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Logical Reasoning

Logical Reasoning. First Order Predicate Logic as a Language Resolution For Propositional Calculus For First Order Predicate Calculus Unification Resolution and PROLOG very short!

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Logical Reasoning

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  1. Logical Reasoning • First Order Predicate Logic as a Language • Resolution • For Propositional Calculus • For First Order Predicate Calculus • Unification • 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. Using Resolution for Propositional Calculus • 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): A1 v…v An v C; B1v…v Bm v ~C _____________________________ A1v…v An v B1v…v Bm • Steps of a resolution proof: • Convert assumption into clauses • Convert negated conclusion into clauses • Determine if the empty clause can  be derived from the clauses generated in steps 1 and 2. • Yes: theorem is proven • No*: theorem is not proven *:= things are more complicated for FOPL

  3. 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. Resolution for Propositional Calculus Example: P v (Q  R), Q  S, P  Q |- RS Clauses: • P v Q • P v R • ~Q v S • ~P v Q • R • ~S • ~Q using 3,6 • ~P v S using 3,4 • ~P using 6,8 • Q using 1,9 •  using 7,10

  5. 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).

  6. PROLOG and Resolution • grandchild(X,Z):-child(X,Y),child(Y,Z) **PROLOG Rule*** • Asserted facts: child(pete,john), child(sally,john), child(john,fred), child(tom,fred) • ?-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): • ~child($x,$y) v ~child($y,$z) v grandchild($x,$z) • ~grandchild($u,Fred) “negated conclusion” • child(Pete,John) • child(Sally,John) • child(John,Fred) • child(Tom,Fred) Resolution in PROLOG: Find all substitutions to the free variables in the query expression ($u in the example) that lead to a contradiction.

  7. PROLOG Example Continued • ~child($x,$y) v ~child($y,$z) v grandchild($x,$z) rule • ~grandchild($u,Fred) “negated conclusion/query” • child(Pete,John) fact • child(Sally,John) fact • child(John,Fred) fact • child(Tom,Fred) fact • ~child($u,$y) v ~child ($y, Fred) using 1, 2 • ~child($u, John) using 5,7 •  for (($u Pete)) using 3,8 First answer! •  for (($u Sally)) using 4,8 Second answer! • ~child($u, Tom) using 6,7 No more empty clauses are found; therefore the PROLOG system returns two answers: Pete and Sally

  8. Example Sentences • Every vegetarian is intelligent. • Every NBA-player owns at least one house in Texas. • There are at least 2 giraffes in the Houston-Zoo.

  9. 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) ))

  10. Answers Sept. 23, 2004 EnglishFOPL Exam • frog(Fred) ^green(Fred) • x (student(x) ^ (not(take(x,AGR2320) v not(take(x,AGR2388)) • 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))

  11. FOPL as a Language SolutionsAnswers to the Un-graded Quiz • frog(Fred) ^green(Fred) • Vx (red(x) ^ car(x)  dangerous(x)) • VxVyVz (person(x) ^ person(y) ^ has-ssn(x,z) ^ has-ssn(y,z)  x=y) • ]x (student(x) ^ registered(x,COSC6367) ^ Vy (registered(y,COSC6367) ^ not (x=y)  send-mail(x,y))) • Vm (man(m) ^ white(m)  not (jump-high(m)) ) • Not possible; no quantifier for “most”!

  12. Answers February 26, 2009EnglishFOPL Exam • m w (man(m)(woman(w) ^ loves(m,w))) • x (student(x) ^ (not(take(x,COSC6340) ^ not(take(x,COSC 6351)) • xyz ((person(x) ^ person(y) ^ has-ssn(x,z) ^ has-ssn(y,z))  x=y) • 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)

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