CMPSCI 250: Introduction to Computation

Solutions to First Midterm Exam

David Mix Barrington

Posted 20 October 2010

• There were five problems for 100 total points plus 10 extra credit. Actual scale was A=90, C=60.
• No books, notes, calculators, or collaboration.
• Question text is in black, solutions in blue.

  Q1: 10 points
  Q2: 20 points
  Q3: 25 points
  Q4: 15 points
  Q5: 30+10 points
 Total: 100+10 points

• Questions 1 and 2 deal with a set of four dogs: Ace, Biscuit, Cardie, and Duncan, and a predicate B(x), where x is a dog, meaning "x is now barking".

• Question 1 (10): Translate the following four statements as indicated, using constants a, b, c, and d to represent the four dogs. (Note: As given, the test confused notation, using "a" to represent the boolean "B(a)" as well as the dog "a". The notation has been corrected here.)
  • Statement I (to English) (B(a) ⊕ B(b)) → B(c)

    If Ace or Biscuit is barking, but not both, then Cardie is barking.

  • Statement II (to symbols) If any of the other three dogs is barking, then so is Duncan.

    (B(a) ∨ B(b) ∨ B(c)) → B(d)

  • Statement III (to English) (B(d) ∨ ¬B(a)) → B(b)

    If Duncan is barking or Ace is not barking (or both of these are true), then Biscuit is barking.

  • Statement IV (to symbols) If Biscuit is barking and at least one of Cardie and Duncan is barking, then Ace is not barking.

    (B(b) ∧ (B(c) ∨ B(d))) → ¬B(a)

• Question 2 (20): Using Statements I, II, III, and IV from Question 1, determine exactly which of the four dogs are barking and which are not. Make sure that your answer satisfies all four of the statements. (You may use truth tables or a deductive sequence proof.)

  One way to do this was a truth table. But remember that to prove that a given set of truth values is the only solution that satisfies all the premises, you need to show not only that the line for that set of truth values has 1's for all four statements, but that each other set of truth values has 0 for at least one premise. If you have a column of your truth table for the compound proposition "I ∧ II ∧ III ∧ IV", this will be automatic as this column will have exactly one 1 and the rest 0's.

  Here we present a deductive sequence proof:

  1. Assume B(a).

  2. By Joining we get B(a) ∨ B(b) ∨ B(c).

  3. By Modus Ponens on statement II we get B(d).

  4. By Joining we get B(d) ∨ ¬B(a).

  5. By MP on Statement III we get B(b).

  6. By Joining from line 3 we get B(c) ∨ B(d).

  7. By Conjuction on lines 5 and 6 we get B(b) ∧ (B(c) ∨ B(d)).

  8. By MP on Statement IV we get ¬B(a).

  9. Lines 1 and 8 form a contradiction, so the assumption B(a) is disproved and we get ¬B(a).

  10. By Joining we get B(d) ∧ ¬B(a).

  11. By MP on Statement III we get B(b).

  12. Since we have ¬B(a) and B(b) we have B(a) ⊕ B(b) by the definition of ⊕.

  13. By MP on Statement I we get B(c).

  14. By Joining from either line 11 or line 13 we get B(a) ∨ B(b) ∨ B(c).

  15. By MP on Statement II we get B(d).

  16. We have proved that ¬B(a) ∧ B(b) ∧ B(c) ∧ B(d) is the only possible solution.

  17. And this setting does satisfy Statements I-IV, which we can see most easily by noting that it makes the conclusion of each statement true.

• Question 3 (25): In this problem we have a set of dogs D, a set of colors C, and a relation IC(x, y) meaning "dog x has color y". We also assume that D contains the two particular dogs Cardie and Duncan (and perhaps others), and that IC(Cardie, brown) and IC(Duncan, gray) are both true.

  • (a,5) Describe, in terms of dogs and colors, the two conditions necessary for IC to be a function.

    Each dog must have a color, and no dog may have more than one distinct color.

  • (b,10) Now assume that IC is a function, and assume that there are an equal number of dogs of each color. (That is, if e and f are any two colors in C, the number of dogs of color e equals the number of dogs of color f.) Prove or disprove: "If there are a prime number of dogs in D, then IC is an injection." (Recall that an "injection" is defined to be the same thing as a "one-to-one function".)

    This statement is true. The number of dogs, |D|, must equal the number of colors, |C|, times the number of dogs of each color, which we will call k. Thus |C| divides |D|. If |D| is prime, then, |C| must be either 1 or |D|. The case of |C| = 1 is ruled out because we are told that there are at least two colors, brown and gray. So |C| = |D| must be true, each color has exactly one dog, there are never two distinct dogs with the same color, and thus IC is an injection.

