CS244 - Lectures 5-6, Sept 28-Oct 5 - Resolution
The Lecture Outline is followed by the detailed Lecture Notes.
LECTURE OUTLINE
** = important material not in the textbook
Ch 3: Representation:
Reification
**Ontologies
**Time
Ch 4: Resolution:
Kinds of proofs
4.1
Conversion to CNF
Resolution derivations
A non-deterministic entailment procedure
Examples propositional resolution proofs
**Size-preserving converson to CNF
Proof of correctness of propositional resolution
**Proof of completeness of propositional resolution
4.2
Conversion of universally-quantified formulas
First-order resolution
4.2.3
Skolemization
**Relation of a formula to its Skolemization
**Most general unifier algorithm
Examples of FOL proofs
Answer extraction
4.2.4
Handling equality
**Paramodulation
4.3
The Herbrand Theorem
**Proof of Herbrand theorem
Semi-decidability of FOL
**Proof that FOL is not decidable
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LECTURE NOTES
** = important material not in the textbook
Ch 3: Representation:
Reification
What is point of example:
Purchase(sam,bike,200)
vs
Purchase(p99) & agent(p99)=sam & object(p99)=bike & price(p99)=200
Suppose you don't want to use any function symbols or equality?
Purchase(p99) & agent(p99,sam) & ...
What does this say about the expressive power of FOL if it is restricted
to (at most) 2 place predicates? (Unchanged)
Compare to relational database versus XML
**Ontologies
ontology = fancy word for a classification system
Example
animal
mammal reptile
human cat
sam bill puff
How should this be represented in FOL?
Note difference between classes (predicate) and individuals (terms)
How to distinguish exhaustive versus non-exhaustive ways of
specializing a concept?
Inheritance of properties
animals have hair
therefore, sam has hair
No special mechanism needed!
Ontologies aren't only about objects, they can be about
abstract things like events.
Exchange
Purchase Gifting Trade
OnlinePurchase BrickAndMotarPurchase
What are the arguments to these predicates?
Purchase(oldowner, newowner, object, price) =>
Exchange(oldowner, newowner, object)
Exchange(oldowner, newowner, object) =>
E price . Purchase(oldowner, newowner, object, price)
v
E occasion . Gifting(oldowner, newowner,
object, occasion)
v
E object2 . Gifting(oldowner, newowner,
object, object2)
Reified approach:
Purchase(e) => Exchange(e)
Exchange(e) => Purchase(e) v Gifting(e) v Trade(e)
**Time
How to represent time?
Discrete time: add extra argument to predicates:
On(box1,box2,time)
Linear time:
On(box1,box2,time) & Pickup(box2,time) => ~On(box1,box2,next(time))
or, allowing integer arithmetic:
On(box1,box2,time) & Pickup(box2,time) => ~On(box1,box2,time+1)
Branching time:
On(box1,box2,time) => ~On(box1,box2, do(pickup(box1),time) )
Here, actions are FUNCTIONS rather than predicates!
A different possible future:
On(box1,box2,time) => On(box1,box2, do(paint(box1),time) )
Time intervals:
On(box1, box2, starttime, endtime)
Reified events:
Exchange(e) & time(e)=3
A e . Exchange(e) =>
Owns(newowner(e), object(e), next(time(e)))
****************************************************************************
Ch 4: Resolution:
Kinds of proofs:
Is a sentence (query) entailed by a knowledge base?
Is a sentence valid?
Is a set of sentences unsatisfiable?
Does a set of sentence entail a false statement?
KB |= S
|= (KB) => S
KB U {~S} is not satisfiable
KB U {~S} |= ~TRUE
4.1
Conversion to CNF, propositional case
1. Eliminate => and <=>
2. Move in ~ LITERAL: a propositional atom or its negation
3. Distribute & over V
4. Eliminate repeated literals in a clause
5. Eliminate clauses containing a literal and its negation
6. Erase & and V signs
((Rich v Talented) & ~Addict) => (Famous & Happy)
Meaning of an empty CNF {} : true
Meaning of an empty clause {{}}: false
Resolution derivations
class on 9/30/2010 starts here ===>
Resolution rule:
{P} U C1 + {~P} U C2
----------------
C1 U C2
Proof that resolution rule is sound using interpretations:
Suppose I is an interpretation where
I|={P} U C1
I|={~P} U C2
Consider cases:
case 1: I|=P
Therefore I|=C2
case 2: I|=~P
Therefore I|=C1
Since one of these cases must hold,
I |= C1 or I |= C2
Thus I |= C1 U C2
A non-deterministic entailment (satisfiability) procedure
Res(S)
if {} in S then return UNSAT
choose C1, C2 containing complementary
literals such that resolve(C1,C2) not in S
if no such pair, return SAT
S = S U resolve(C1,C2)
Examples propositional resolution proofs
If the unicorn is mythical, then it is immortal.
