Tutorial 2 Logic of Quantified Statements

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1 Tutorial 2 Logic of Quantified Statements 1. Let V be the set of all visitors to Universal Studios Singapore on a certain day, T (v) be v took the Transformers ride, G(v) be v took the Battlestar Galactica ride, E(v) be v visited the Ancient Egypt, and W (v) be v watched the Water World show. Express each of the following statements using quantifiers, variables, and the predicates T (v), G(v), E(v) and W (v). The statements are not related to one another. a. Every visitor watched the Water World show. b. Every visitor who took the Battlestar Galactica ride also took the Transformers ride. c. There is a visitor who took both the Transformers ride and the Battlestar Galactica ride. d. No visitor who visited the Ancient Egypt watched the Water World show. e. Some visitors who took the Transformers ride also visited the Ancient Egypt but some did not (visit the Ancient Egypt). a. v V, (W (v)). b. v V, (G(v) T (v)). (In words: For any v V, if G(v) then T (v).) c. v V such that (T (v) G(v)). d. v V, (E(v) W (v)). (Alternatively: v V, ( E(v) W (v)).) e. ( v V such that (T (v) E(v))) ( u V such that (T (u) E(u))). Note that the above is a conjunction (AND) of two existential clauses. Each existential variable v or u has a scope only within its own clause. Thus, there is no ambiguity to use the same variable in both clauses, as in: ( v V such that (T (v) E(v))) ( v V such that (T (v) E(v))). Nevertheless, to avoid confusion, using different variable names is still preferred. Also, note that it is incorrect to write v V such that (T (v) E(v)) (T (v) E(v)). because the scope of v covers both T (v) E(v) and T (v) E(v). This means that there exists a person v who visited Transformer and Ancient Egypt, and the same person visited Transformer but not Ancient Egypt! 2. For each of the following statements write its negation, converse, inverse, and contrapositive. Indicate which among the statement, its converse, its inverse, and its contrapositive are true and which are false. Give a counterexample for each that is false. a. x R, if x(x + 1) > 0 then x > 0 or x < 1. b. x R, if x > 3 then x 2 > 9. a. Statement: x R, if x(x + 1) > 0 then x > 0 or x < 1. Negation: x R such that x(x + 1) > 0 and x 0 and x 1. Page 1

2 Converse: x R, if x > 0 or x < 1 then x(x + 1) > 0. Inverse: x R, if x(x + 1) 0 then x 0 and x 1. Contrapositive: x R, if x 0 and x 1 then x(x + 1) 0. The statement, its converse, inverse and contrapositive are all true. b. Statement: x R, if x > 3 then x 2 > 9. Negation: x R such that x > 3 and x 2 9. Converse: x R, if x 2 > 9 then x > 3. Inverse: x R, if x 3 then x 2 9. Contrapositive: x R, if x 2 9 then x 3. The statement and its contrapositive are true. The converse and the inverse are false. A counterexample for the converse: Let x 2 = 25 > 9, but x = 5 3. A counterexample for the inverse: Let x = 5 3, then x 2 = Prove or disprove the following statement: a, b, c Z, if a b is even and a c is even, then b c is even. Proof. (Direct proof) 1. For any a, b, c Z: 1.1. If a b is even and a c is even: Let s Z such that a b = 2s [by definition of even ] Let t Z such that a c = 2t [by definition of even ] Then b c = (a c) (a b) = 2t 2s = 2(t s) [by basic algebra] Also, (t s) Z [by closure of integers under subtraction] Therefore b c = 2k for some k Z. [We usually omit this line.] Thus b c is even [by definition of even ]. Alternate proof: Proof (Direct proof). 1. Take any integers a, b, c. 2. If a b is even and a c is even: 2.1. Then 2 a b [by definition of divisibility] Then 2 a c [by definition of divisibility] Then 2 (a b)( 1) + (a c)(1) [by Theorem Linear Combination] And thus 2 b c [by basic algebra] Therefore b c is even [by definition of divisibility]. 4. Refer to Figure (in page 120 of the book and in slide 51 of the lecture slides of The Logic of Quantified Statements ). Determine whether each of the following statements is true or false. a. students S, a dessert D such that S chose D. b. students S, a salad T such that S chose T. c. a beverage B such that students S, S chose B. d. an item I such that S, S did not choose I. e. a station Z such that students S, an item I such that S chose I from Z. Page 2

