203 EECS Homework 2
Section 1.3
1. (E) 6bef and 10cde 6a) Every student in school has visited North Dakota 6b) Not every student in the school has visited North Dakota 6c) No student in the school has visited North Dakota 10c) x(C(x) F (x) D(x)) 10d) x(C(x) D(x) F (x)) 10e) (xC(x)) (xD(x)) (xF (x)) 2. (M) 42abc 42a) Let A(x) mean "user x has access to an electronic mailbox" xA(x) 42b) Let S(x, y) mean "system x is in y state" xS(f ilesystem, locked) = A(x) 42c) Using prepositional function from part 42b S(f irewall, diagnostic) = S(proxyserver, diagnostic) 3. (C) 44 We want P and Q that are sometimes false and sometimes true. Let P(x) mean "x is even" Let Q(x) mean "x is multiple of 3" for two values in domain of integers, x = 4 and x = 9. (x(P (x)) (x(Q(x))) reduces to F alse F alse and hence true. But x(P (x) Q(x)) is false. Hence two are not logically equivalent.
Section 1.4
4. (E) 6def 6d) Some student is enrolled in both Math222 and CS252. 6d) There are two distinct students such that second one is enrolled in all the courses that first one is enrolled in. 6d) There are two distinct students that are enrolled in the same courses. 5. (M) 10hij 10h) y((x(F (x, y)) (z(wF (w, z) = z = y))) 10i) xF (x, x) 10j) xy(x = y F (x, y) z((F (x, z) z = x) = z = y)) (We don't assume the sentence is asserting that person can or can't fool him/herself) 6. (E) 16cd 16c) Let P(s, c, m) denote that student s has class standing c and is majoring in m. Domain of s is all students in class, of c all four standings and of m all possible majors. scm(P (s, c, m) c = junior m = mathematics).
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True because there is a sophomore majoring in computer science. 16d) x(mP (x, sophomore, m) cP (x, c, computer)). False, as there is a freshman mathematics major. 7. (M) 22 pabc(p > 0 p = (a2 + b2 + c2 )) 8. (E) 24bcd 24b) A nonnegetive number minus a negative number is positive. 24c) Difference of two nonpositive numbers is not necassarily nonpositive. 24d) The product of two numbers is non-zero if and only if the two numbers are non zero. 9. (E) 36d 36d) Let S(x, y) mean that student x solved problem y. Let domain of x be all students in this class and of y, all exercises in this book. Then statement means xyS(x, y). Negation is xyS(x, y). In English, for every student there is some exercise that he or she has not solved.
Section 1.5
10. (M) 14acd 14a) Let P(x) mean that "x is in the class". Q(x) mean "x has got a convetible". R(x) mean "x has gotten a speeding ticket". 1)x(Q(x) = R(x)) Premise = 2)Q(Linda) R(Linda) Universal instantiation of 1 3)P(Linda) Premise 4)Q(Linda) Premise 5)R(Linda) Modus Ponens using 2 and 4 6)P (Linda) R(Linda) Conjunction using 3 and 5 7)xP (x) R(x) Existential quantification from 6 14c) Let P(x) mean "x was a movie produced by John Sayles"; Q(x) mean "movie x was about Coal Miners"; R(x) mean that "movie x is wonderful". Let domain be all movies and a be the unspecified movie. 1)x(P (x) = R(x)) Hypothesis 2)x(P (x) Q(x)) Hypothesis 3)P (a) Q(a) Existential Instantiation from 2 4)P (a) = R(a) Universal Instantiation from 1 5)P(a) Simplification from 3 6)Q(a) Simplification from 3 7)R(a) Modus ponens with 4 and 5 8)Q(a) R(a) Conjunctions using 6 and 7 9)xQ(x) R(x) Existential generalization from 8 14d) Let P(x) mean "x is in the class"; Q(x) mean "x has been to france"; R(x) mean "x has visited Louvre". Let domain be all people and a be the unspecified person.
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1)x(P (x) Q(x)) 2)x(Q(x) = R(x)) 3)P (a) Q(a) 4)Q(a) = R(a) 5)Q(a) 6)R(a) 7)P(a) 8)P (a) R(a) 9)x(P (x) R(x))
Premise Premise Existential instantiation from 1 universalll instantiation from 2 Simplification from 3 Modus ponens with 4 and 5 Simplification from 3 Conjunction of 6 and 7 Existential generalization from 8
11. (E) 18 We know some s exists that makes the existential in premise true but we can't conclude that this s is Max. 12. (E) 24 24) Steps 3 and 5 are wrong as the simplification doesn't apply to disjunctions. 13. (M) 28 1)x(P (x) Q(x)) 2)x((P (x) Q(x)) = R(x)) 3)P (a) Q(a) 4)((P (a) Q(a)) = R(a)) 5)(P (a) Q(a)) R(a) 6)(P (a) Q(a)) R(a) 7)(Q(a) P (a)) R(a) 8)Q(a) (P (a) R(a)) 9)P (a) (P (a) R(a)) 10)(P (a) P (a)) R(a) 11)P (a) R(a) 12)R(a) = P (a) 13)x(R(x) = P (x))
premise premise universal instantiation from 1 universal instantiation from 2 simplification of implification De Morgan's and simplification Commutativity Associativity Resolution on 3 and 8 Associativity Idempotence Implication universal generalization
14. (C) 34bd 34b) Let l mean "logic is difficult"; m mean "mathematics is not easy";s mean "many students like logic". Then: 1)l s Premise 2)m = l Premise To check m = s It is false when m is false and s is true. If s is true then l has to be true for first premise to hold. If l is true then second premise is true. So premises hold even when this statement is false. So we can't conclude it. 34d) This is equivalent to m l. If we take contrapositive of premise 2 and simplify we get the same. So this is true.
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