For Wednesday

• Read chapter 10• Prolog Handout 4

Exam 1

• Monday• Take home due at the exam

Try It

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

The Anonymous Variable

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

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

Arithmetic Comparison

• Comparison operators:><>==< (note the order: NOT <=)=:= (equal values)=\= (not equal values)

Arithmetic Examples

• Retrieving people born 1950-1960:?- born(Name, Year), Year >= 1950, Year =< 1960.

• Difference between = and =:=?- 1 + 2 =:= 2 + 1.yes?- 1 + 2 = 2 + 1.no?- 1 + A = B + 2.A = 2B = 1

Length of a List

• Definition of length/2

length([], 0).length([_ | Tail], N) :-

length(Tail, N1),N is 1 + N1.

• Note: all loops must be implemented via recursion

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 < End,Next is Begin + 1,Sum2 is Sum1 + Next,sum(Next, End, Sum2, Sum).

The Cut (!)

• A way to prevent backtracking.• Used to simplify and to improve efficiency.

Negation

• Can’t say something is NOT true• Use a closed world assumption• Not simply means “I can’t prove that it is

true”

Dynamic Predicates

• A way to write self-modifying code, in essence.

• Typically just storing data using Prolog’s built-in predicate database.

• Dynamic predicates must be declared as such.

Using Dynamic Predicates

• assert and variants• retract

– Fails if there is no clause to retract

• retractall– Doesn’t fail if no clauses

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 • First order form

– For two literals pj and qk in two clauses • p1 Ú ... pj ... Ú pm

• q1 Ú ... qk ... Ú qn

such that q=UNIFY(pj , ¬qk), derive

SUBST(q, p1Ú...pj 1Úpj+1...ÚpmÚq1Ú...qk 1 qk+1...Úqn)

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...Ú¬pmÚq1Ú...Úqn Û

p1Ù... Ù pm Þ q1Ú ...Ú qn

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).

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)

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) -> S(A)

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)

Refutation Proofs• Unfortunately, resolution proofs in this form

are still incomplete. • For example, it cannot prove any tautology

(e.g. PÚ¬P) 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

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

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.

Conversion to Clausal Form• Eliminate implications and biconditionals by

rewriting them. p Þ q -> ¬p Ú q

p Û q > (¬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) -> ¬p Ù ¬q

¬(p Ù q) -> ¬p Ú¬q

¬"x p -> $x ¬p

¬$x p -> "x ¬p

¬¬p -> p

Conversion continued

• Standardize variables to avoid use of the same variable name by two different quantifiers.

"x P(x) Ú $x P(x) -> "x1 P(x1) Ú $x2 P(x2) • Move quantifiers left while maintaining

order. Renaming above guarantees this is a truth preserving transformation.

"x1 P(x1) Ú $x2 P(x2) -> "x1 $x2 (P(x1) Ú P(x2))

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)) -> P(C1) Ù Q(C1)

– If it is within the scope of n universally quantified variables, then replace it with a unique n ary function over these universally quantified variables.

"x1$x2(P(x1) Ú P(x2)) -> "x1 (P(x1) Ú P(f1(x1)))

"x(Person(x) Þ $y(Heart(y) Ù Has(x,y))) ->

"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.

Conversion continued• Distribute Ù over Ú to convert to conjunctions of

clauses (aÙb) Ú c -> (aÚc) Ù (bÚc)

(aÙb) Ú (cÙd) -> (aÚc) Ù (bÚc) Ù (aÚd) Ù (bÚd) – Can exponentially expand size of sentence.

• Flatten nested conjunctions and disjunctions to get final CNF (a Ú b) Ú c -> (a Ú b Ú c)

(a Ù b) Ù c -> (a Ù b Ù c)

• Convert clauses to implications if desired for readability

(¬a Ú ¬b Ú c Ú d) -> a Ù b Þ c Ú d

Sample Clause Conversion"x((Prof(x) Ú Student(x)) Þ ($y(Class(y) Ù Has(x,y)) Ù

$y(Book(y) Ù Has(x,y))))

"x(¬(Prof(x) Ú Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù $y(Book(y) Ù Has(x,y))))

"x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù $y(Book(y) Ù Has(x,y))))

"x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù $z(Book(z) Ù Has(x,z))))

"x$y$z((¬Prof(x)Ù¬Student(x))Ú ((Class(y) Ù Has(x,y)) Ù (Book(z) Ù Has(x,z))))

(¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù Book(g(x)) Ù Has(x,g(x))))

Clause Conversion(¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù

Book(g(x)) Ù Has(x,g(x))))

(¬Prof(x) Ú Class(f(x))) Ù

(¬Prof(x) Ú Has(x,f(x))) Ù

(¬Prof(x) Ú Book(g(x))) Ù

(¬Prof(x) Ú Has(x,g(x))) Ù

(¬Student(x) Ú Class(f(x))) Ù

(¬Student(x) Ú Has(x,f(x))) Ù

(¬Student(x) Ú Book(g(x))) Ù

(¬Student(x) Ú Has(x,g(x))))

Sample Resolution Problem

• Jack owns a dog. • Every dog owner is an animal lover. • No animal lover kills an animal. • Either Jack or Curiosity killed Tuna the cat. • Did Curiosity kill the cat?

In Logic Form

A) $x Dog(x) Ù Owns(Jack,x)

B) "x ($y Dog(y) Ù Owns(x,y)) Þ AnimalLover(x))

C) "x AnimalLover(x) Þ ("y Animal(y) Þ ¬Kills(x,y))

D) Kills(Jack,Tuna) Ú Kills(Cursiosity,Tuna)

E) Cat(Tuna)

F) "x(Cat(x) Þ Animal(x))

Query: Kills(Curiosity,Tuna)

In Normal Form

A1) Dog(D)

A2) Owns(Jack,D)

B) Dog(y) Ù Owns(x,y) Þ AnimalLover(x)

C) AnimalLover(x) Ù Animal(y) Ù Kills(x,y) Þ False

D) Kills(Jack,Tuna) Ú Kills(Curiosity,Tuna)

E) Cat(Tuna)

F) Cat(x) Þ Animal(x)

Query: Kills(Curiosity,Tuna) Þ False

Resolution Proof