If it is not mythical, then it is an animal.
If the uniforn is either immortal or an animal,
then it is horned.
Proved: the unicorn is horned.
**Size-preserving converson to CNF
Bad case for direct conversion to cnf:
(P1 & P2 & P3) v (Q1 & Q2 & Q3)
If each conjunction is of length n, what
is number of clauses of result?
n**2
Even worse:
(P1 & Q1) v (P2 & Q2) v ... v (Pn & Qn)
What is number of clauses?
2**n
How to prevent blow up? Introduce new propositions!
Show how this is done:
Instead of distributing v over &:
1. Introduce a new proposition Ci for each
conjunction
2. Add clauses making Ci equivalent to the
corresponding conjunction
3. Replace original disjunction by
disjunction of Ci
Example:
(P1 & Q1) v (P2 & Q2) v ... v (Pn & Qn)
~C1 v P1
~C1 v Q1
~P1 v ~ Q1 v C1
same for other i
C1 v C2 v ... Cn
Explain relationship between original and augmented
theory:
They are not *quite* equivalent because
new theory contains extra propositions.
However:
New |= Old
And
New is UNSAT
iff
Old is UNSAT
and for any sentence S not containing
the new propositions,
New |= S
iff
Old |= S
**Proof of completeness of propositional resolution
Theorem: A set S of clauses is unsatisfiable iff there
is a deduction of the empty clause from S.
Correctness is <---, and completeness is --->
Suppose S is unsat. We will build a binary tree whose nodes
"split" on the truth value assigned to a proposition. Thus:
- each leaf is an interpretation
- each internal node is a *partial* interpretation
At each leaf, all the clauses must be false, otherwise there
would be a satisfying interpretation.
Put each clause on the highest node in the tree where it is
false. The result is an (upside down) resolution proof
of the empty clause!
Example: P, ~P v Q, ~P v ~Q
Create tree
P/\~P
/ [P]
Q/\~Q
[~P,~Q] [~P,Q]
4.2
Conversion of universally-quantified formulas
1. Eliminate => and <=>
2. Move in ~
3. Rename quantified variables, so all are distinct
4. Move out quantifiers
5. Distribute & over V
6. Skolemization: replace existentially-quantified
variables by a new FUNCTION symbol, whose
arguments are all the universally quantified
variables to its left
Example:
Everyone who loves everyone loves someone who loves no one.
A x . (A y . L(x,y)) => E z . L(x,z) & A w . ~L(z,w)
**Relation of a formula to its Skolemization
Is the converted formula EQUIVALENT to the original?
No, because of the new function symbols.
However, as in the case with introducing new
propositions, they are closely linked:
Skolem Formula |= Original Formula
Skolem Formula is UNSAT
iff
Original Formula is UNSAT
and for any sentence S not containing
the skolem functions,
New |= S
iff
Old |= S
Class on 10/5 begins here:
Most general unifier algorithm (Sec. 4.3.6)
** First-order resolution:
Need to UNIFY the complementary literals,
they might not be IDENTICAL.
MGU(x,y,THETA):
if THETA = FAIL then return FAIL
if x = y return THETA
if x is a variable then
if x appears in y then return FAIL
THETA = THETA U {x/y}
if y is a variable then
if y appears in x then return FAIL
THETA = THETA U {y/x}
/* x and y must both be functional terms */
if head(x) != head(y) then FAIL
for x' in body(x) and y' in body(y)
THETA = MGU(x,y,THETA)
if THETA = FAIL return FAIL
return THETA
MGU(Knows(John,x), Knows(John, Jane)}
MGU(Knows(John,x), Knows(y, Bill)}
MGU(Knows(John,x), Knows(y, mother(y))}
MGU(Knows(John,x), Knows(x,Elizabeth)} == fail
Before computing Most General Unifier: rename variables
in the two clauses if necessary to make them distinct.
