ON1 Big Mock Test 4 October 2013 Answers This test counts 4% of assessment for the course. Time allowed: 10 minutes

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1 ON1 Big Mock Test 4 October 2013 Answers This test counts 4% of assessment for the course. Time allowed: 10 minutes The actual test will contain only 2 questions. Marking scheme: In each multiple choice question, 2 marks are given for a complete correct answer, 1 mark for an incomplete correct answer, 0 for an incorrect or partially incorrect answer or no answer. A correct answer might contain more than one choice. Tick the correct box (or boxes): 1. Which of the following subsets of R is non-empty? (A) [2, 3] [0, 2] (B) ( 1, 1) ( 2, 2) (C) { 1, 1} { 2, 2} (D) None of the above Solution: (A) and (B). Indeed, [2, 3] [0, 2] = {2} =, ( 1, 1) ( 2, 2) = ( 1, 1), but { 1, 1} { 2, 2} =. 2. Which of the following sets is empty? (A) [1, 2] [2, 3] (B) (1, 2) (2, 3) (C) {1, 2} {2, 3} (D) All of the above Solution: (B). Two other sets are not empty: [1, 2] [2, 3] = {2} and Boolean algebra of sets {1, 2} {2, 3} = {2}. 3. Let X, Y and Z be sets such that X Y. Which of the following must be true? (A) X Z Y Z (B) X Y (C) X (Y Z) = X (Y Z)

2 Solution: (A) Use the following Venn diagram: 4. Let X, Y and Z be sets such that Y X. Which of the following must be true? (A) X Z Y Z (B) Y X (C) Y Z X Z (D) X (Y Z) = Y (X Z) Solution: (D). Indeed, by distributivity since Y X, we have X Y = Y and X (Y Z) = (X Y ) (X Z); X (Y Z) = Y (X Z). The falsity in the general situation of (A), (B), and (C), and necessary validity of (D) can be be easily seen from a Venn diagram (similar to the one in the solution to the previous question) which reflects the assumption that Y X. Try to do that! 5. Let X, Y and Z be sets such that Y X. Which of the following must be true? (A) Y X (B) Y X (C) Y Z X Z Solution: (C). Draw Venn diagrams. 2

3 6. Let X and Y be subsets of an universal set U such that Y X. Which of the following can never be true? (A) Y X (B) X Y (C) X Y (D) None of the above Solution: (D). Indeed (A) and (C) are both true when Y = X (actually, (C) is always true when Y X). (B) is true when Y X. Since each of (A), (B), (C) can happen be true, (D) is the answer. If you are still feel confused, consider the following rewording of the problem which is equivalent to the original one, that is, retains exactly the same meaning: Assume that X and Y are subsets of an universal set U such that Y X. Which of the following statements (A), (B), or (C) will always contradict the assumptions made? (A) Y X (B) X Y (C) X Y (D) None of the above 7. [Omitted, repeated the previous problem] Tautologies 8. Which of the following statements is a tautology? (A) (p q) ( p q) (B) (p q) ( p q) (C) (q p) ( p q) Solution: (A). There are several approaches to solution. The first two are universal they work for every formula. (I) The simplest (but not the shortest) approach just make truth tables for each of the three formulae. There is no need to compute the entire table; as soon as you got truth value F for the formula for some combination of values of p and q, you know 3

4 that the formula is not a tautology. (II) Similarly, you can use logical equivalences. For example, for (C) you have (q p) ( p q) ( q p) ( p q) ( q p) (p q) ( q p) ( q p) q p. But q p is not a tautology (why?). (III) Occasionally, a direct inspection of the formula yields the answer. For example, to see that (A) is a tautology it suffices to observe that one of the statements p and p is false, hence the corresponding conditional statement p q or p q is true; therefore, the disjunction (p q) ( p q) of two conditionals is true. (IV) Sometimes, a formulation of a statement in natural language helps. For example, to see that (B) is not a tautology, make a sentence of the same logical construction (p q) ( p q), say Apple is red and sweet or apple is not red and not sweet. But an apple can happen to be golden and sweet, thus making the sentence false. Hence (B) is not a tautology. Of course, in a multiple choice test you can use any method that works. 4

5 Logical equivalencies 9. Which of the following statements is NOT logically equivalent to (p p) (p q)? (A) (p q) (p p) (B) q (p q) (C) (q q) (q r) (D) None of the above Solution: (A), because it is F when p and q are T, while all other statements are tautologies. 10. Which of the following statements is logically equivalent to p (p q)? (A) p (p q) (B) q (p q) (C) (D) p q None of the above Solution: (B), because the formulae p (p q) and q (p q) are both tautologies, while the formulae p (p q) and are not. p q 5

6 Predicates 11. For real numbers x and y, let p(x, y) denote the predicate x > y + 1. Which of the following statements is true? (A) p(2, 1) p(1, 1) (B) p(0, 0) p(1, 2) (C) p(π, 1) p(1, π) (here π = ). Solution: (A), Because p(2, 1) is F. 12. For a real number x, let p(x) denote the predicate x > 1 and let q(x) denote the predicate x {0, 1, 2}. Which of the following statements is true? (A) p(1) q( 1) (B) p(1) p( 1) (C) (p(2) q(2)) Solution: (B), because (A), (B), and (C) become, correspondingly, T F, T F, and (T T ); of these, only (B) is T. 13. For a real number x, let p(x) denote the predicate x 2 > 0 x 0 and let q(x) denote the predicate x {0, 1, 2}. Which of the following statements is true? (A) p(1) q( 1) (B) p(1) p( 1) (C) (p(2) q(2)) (D) None of the above. Solution: (D). First of all notice that p(x) is T for all real x, while q( 1) is F and q(2) is T. Therefore statements (A), (B), and (C) become (A) T F, (B) T T, and (C) (T T ), none of which is T. 6

Philosophy 220. Truth-Functional Equivalence and Consistency

Philosophy 220. Truth-Functional Equivalence and Consistency Philosophy 220 Truth-Functional Equivalence and Consistency Review Logical equivalency: The members of a pair of sentences [of natural language] are logically equivalent if and only if it is not [logically]