372 CS Homework 3 Due date: In class Thusrday Feb. 28, 2008 By email Sunday 12:59 p.m. to Robert Xiao ([e-mail]) use subject line (HWK#3 CS372)
SHOW YOUR WORK FOR ALL QUESTIONS
1 Determine the truth value of each of these statements if the domain of each variable consists of all real numbers a) !x"y (x2 =y) True (y = x2) b) !x"y (x = y2) False (if x is negative no such y exists) c) "x!y (xy = 0) True (x=0) d) "x"y (x + y # y + x) False (communtative law for addition alwyas holds) e) !x ( x # 0 $ "y (xy = 1)) True (let y = 1/x) f) "x!y (y # 0 $ xy = 1) False (the reciprocal of y depends on y; there is not one x that works for all y) g) !x"y ( x + y = 1) True (y = 1-x) h) "x"y (x + 2y = 2 % 2x + 4y = 5) False (inconsistent system of equations) i) !x"y (x+ y = 2 % 2x y = 1)
False (this system of equations has only one solution; e.g., if x=0n no y satisfies y =2 and y = 1) j) !x !y "z (z = (x+y)/2) True (z = (x+y) /2) 2 Let F(x,y) be the statement "x can fool y", where the domains consists of all people in the world. Use quantifiers to express each of these statements: a) Everybody can fool Fred !(x) F(x Fred) b) Evelyn can fool everybody !(y) F(Evelyn, y) c) Everybody can fool somebody !x"y F(x y) d) There is no one who can fool everybody &"y !x F(x y) e) Everyone can be fooled somebody !y"x F(x y) f) No one can fool Fred and Jerry &"y [F(y Fred) % F(y Jerry)] g) Nancy can fool exactly two people "y1 "y2 (F(Nancy, y1) % F(Nancy, y2) % y1 # y2 % !y (F(Nancy,y) $ (y=y1 ' y =y2) h) There is exactly one person whom everybody can fool "y(!x F(x,y) !z (!x F(x,z) $ z=y)) i) No one can fool himself or herself &"F(x,x) j) There is someone who can fool exactly one person besides himself or herself "x"y (x#y%F(x,y) %!z ((F(x,z) % z#x)$ z=y)) (we do not assume that this sentence is asserting that this person can or cannot fool her/himself)
3. Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logicalconnectives). a) !x!yP(x, y) ----- "x"y P(x, y) b) !y"xP(x, y) ------ "y!x P(x, y) \ c) !y!x(P(x, y) % Q(x, y)) ----- "y "x (P(x, y)' Q(x, y)) d) ("x"yP(x, y) % !x!yQ(x, y)) ------ !x!y P(x, y)' "x"y Q(x, y) e) !x("y!zP(x, y, z) % "z!yP(x, y, z)) ---- "x(!y"z P(x, y, z)' !z"y P(x, y, z)) 4. Show that !x P(x) '!x Q(x) and !(x) (P(x) ' Q(x)) are not logically equivalent. It is intuitively clear that the first asserts statement more than the second. The first statement says that one of the predicates, P or Q, is universally true. The second statement says that for every x, one of P or Q hold. In order to show that they are indeed different lret's give a counterexample; consider the domain of discourse the positive integers; let P(x) x is odd; Q(x) x is even. While !(x) (P(x) ' Q(x)) is true; since every number is odd or even, the statement !x P(x) '!x Q(x) is false (it is not the case that every number is odd or every number is even. 5. For each of these arguments, explain which rules of inference are used for each step. a) Linda, a student in this class, owns a red convertible. Everyone who owns a red convertible has gotten at least one speeding ticket. Therefore, someone in this class has gotten a speeding ticket. Let c(x) be "x is in this class"; r(x) be "x owns a red convertible" and let t(x) be "x has gotten a speeding ticket". We are given the premise c(Linda); r(Linda); !x (r(x) $ t(x))) and we want to conclude "x (c(x) %t(x)). Step 1 !x (r(x) $ t(x))) 2 r(Linda) $t(Linda) 3 r(Linda) 4 t(Linda) 5 c(Linda) 6 c(Linda) %t(Linda) 7 "x(c(x) %t(x)) Reason Hypothesis Universal instantiation using (1) Hypothesis Modus ponens using (2) and (3) Hypothesis Conjunctions using (4) and (5)] Existential generalization using (6)
b) There is someone in this class who has been in Barcelona. Everyone who goes to Barcelona visits Picasso Museum. Therefore someone in this class has visited Picasso Museum. Let c(x) "x is in this class"; b(x) " x has been to Barcelona"; p(x) "x has been to the Picxasso Museum". We are given premises "x (c(x) %b(x)), !x (b(x) $p(x)) and we want to conclude "x (c(x) %p(x)). In our proof y represents an unspecified particular person. Step 1 "x (c(x) %b(x)), 2 c(y) % b(y) 3 b(y) 4 c(y) 5 !x (b(x) $p(x)) 6 b(y) $p(y) 7 p(y) 8. c(y) % p(y) 9 "x (c(x) %p(x)). Reason Hypothesis Existential instantiation using (1) Simplification using (2) Simplification using (2) Hypothesis Universal instantiation using (5) Modus ponens using (3) and (6) Conjunction using (4) and (7) Existential generalization using (8)
5. Use resolution refutation to prove: (Q ! P) ) ! ((Q ) !P) ! Q) {(Q ! ~p ) ! [( Q ! P) ! ~Q]} Negate it: ~{(Q ! ~p ) ! [( Q ! P) ! ~Q]} Convert to CNF ~{ ~(~Q or ~p) or [ ~ ( ~Q or P) or ~Q]} ~{ (Q and P) or [ (Q and ~P) or ~Q] } ~ (Q and P) and ~ (Q and ~P) and Q (~Q or ~ P ) and (~Q or P) and Q Apply resolution: (1) (2) (3) (4) (~Q or ~ P ) (~Q or P) (Q) ~Q ( Resolving 1 and 2 )
(5) Nil (resolving 3 and 4)