3 a. The statement in English: Every student chose a dessert. This is true. Every student chose at least one dessert: Uta chose pie, Tim chose both pie and cake, and Yuen chose pie. b. The statement in English: Every student chose a salad. False. Yuen did not choose any salad. c. The statement in English: There is a beverage chosen by all students. False. None of the beverages was chosen by all students. Milk was chosen by Uta and Tim, soda was chosen by Yuen, and coffee was chosen by Tim. d. The statement in English: There is an item that nobody chose. False. Every one of the nine items was chosen by at least one student. e. The statement in English: There is a station from which every student chose something. True. The station can be the main courses station, the desserts station or the beverages station. Every student chose at least an item from each of these three stations. 5. Some of the arguments below are valid by universal modus ponens or universal modus tollens; others are invalid and exhibit the converse error or the inverse error. State which are valid and which are invalid. Justify your answers. a. If a person likes coffee, he likes toffee. Jack does not like toffee. Jack does not like coffee. b. All positive odd integers greater than 1 are primes. 4 is not a positive odd integer greater than 1. 4 is not a prime. c. Anyone above 18 and passed the driving test gets a driving licence. Sharon gets a driving licence. Sharon is above 18 and passed the driving test. a. Valid by universal modus tollens. b. Invalid; inverse error. c. Invalid; converse error. 6. Indicate which of the following statements are true and which are false. Justify your answers. a. a Z, b Z such that a + b = 0. (AY2015/6 Sem2 Midterm test) b. a Z such that b Z, a + b = 0. (AY2015/6 Sem2 Midterm test) c. x R +, y R + such that their arithmetic mean is equal to their geometric mean. a. True. Proof. (by Construction) a.1. For any a Z: a.1.1. Let b = 1 a = a. a.1.2. Then b Z [by closure property of integers under multiplication]. a.1.3. Then a + b = a + ( a) = 0 [by basic algebra]. b. False. Page 3

4 Proof. (Counter-example) b.1. Take any a Z. b.2. Let b = a + 1. b.3. b Z [by closure property of integers under multiplication and addition]. b.4. Thus a + b = a + ( a + 1) = 1 0 [by basic algebra]. c. True. Proof. (by Construction) c.1. For any x R + : c.1.1. Let y = x. c.1.2. Arithmetic mean of x and y = (x + y)/2 = 2x/2 = x. c.1.3. Geometric mean of x and y = xy = x 2 = x. c.1.4. Thus arithmetic mean is equal to geometric mean. Note: The proof in (c) uses Construction. But instead of finding a fixed number for y (which you can t because it depends on x), you are showing a strategy for finding y which will work for any x. This technique is also used in the proof for (a). 7. Given the following argument: 1. If an object is above all the triangles, then it is above all the blue objects. 2. If an object is not above all the gray objects, then it is not a square. 3. Every black object is a square. 4. Every object that is above all the gray objects is above all the triangles. If an object is black, then it is above all the blue objects. a. Reorder the premises in the argument to show that the conclusion follows as a valid consequence from the premises. It may be helpful to rewrite the statements in if-then form and replace some (if any) statements by their contrapositives. b. Rewrite your answer in part (a) using predicates and quantified statements. a. 3. If an object is black, then it is a square. 2. (Contrapositive form) If an object is a square, then it is above all the gray objects. 4. If an object is above all the gray objects, then it is above all the triangles. 1. If an object is above all the triangles, then it is above all the blue objects. If an object is black, then it is above all the blue objects. b. Let O, the domain, be the set of objects. 3. x O, Black(x) Square(x). 2. (Contrapositive form) x, y O, Square(x) (Gray(y) Above(x, y)). 4. x, y, z O, (Gray(y) Above(x, y)) (T riangle(z) Above(x, z)). 1. x, z, w O, (T riangle(z) Above(x, z)) (Blue(w) Above(x, w)). x, w O, Black(x) (Blue(w) Above(x, w)). Page 4

5 8. (AY2015/6 Sem2 Midterm test) You are given the following English statements: 1. All swimmers are able to swim across the river. 2. No archers are short-sighted. 3. Patrick wears glasses. 4. Everybody is either an archer or a swimmer. a. Rewrite each of the above sentences into formal statements, using quantifiers wherever appropriate, and well-named predicates. You may assume that the domain is the set of people, which may be omitted in your statements. You may use the logically equivalent form in your statements. You may use Patrick as an instance instead of creating a predicate called Patrick. b. There is a missing statement above. Adding that missing statement would allow you to answer this question Is Patrick able to swim across the river? Write down the missing statement (as a formal quantified statement) and the conclusion (in English) about Patrick. c. Show your proof to derive the conclusion about Patrick in part (b) above. a. a.1. x, Swimmer(x) CrossRiver(x). [rule 1] a.2. x, ShortSighted(x) Archer(x). (or its contrapositive statement) [rule 2] a.3. W earglasses(p atrick). [fact 3] a.4. x, Archer(x) Swimmer(x). [rule 4] a.5. (Alternatively, to express the fact that one is either Archer or Swimmer but not both:) x, Archer(x) Swimmer(x). [rule 4 alt]. b. Missing statement: x, W earglasses(x) ShortSighted(x). [rule 5] Conclusion: Patrick is able to swim across the river. c. c.1. ShortSighted(P atrick). [fact 6: Universal Modus Ponens using fact 3 and rule 5] c.2. Archer(P atrick). [fact 7: Universal Modus Ponens using fact 6 and rule 2] c.3. Swimmer(P atrick). [fact 8: Elimination using fact 7 and rule 4 (or rule 4 alt)] c.4. CrossRiver(P atrick). [Universal Modus Ponens using fact 8 and rule 1] Page 5

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