Pompeian(x) ? Roman(x) o.k. !
• 6. ?x ?y loyal_to(x,y)

• 7. ?x?y man(x) ? ruler(y) ? try_assassinate(x,y)
  ? loyal_to(x,y)

?x?y (man(x) ? ruler(y) ? try_assassinate(x,y))
? loyal_to(x,y) ?x?y man(x) ? ruler(y) ?
try_assassinate(x,y) ? loyal_to(x,y) false ?
man(x) ? ruler(y) ? try_assassinate(x,y) ?
loyal_to(x,y)
32
Axioms in Normal form

• 1. man(Marcus)
• 2. Pompeian(Marcus)
• 3. Roman(x) ? Pompeian(x)
• 4. ruler(Caesar)
• 5. loyal_to(x,Caesar) ? hates(x,Caesar) ?
  Roman(x)
• 6. loyal_to(x,f(x))
• 7. false ? man(x) ? ruler(y) ? try_assassinate(x,y
  ) ? loyal_to(x,y)
• 8. try_assassinate(Marcus,Caesar)

To show hates(Marcus,Caesar) Negation
hates(Marcus,Caesar) Normal Form false ?
hates(Marcus,Caesar)
33
Resolution proof (1)
false ? hates(Marcus,Caesar)
34
Resolution proof (2)
loyal_to(Marcus,Caesar)
35
Example from Group Theory

• Let ? be a group operation.

• Prefix notation p(x,y,z) ? x ? y z

• Definition of a monoid, with left-neutral and
  left inverse element

• ? is defined for all elements of the set

?x?y?z p(x,y,z)
(1)

• ? is associative (x ? y) ? z x ? (y ? z)

?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ? (p(u,z,w) ?
p(x,v,w))
(2)
36
Example from Group Theory (2)

• ? has a left neutral and a left inverse element

?x (?y p(x,y,y) ? ?y?z p(z,y,x)
(3)

• Theorem there exists also a right inverse !

?x (?y p(x,y,y) ? ?y?z p(y,z,x)
(4)

• To be proved automatically by resolution.

37
Normalization
(1) ?x?y?z p(x,y,z)

• Steps 1,2,3,4 o.k.
• Step 5 skolemization

• Steps 6,7 o.k.
• Step 8 clausal form

p(x,y,m(x,y))
38
Normalization (continued)
(2) ?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
(p(u,z,w) ? p(x,v,w))

• Step 1 eliminate ? and ?

?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
((p(u,z,w) ? p(x,v,w)) ? (p(x,v,w) ? p(u,z,w)))
?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
((p(u,z,w) ? p(x,v,w)) ? (p(x,v,w) ? p(u,z,w)))
?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
((p(u,z,w) ? p(x,v,w)) ? (p(x,v,w) ? p(u,z,w)))
39
Normalization (continued)
?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
((p(u,z,w) ? p(x,v,w)) ? (p(x,v,w) ? p(u,z,w)))

• Step 2 move negation inside

?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v)) ?
((p(u,z,w) ? p(x,v,w)) ? (p(x,v,w) ? p(u,z,w)))

• Steps 3,4,5 o.k.
• Step 6 move disjunctions inside

A ? (B ? C) (A ? B) ? (A ? C)
?x?y?z?u?v?w ((p(x,y,u) ? p(y,z,v)) ?
(p(u,z,w) ? p(x,v,w))) ? ((p(x,y,u) ?
p(y,z,v)) ? (p(x,v,w) ? p(u,z,w)))
40
Normalization (continued)
?x?y?z?u?v?w ((p(x,y,u) ? p(y,z,v)) ?
(p(u,z,w) ? p(x,v,w))) ? ((p(x,y,u) ?
p(y,z,v)) ? (p(x,v,w) ? p(u,z,w)))
?x?y?z?u?v?w ((p(x,y,u) ? p(y,z,v)) ?
(p(u,z,w) ? p(x,v,w))) ? ((p(x,y,u) ?
p(y,z,v)) ? (p(x,v,w) ? p(u,z,w)))

• remove redundant parentheses ( )

?x?y?z?u?v?w (p(x,y,u) ? p(y,z,v) ?
p(u,z,w) ? p(x,v,w)) ? (p(x,y,u) ? p(y,z,v) ?
p(x,v,w) ? p(u,z,w))

• Steps 7,8 eliminate ? en ?