Logical Reasoning 1 2 3 First Order Predicate

Logical Reasoning 1. 2. 3. First Order Predicate Logic as a Language Resolution a) For Propositional Calculus b) For First Order Predicate Calculus c) Unification Resolution and PROLOG very short! Ch. Eick: FOPL, Resolution and PROLOG

Using Resolution for Propositional Calculus o o Makes a proof by contradiction Works with clauses --- knowledge is represented as a conjunct of disjuncts (conjunctive normal form) Does not use modus ponens; uses the inference rule of resolution (m, n 0): A 1 v…v An v C; B 1 v…v Bm v ~C _______________ A 1 v…v An v B 1 v…v Bm Steps of a resolution proof: 1. Convert assumption into clauses 2. Convert negated conclusion into clauses 3. Determine if the empty clause be derived from the can clauses generated in steps 1 and 2. o Yes: theorem is proven o No*: theorem is not proven *: = things are more complicated for FOPL Ch. Eick: FOPL, Resolution and PROLOG

Proof by Contradiction • Assume you want to prove: A 1, …, An |- B then the truth of this statement is verified as follows: – We assume that A 1, . . , An is true – We assume that ~B is true – We show that it can never be the case that A 1, . . , An are true and B is false…; that is, we look for a contraction (e. g. P and ~P are both true). 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 Pv. 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 (2) Ch. Eick: FOPL, Resolution and PROLOG

Resolution for Propositional Calculus Homework Example: Pv. Q R R S T T U v (X Y) ~U |P (X Y) 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 (q denotes a substitution): § P v R, ~R’ v S, R and R’ unify, unify(R, R’)=q § q(P) v q(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). Ch. Eick: FOPL, Resolution and PROLOG

Ch. Eick: FOPL, Resolution and PROLOG

Ch. Eick: FOPL, Resolution and PROLOG

Assignment 3: 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/pro 74 b-propertiesmult. html Properties of Addition: http: //www. aaastudy. com/pro 74 ax 2. htm Expression Evaluation: (+ 2 (+ 5 a)) unifies (+ 7 a) Ch. Eick: FOPL, Resolution and PROLOG

PROLOG and Resolution 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” 1. 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. Ch. Eick: FOPL, Resolution and PROLOG

PROLOG Example Continued ~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($u, John) using 5, 7 9. (($u Pete)) using 3, 8 First answer! for 10. (($u Sally)) using 4, 8 Second answer! for 11. ~child($u, Tom) using 6, 7 1. No more empty clauses are found; therefore the PROLOG system returns two answers: Pete and Sally Ch. Eick: FOPL, Resolution and PROLOG

Example Sentences Every vegetarian is intelligent. 2. Every NBA-player owns at least one house in Texas. 3. There at least 2 giraffes in the Houston-Zoo. 1. 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 at least 2 giraffes in the Houston-Zoo. ]g 1]g 2 (giraffe(g 1) ^ giraffe(g 2) ^ not(g 1=g 2)) ^ lives(g 1, HOU_Zoo) ^ lives(g 2, HOU_Zoo) )) Ch. Eick: FOPL, Resolution and PROLOG

Answers Sept. 23, 2004 English FOPL Exam (1) (2) (3) (4) (5) (6) frog(Fred) ^ green(Fred) x (student(x) ^ (not(take(x, AGR 2320) v not(take(x, AGR 2388)) x y z ((person(x) ^ person(y) ^ has-ssn(x, z) ^ has-ssn(y, z)) x=y) x ((red(x) ^ car(x)) dangerous(x)) x y(brother(x, Fred) ^ brother(y, Fred) ^ not(x=y)) ^ ~ s sister(s, Fred) a b c((ontop(a, b) ^ ontop(b, c)) ontop(a, c)) same s not(sister(s, Fred)) Ch. Eick: FOPL, Resolution and PROLOG

FOPL as a Language Solutions Answers to the Un-graded Quiz (1) (2) (3) (4) (5) (6) frog(Fred) ^ green(Fred) Vx (red(x) ^ car(x) dangerous(x)) Vx. Vy. Vz (person(x) ^ person(y) ^ has-ssn(x, z) ^ hasssn(y, z) x=y) ]x (student(x) ^ registered(x, COSC 6367) ^ Vy (registered(y, COSC 6367) ^ not (x=y) send-mail(x, y))) Vm (man(m) ^ white(m) not (jump-high(m)) ) Not possible; no quantifier for “most”! Ch. Eick: FOPL, Resolution and PROLOG