1. 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 Order Predicate Calculus c)Unification 3. Resolution and PROLOG very short! Textbook’s Perspective: knowledge based systems; giving a somewhat comprehensive introduction to logic; cover 7,[9],[[10]] 6368 Perspective: Using FOPL to make statements about the world; use computers to prove theorems and for question-answering.

2. Ch. Eick: FOPL, Resolution and PROLOG Using Resolution for Propositional Calculus o Makes a proof by contradiction o 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,n  0): A 1 v…v A n v C; B 1 v…v B m v ~C _____________________________ A 1 v…v A n v B 1 v…v B m o Steps of a resolution proof: 1.Convert assumption into clauses 2.Convert negated conclusion into clauses 3.Determine if the empty clause can  be derived from the clauses generated in steps 1 and 2. oYes: theorem is proven oNo*: theorem is not proven *:= things are more complicated for FOPL

3. Ch. Eick: FOPL, Resolution and PROLOG 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).

4. Ch. Eick: FOPL, Resolution and PROLOG Resolution for Propositional Calculus Example: P v (Q  R), Q  S, P  Q |- R  S 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

5. Ch. Eick: FOPL, Resolution and PROLOG 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).

6. Ch. Eick: FOPL, Resolution and PROLOG

8. PROLOG and Resolution 1. 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.

9. Ch. Eick: FOPL, Resolution and PROLOG PROLOG Example Continued 1. ~child($x,$y) v ~child($y,$z) v grandchild($x,$z) rule 2. ~grandchild($u,Fred) “negated conclusion/query” 3. child(Pete,John) fact 4. child(Sally,John) fact 5. child(John,Fred) fact 6. child(Tom,Fred) fact 7. ~child($u,$y) v ~child ($y, Fred) using 1, 2 8. ~child($u, John) using 5,7 9.  for (($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

10. Ch. Eick: FOPL, Resolution and PROLOG 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.

11. Ch. Eick: FOPL, Resolution and PROLOG 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) ))

12. Ch. Eick: FOPL, Resolution and PROLOG Answers Sept. 23, 2004 English  FOPL Exam (1) frog(Fred) ^ green(Fred) (2) x (student(x) ^ (not(take(x,AGR2320) v not(take(x,AGR2388)) (3) xyz ((person(x) ^ person(y) ^ has-ssn(x,z) ^ has-ssn(y,z))  x=y) (4) x ((red(x) ^ car(x))  dangerous(x)) (5) xy(brother(x,Fred) ^ brother(y,Fred) ^ not(x=y)) ^ ~s sister(s,Fred) (6) abc((ontop(a,b) ^ ontop(b,c))  ontop(a,c))  s not(sister(s,Fred)) same

13. Ch. Eick: FOPL, Resolution and PROLOG FOPL as a Language Solutions Answers 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”!

14. Ch. Eick: FOPL, Resolution and PROLOG Answers February 26, 2009 English  FOPL Exam (1) m w (man(m)  (woman(w) ^ loves(m,w))) (2) x (student(x) ^ (not(take(x,COSC6340) ^ not(take(x,COSC 6351)) (3) xyz ((person(x) ^ person(y) ^ has-ssn(x,z) ^ has- ssn(y,z))  x=y) (4) x (takes(x,COSC 6368) ^ esg1g2(((exam(e,COSC6368,s,g1) ^ (exam(e,COSC6368,x, g2)) ^ not(e=s))  g2>g1)) No answer (“most” is a fuzzy quantifier) for (5)!! Remark: Exam(name,course,student,grade) is a predicate that captions the grades of students in courses for a particlar semester; e.g. exam(Mt,COSC6368,Fred,3.0)