MGU(Knows(John,x), Knows(z,Elizabeth)}
MGU(Knows(father(x),x), Knows(z,sister(John)))
Resolve: [Knows(father(x),x)], [~Knows(z,sister(John)) v Happy(z)]
[Happy(father(sister(John)))]
MGU(Knows(father(x),x), Knows(z,sister(z)))
FAILS!
"Occurs check": block unification if you'd get a cycle in
substitutions:
Examples of FOL proofs
Everyone who loves all animals is loved by someone
Anyone who kills an animal is loved by no one
Jack loves all animals
Jack or Curiousity killed Tuna
Tuna is a cat
Did Curiousity kill the cat Tuna?
Handling equality
Note that MGU doesn't understand =.
So:
[A=B]
[Friend(A,C)] should resolve with [~Friend(B,C)]
How to handle?
Explicit axioms:
x=y & Friend(x,z) => Friend(y,z)
Modify Resolution adding final case:
if a unit clause [x=y] or [y=x] occurs, return THETA
Incomplete. Really should create sets of equated literals.
if x and y are in the same equivalence class, return THETA
Still incomplete, since only takes information from unit clauses into
account.
Class on 10/7 begins here
**Paramodulation
If John is rich, his father is bill.
Anyone's mother is married to their father.
[~rich(john) v father(john)=bill] [Married(father(z),mother(z))]
Paramodulation: for any terms x,y,z, where Unify(x,z)=THETA
and z appears in the clause m1 v ... mn,
l1 v ... v lk v x=y m1 v ... mn[z]
-------------------------------------------
SUBST(THETA, l1 v ... v lk v m1 ... mn[y])
[~rich(john) v Married(bill,mother(john))]
If John is rich, bill is married to his mother.
Answer extraction
Consider simplified example of "did Curiousity kill the cat?"
Suppose we want to know WHO killed the cat.
Want to "trace binding" of original existential quantifier
in question (which becomes a universal).
E x . Kill(x,Tuna)
-->
~Kill(x,Tuna) v Answer(x)
Stop when clause containing only Answer is derived.
Completeness of FOL Resolution
The Herbrand Theorem
Define Herbrand universe
All the ground terms that can be constructed from
the function symbols in the theory.
(Note: if no constants (0-ary function symbols),
add one.)
Define Saturation (of a clausal theory)
All ground (variable free) clauses that can
be created by subsituting some set of ground terms
for the variables in an FOL CNF theory.
Define Herbrand base: the saturation of a FOL CNF theory
by its Herbrand Universe. Notation: H(S)
Herbrand theorem: if a set S of clauses is unsatisfiable, then
there is a finite subset of its Herbrand base H(S) that is unsatisfiable.
Completeness of resolution for FOL based on LIFTING LEMMA:
C1, C2 FOL clauses
C1', C2' ground instances
if C' in resolve(C1', C2'),
then exists C in resolve(C1,C2)
such that C' is a ground instance of C.
Proof that FOL is not decidable
Completeness means that if something is entailed by (is a
theorem of) a theory, there exists a resolution proof.
You can exhaustively search for longer and longer proofs
until one is found.
So, FOL is at least semi-decidable.
If the query is not a theorem, resolution might run forever,
generating longer and longer clauses with larger and
larger terms -- infinitely many different clauses!
Is there a proof system that is guaranteed to tell us
that something is NOT a theorem? I.e., is FOL
decidable?
No!
You can reduce the Halting Problem to that of determining if
a formula is a theorem.
Proven by Alonzo Church (1936) and Alan Turing (1937).
Idea: encode a Turning machine in FOL.
E.g.
FSM state of Turning machine = {S1, ...., Sn, HALT}
tape symbols = {A, B, C, SOT, EOT}
use functions B, N to concatenate symbols
tape:
V
a b c b a c
tape to left of head
B(b, B(a, sot))
tape to right of head
N(b, N(a, N(c, eot)))
Config(state, left_of_head, head, right_of_head)
Initial configuration
Operation of machine: if you read "A" in
state s1, replace it with "B", go to state s3, and move head right:
Config(S1, l, A, N(x,y)) => Config(S3, B(A,l), x, y)
Halting: prove E l,h,r . Config(HALT, l, h, r)