1) Let h = John is healthy, w = John is wealthy and s = John is wise Write the following statement is symbolic form

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1 Math 378 Exam 1 Spring 2009 Show all Work Name 1) Let h = John is healthy, w = John is wealthy and s = John is wise Write the following statement is symbolic form a) In order for John to be wealthy it is necessary that he is wise. ~s ~ w (or w s) b) John is wealthy only if he is healthy. ~h ~w (or w h) 2) Show the logical equivalence of the following statement in TWO ways. A) by truth table be sure to mention why your truth table actually shows an equivalence. B) by using the rules of logical equivalence given in the text. ( p ( p q) q p q ~p ~q p ~q ~p ~q (p ~q) (~ p ~q) ~q T T F F T F F F T F F T T T T T F T T F F T F F F F T T T T T T * * * The last two columns have the same truth values so they are equivalent. ( p (~ p ( p ~ q Distributive Law c ~ q Negation Law ~ q Identity Law 3) Determine if the following are logically equivalent. You may use either method you wish, but be sure to give a reason for your answer. (note: You did not have to have all the reasons correct for full credit.) ( p q) (~ p ( p ) and t ( p q) (~ p ) (( p q) associative a ( b c) ( a b) c [( p ( q ] distributive [ t ( q ] negation law ( q identity law ( q ~ p p) ( q ~ p distributive ( q ~ p p) ( q ~ q associative on second group ( q t) ( t negation law twice t t idempotent law t identity

2 4) Rewrite p q using only,, and ~ One possibility from the definition: ( q ( p A second possibility: ( p q) (~ p (both are true or both false) 5) Write the following statements in if-then form. a) Catching the 8:05 bus is a necessary condition for my being on time for work If I don t catch the 8:05 bus then I won t be on time for work. If I am on time for work then I caught the 8:05 bus. b) A sufficient condition for Jon s team to go to the playoffs is that it wins its next game. If Jon s team wins its next game then it will go to the playoffs. 6) You are on the island of knight in knaves where Knights always tell the truth and Knaves always lie. Two natives, Mel and Belle, approach you. Belle says Mel and I are not the same. Mel tells you that Of Belle and I exactly one is a knight. What are Mel and Belle? Suppose Mel is a knight. What Mel says is true. Therefore by what Mel says, Belle must be a knave. Thus what Belle says must be a lie. But Belle tells the truth when she says Mel and I are not the same. Contradiction! Therefore Mel is a knave. What Mel says is false. Therefore it is not the case that Exactly one of them is a knight. Thus there are either two knaves or two knights. Since Mel is a knave we must also have that Belle is a knave. Note: If you start with one a knave and don t get a contradiction you must still show that the other assumption leads to a contradiction. Alternate proof : Mel Belle Comment T T Contradicts both Mel and Belle s statements since they would both be lying. T F Contradiction to what Belle said. Belle is supposed to be lying. F T Contradiction to what Mel said. Mel is supposed to be lying. F F OK. Mel and Belle are both lying (as knaves are supposed to do). Mel and Belle are both Knaves. 7) Express the following statement using both and quantifiers. In every forest there is a Hobbit. Forests F, Hobbit H such that H is in F F { forests}, H {Hobbits} such that H is in F

3 8) Let D be the set of all students at WVU Tech. Let M(s) be s is in Calc I, let C(s) be s is in Comp Sci I, and let E(s) be s is in English II. Express the following statements using quantifiers, variables and the above predicates. Note: M(s), C(s), and E(s) are predicates NOT SETS a) Some students in English II are also in Calc I. s D such that E(s) M(s) b) The negation of All students in Calc I are in Comp Sci I. Original Statement s D, M(s) C(s) s D, (C(s) ~M(s) (getting rid of to help negation later) Negation: s D such that ~(C(s) ~M(s)) Better: s D such that ~C(s) M(s) (uses DeMorgan to get rid of extra negative) 9) Consider the statement : animals A, if A is a Duck then A quacks and A waddles. USE DE MORGANS where applicable. Solutions: First think: a) Write the contrapostive. animals A, D(A) Q(A) W(A) animals A, ~[ Q(A) W(A)] D(A) now use De Morgan animals A, ~ Q(A) ~W(A) D(A) animals A, If A does not quack or A does not waddle then A is not a duck b) Write the converse. animals A, Q(A) W(A) D(A) animals A, If A quacks and A waddles, then A is a Duck c) Write the inverse. animals A, ~D(A) ~[ Q(A) W(A)] now use De Morgan animals A, ~D(A) ~ Q(A) ~W(A) animals A, If A is not a Duck then A does not quack or A does not waddle 10) Use symbols to write the logical form of each argument, and determine whether the argument is valid or invalid. Justify your answer (Valid by universal modus ponens or universal modus tollens; invalid by the converse or inverse error, etc) If this number is larger than 2, then its square is larger than 4. This numbers square is larger than 4. Therefore this number is larger than 2. Invalid. Converse error. p q, q therefore p.

4 11) Indicate whether the argument below is valid or invalid. Support you answer with a diagram. No good food is wasted. No college cafeteria food is good. No college cafeteria food is wasted. Good Food Wasted Food Invalid. Inverse Error College Cafeteria Food 12) Reorder (unjumble) the following premises so that the argument makes logical sense. (Ie. give the correct order a,c,e,b,d etc.) Note: this is not a proof, but it has the structure of a proof. a) If m = 2r and n = 2s then m + n = 2r + 2s = 2(r + s). b) If r and s are integers then r + s is an integer. c) m and n are even integers. d) If r + s is an integer then 2(r + s) is even. e) If m and n are even then m = 2r and n = 2s for some integers r and s. m + n = 2(r +s) is even c) Given fact: m and n are even integers e) combine with c m = 2r and n = 2s for some integers r and s. (modus ponens) a) combine with e m + n = 2(r + s) (modus ponens) b) combine with a r + s is an integer (modus ponens) d) combine with b 2(r + s) is even (modus ponens) combine a and d to get conclusion. (transitivity) Only answer REQUIRED was c, e, a, b, d

5 13) Prove or find a counter example. a) The sum of two odd integers is even. Proof: Suppose m and n are two particular but arbitrarily chosen odd integers. By definition of odd we know m = 2j + 1 and n = 2k + 1 for some integers j and k. Hence by substitution and simplification we see that m + n = (2j + 1) + (2k + 1) = 2j + 2k + 2 = 2(j + k + 1). But j + k + 1 is an integer since it is the sum of integers. Let p be the integer j + k + 1. Thus m + n = 2p where p is an integer. Hence m + n is even by definition. Q.E.D. b) If p and q are rational numbers then (p + q)/ 2 is rational. Proof: Suppose p and q are rational numbers. By definition, since p and q are rational numbers, there exist integers a, b, c, and d with b 0 and d 0 such that p = a/b and q = c/d. Thus by substitution and simplification ad + bc p + q a / b + c / d bd ad + bc = = =. But ad + bc and 2bd are integers since they are the bd product and sum of integers. In addition 2bd 0 since neither b nor d equal zero. Let m be the integer ad + bc and let n be the non-zero integer 2bd. Then by definition, since (p+q)/2 = m/n, we know (p + q)/2 is rational.

6 c) BONUS: If p and q are rational numbers and p < q, then (p + q)/2 < q.