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For Monday

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Arithmetic in Prolog. Basic arithmetic operators are provided for by built-in procedures: ... xv) Representable by a set of clauses (disjunctions of literals) ... – PowerPoint PPT presentation

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Title: For Monday

1
For Monday

Read chapter 10, sections 1-3
Prolog handout 4
Consider advanced topics youre interested in
well discuss the second half of the course.

2
Try It

reverse(List,ReversedList)
evenlength(List)
oddlength(List)

3
The Anonymous Variable

Some variables only appear once in a rule
Have no relationship with anything else
Can use _ for each such variable

4
Arithmetic in Prolog

Basic arithmetic operators are provided for by
built-in procedures , -, , /, mod, //
Note carefully ?- X 1 2. X 1 2 ?- X
is 1 2. X 3

5
Arithmetic Comparison

Comparison operators gt lt gt lt (note the
order NOT lt) (equal values) \ (not
equal values)

6
Arithmetic Examples

Retrieving people born 1950-1960 ?- born(Name,
Year), Year gt 1950, Year lt 1960.
Difference between and ?- 1 2 2
1. yes ?- 1 2 2 1. no ?- 1 A B
2. A 2 B 1

7
Length of a List

Definition of length/2length(, 0).length(_
Tail, N) - length(Tail, N1), N is 1 N1.
Note all loops must be implemented via recursion

8
Counting Loops

Definition of sum/3sum(Begin, End, Sum)
- sum(Begin, End, Begin, Sum).sum(X, X, Y,
Y).sum(Begin, End, Sum1, Sum) - Begin lt
End, Next is Begin 1, Sum2 is Sum1
Next, sum(Next, End, Sum2, Sum).

9
Negation

Cant say something is NOT true
Use a closed world assumption
Not simply means I cant prove that it is true

10
And much more . . .
11
Resolution

Propositional version.
a Ú b, b Ú c - a Ú c OR aÞ b, b Þ c - a
Þ c
Reasoning by cases OR transitivity of
implication
Firstorder form
For two literals pj and qk in two clauses
p1 Ú ... pj ... Ú pm
q1 Ú ... qk ... Ú qn
such that qUNIFY(pj , qk), derive
SUBST(q, p1Ú...pj1Úpj1...ÚpmÚq1Ú...qk1
qk1...Úqn)

12
Implication form

Can also be viewed in implicational form where
all negated literals are in a conjunctive
antecedent and all positive literals in a
disjunctive conclusion.
p1Ú...ÚpmÚq1Ú...Úqn Û
p1Ù... Ù pm Þ q1Ú ...Ú qn

13
Conjunctive Normal Form (CNF)

For resolution to apply, all sentences must be in
conjunctive normal form, a conjunction of
disjunctions of literals
(a1 Ú ...Ú am) Ù
(b1 Ú ... Ú bn) Ù
..... Ù
(x1 Ú ... Ú xv)
Representable by a set of clauses (disjunctions
of literals)
Also representable as a set of implications
(INF).

14
Example

Initial CNF INF
P(x) Þ Q(x) P(x) Ú Q(x) P(x) Þ Q(x)
P(x) Þ R(x) P(x) Ú R(x) True Þ P(x) Ú R(x)
Q(x) Þ S(x) Q(x) Ú S(x) Q(x) Þ S(x)
R(x) Þ S(x) R(x) Ú S(x) R(x) Þ S(x)

15
Resolution Proofs

INF (CNF) is more expressive than Horn clauses.
Resolution is simply a generalization of modus
ponens.
As with modus ponens, chains of resolution steps
can be used to construct proofs.
Factoring removes redundant literals from clauses
S(A) Ú S(A) -gt S(A)

16
Sample Proof

P(w) ? Q(w) Q(y) ? S(y)
y/w
P(w) ? S(w) True ? P(x) ? R(x)
w/x
True ? S(x) ? R(x) R(z) ? S(z)
x/A, z/A
True ? S(A)

17
Refutation Proofs

Unfortunately, resolution proofs in this form are
still incomplete.
For example, it cannot prove any tautology (e.g.
PÚP) from the empty KB since there are no
clauses to resolve.
Therefore, use proof by contradiction
(refutation, reductio ad absurdum). Assume the
negation of the theorem P and try to derive a
contradiction (False, the empty clause).
(KB Ù P Þ False) Û KB Þ P

18
Sample Proof

P(w) ? Q(w) Q(y) ? S(y)
y/w
P(w) ? S(w) True ? P(x) ? R(x)
w/x
True ? S(x) ? R(x) R(z) ? S(z)
z/x
S(A) ? False True ? S(x)
x/A
False

19
Resolution Theorem Proving

Convert sentences in the KB to CNF (clausal form)
Take the negation of the proposed theorem
(query), convert it to CNF, and add it to the KB.
Repeatedly apply the resolution rule to derive
new clauses.
If the empty clause (False) is eventually
derived, stop and conclude that the proposed
theorem is true.

20
Conversion to Clausal Form

Eliminate implications and biconditionals by
rewriting them.
p Þ q -gt p Ú q
p Û q gt (p Ú q) Ù (p Ú q)
Move inward to only be a part of literals by
using deMorgan's laws and quantifier rules.
(p Ú q) -gt p Ù q
(p Ù q) -gt p Úq
"x p -gt x p
x p -gt "x p
p -gt p

21
Conversion continued

Standardize variables to avoid use of the same
variable name by two different quantifiers.
"x P(x) Ú x P(x) -gt "x1 P(x1) Ú x2 P(x2)
Move quantifiers left while maintaining order.
Renaming above guarantees this is a
truthpreserving transformation.
"x1 P(x1) Ú x2 P(x2) -gt "x1 x2 (P(x1) Ú P(x2))

22
Conversion continued

Skolemize Remove existential quantifiers by
replacing each existentially quantified variable
with a Skolem constant or Skolem function as
appropriate.
If an existential variable is not within the
scope of any universally quantified variable,
then replace every instance of the variable with
the same unique constant that does not appear
anywhere else.
x (P(x) Ù Q(x)) -gt P(C1) Ù Q(C1)
If it is within the scope of n universally
quantified variables, then replace it with a
unique nary function over these universally
quantified variables.
"x1x2(P(x1) Ú P(x2)) -gt "x1 (P(x1) Ú P(f1(x1)))
"x(Person(x) Þ y(Heart(y) Ù Has(x,y))) -gt
"x(Person(x) Þ Heart(HeartOf(x)) Ù
Has(x,HeartOf(x)))
Afterwards, all variables can be assumed to be
universally quantified, so remove all quantifiers.