Ch. Eick: FOPL, Resolution and PROLOG Logical Reasoning 1. First Order Predicate Logic as a Language 2. Resolution a)For Propositional Calculus b)For First

Ch. Eick: FOPL, Resolution and PROLOG

Logical Reasoning1. First Order Predicate Logic as a Language2. Resolution

a) For Propositional Calculusb) For First Order Predicate Calculusc) Unification

3. Resolution and PROLOG very short!


Using Resolution for Propositional Calculuso Makes a proof by contradictiono Works with clauses --- knowledge is represented as a conjunct

of disjuncts (conjunctive normal form)o Does not use modus ponens; uses the inference rule of

resolution (m,n0): A1 v…v An v C; B1v…v Bm v ~C _____________________________

A1v…v An v B1v…v Bm

o Steps of a resolution proof:1. Convert assumption into clauses2. Convert negated conclusion into clauses3. Determine if the empty clause � can be derived from the

clauses generated in steps 1 and 2.o Yes: theorem is proveno No*: theorem is not proven

*:= things are more complicated for FOPL


Proof by Contradiction• Assume you want to prove: A1,…,An |- B then the truth of

this statement is verified as follows:– We assume that A1,..,An is true– We assume that ~B is true– We show that it can never be the case that A1,..,An are

true and B is false…; that is, we look for a contraction (e.g. P and ~P are both true).


Resolution for Propositional Calculus

Example: P v (Q R), Q S, P Q |- RS

Clauses:

(1) P v Q

(2) P v R(3) ~Q v S(4) ~P v Q(5) R(6) ~S(7) ~Q using 3,6(8) ~P v S using 3,4(9) ~P using 6,8(10) Q using 1,9(11) � using 7,10


Resolution for Propositional Calculus Homework

Example:

P v Q R

RS TT U v (X Y)

~U

|-

P (X Y)


Key Ideas --- Resolution for FOPL

All-quantifiers are replaced by match variables Existential quantifiers are replaced by Skolem functions that

depend on the variables of the surrounding all-quantifiers. All match-variables in different clauses are renamed. A generalized resolution inference rule is used ( denotes a

substitution):

P v R, ~R’ v S, R and R’ unify, unify(R,R’)= (P) v (S)

Similar to resolution for prepositional calculus a proof by contradiction is conducted whose goal is to reach the empty clause (which represents a contradiction).




Assignment3: Resolution for Formulas Containing + and *

Examples: (+ 1 b) and (+ $a b) unify for (($a 1)) (+ (* a b) 5) (+ $a $b) unify for (($a (* a b)) ($b 5)) (+ a (+ $a c)) and (+ a d) do not unify; return nil (+ $a (+ a $b)) and (+ (* $b 5) (+ 5 a)) unify for (($a (+$b 5))

($b a)) (+ a b) and (+ b a) unify for ()

Properties of Multiplication:

http://www.aaamath.com/pro74b-propertiesmult.html

Properties of Addition: http://www.aaastudy.com/pro74ax2.htm

Expression Evaluation: (+ 2 (+ 5 a)) unifies (+ 7 a)


PROLOG and Resolution1. grandchild(X,Z):-child(X,Y),child(Y,Z) **PROLOG Rule***2. Asserted facts: child(pete,john), child(sally,john), child(john,fred),

child(tom,fred)3. ?-grandchild(U,fred) “Who are the grandchildren of Fred?”U=pete…U=sally… “Answers returned by the PROLOG runtime system”

How does PROLOG do it? It uses resolution as follows:In clause form (using “our” syntax): 1. ~child($x,$y) v ~child($y,$z) v grandchild($x,$z)2. ~grandchild($u,Fred) “negated conclusion”3. child(Pete,John)4. child(Sally,John) 5. child(John,Fred) 6. child(Tom,Fred)

Resolution in PROLOG: Find all substitutions to the free variables in the query expression ($u in the example) that lead to a contradiction.


PROLOG Example Continued1. ~child($x,$y) v ~child($y,$z) v grandchild($x,$z) rule2. ~grandchild($u,Fred) “negated conclusion/query”3. child(Pete,John) fact4. child(Sally,John) fact5. child(John,Fred) fact6. child(Tom,Fred) fact7. ~child($u,$y) v ~child ($y, Fred) using 1, 28. ~child($u, John) using 5,79. for (($u Pete)) using 3,8 � First answer!10. for (($u Sally)) using 4,8 � Second answer!11. ~child($u, Tom) using 6,7

No more empty clauses are found; therefore the PROLOG system returns two answers: Pete and Sally


Example Sentences

1. Every vegetarian is intelligent.

2. Every NBA-player owns at least one house in Texas.

3. There are at least 2 giraffes in the Houston-Zoo.


Example Sentences in FOPL

Every vegetarian is intelligent.

Vx (vegetarian(x) intelligent(x))Every NBA-player owns at least one house in

Texas.

Vx (nba(x) ]h(house(h) ^ owns(x,h) ^ location(h,Texas) ))

There are at least 2 giraffes in the Houston-Zoo.

]g1]g2 (giraffe(g1) ^ giraffe(g2) ^ not(g1=g2))

^ lives(g1, HOU_Zoo) ^ lives(g2, HOU_Zoo) ))


Answers Sept. 23, 2004 EnglishFOPL Exam

(1) frog(Fred) ^ green(Fred)(2) x (student(x) ^ (not(take(x,AGR2320) v

not(take(x,AGR2388))(3) xyz ((person(x) ^ person(y) ^ has-ssn(x,z) ^

has-ssn(y,z)) x=y)(4) x ((red(x) ^ car(x)) dangerous(x))(5) xy(brother(x,Fred) ^ brother(y,Fred) ^ not(x=y))

^ ~s sister(s,Fred)(6) abc((ontop(a,b) ^ ontop(b,c)) ontop(a,c))

s not(sister(s,Fred))

same


FOPL as a Language SolutionsAnswers to the Un-graded Quiz

(1) frog(Fred) ^ green(Fred)(2) Vx (red(x) ^ car(x) dangerous(x))(3) VxVyVz (person(x) ^ person(y) ^ has-ssn(x,z) ^ has-ssn(y,z)

x=y)(4) ]x (student(x) ^ registered(x,COSC6367) ^ Vy

(registered(y,COSC6367) ^ not (x=y) send-mail(x,y)))(5) Vm (man(m) ^ white(m) not (jump-high(m)) )(6) Not possible; no quantifier for “most”!

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Ch. Eick: FOPL, Resolution and PROLOG Logical Reasoning 1. First Order Predicate Logic as a Language 2. Resolution a)For Propositional Calculus b)For First