  • (c,10) With the same two conditions as in part (b), prove or disprove: "If IC is an injection, then the number of dogs in D is a prime number."

    This is false. Let |D| = |C| = 4 and let each dog have its own color. The other premises are satisfied, IC is an injection, and yet 4 is not prime. So the premise is true and the conclusion false, making the implication false.

• The remaining questions on the test deal with a set D of dogs that includes three particular dogs, Biscuit, Cardie, and Duncan, denoted by constant symbols b, c, and d, and perhaps other dogs as well. You are given that no two of the three named dogs equal each other. We have a binary relation SC(x, y) on D, defined to mean "dog x sometimes chases dog y". Note that it is possible for a dog to sometimes chase itself.

• Question 4 (15): Translate the following statements as indicated:

  • (a,5) Statement V (to English): ∃x: ∃y: ∀z: SC(x, z) ∧ SC(z, y)

    There exist two dogs such that for any third dog, the first sometimes chases the third and the third sometimes chases the second.

  • (b,5) Statement VI (to symbols): Any dog sometimes chases Biscuit if and only if it (that dog) is not Duncan, and (that dog) is sometimes chased by Biscuit if and only if it is not Cardie.

    ∀x: (SC(x, b) ↔ (x ≠ d)) ∧ (SC(b, x) ↔ (x ≠ c))

  • (c,5) Statement VII (to symbols): The relation SC is reflexive and transitive.

    [∀x: SC(x, x)] ∧ [∀x: ∀y: ∀z: (SC(x, y) ∧ SC(y, z)) → SC(x, z)

• Question 5 (30+10): Recall the following definitions:

  • SC is symmetric: ∀x: ∀y: SC(x, y) → SC(y, x)

  • SC is antisymmetric: ∀x: ∀y: [SC(x, y) ∧ SC(y,x)] → (x = y)

  • (a,10) Prove that if SC satisfies Statements V, VI, and VII, it is not an equivalence relation. (You may not need to use all three statements in your proof.)

    Statement VII says that SC is reflexive and transitive, so we will need to show that SC is not symmetric. Specify Statement VI to c, getting (SC(c, b) ↔ (c ≠ d)) and (SC(b, c) ↔ (c ≠ c)). Since we know that c ≠ d is true and that c ≠ c is false, we know that SC(c,b) is true and that SC(b, c) is false. By Existence, twice, we get that ∃x: ∃y: SC(x, y) ∧ ¬ SC(y, x), which is the negation of the definition of "SC is symmetric".

  • (b,10) Assuming that Statements V, VI, and VII are true, prove the following Statement VIII: ∀w: SC(w, d) ∧ SC(c, w).

    The overwhelming majority of you tried to prove VIII directly from V by the rule of Instantiation, not seeing that the arbitrary x and y from V cannot be declared to be equal to c and d without justification. This is on a par with "There exists a person who is the Pope. Therefore I am the Pope." Only one person had a full-credit proof for this part. Here is one, more formal than you needed to be:

    1. Let u be an arbitrary dog.

    2. By Specification from Statement VI, we get SC(u, b) ↔ (u ≠ d) and SC(b, u) ↔ (u ≠ c). (Corrected 21 October.)

    3. Assume u ≠ d.

    4. By Separation from line 2, and the definition of ↔, we get SC(u, d).

    5. By Specification (three times) on the Transitive part of VII, we get (SC(u, b) ∧ SC(b, d)) → SC(u, d).

    6. By Specification from VI to x = d and some boolean rules, we get SC(b, d).

    7. By Conjunction and by MP on line 6, we get SC(u, d).

    8. Now assume u = d.

    9. By Specification from the Reflexive part of VII, SC(d, d) and therefore SC(u, d).

    10. We have proved SC(u, d) by cases.

    11. Now assume u ≠ c.

    12. By Separation on line 2 and the definition of ↔, we get SC(b, u).

    13. By Specification (three times) on the Transitive part of VII we get (SC(c, b) ∧ SC(b, u)) → SC(c, u).

    14. By Specification from VI to x = c and some boolean rules, we get SC(c, b).

    15. By Conjunction and MP on line 13 we get SC(c, u).

    16. Now assume u = c.

    17. By Specification on the Reflexive part of VII, we get SC(c, c) and thus SC(c, u).

    18. We have proved SC(c, u) by cases.

    19. By Conjunction on lines 10 and 18 we get SC(c, u) ∧ SC(u, d).

    20. Since u was arbitrary, by Generalization we have ∀w: SC(c, w) ∧ SC(w, d).

  • (c,10) For this part only, assume that D contains a fourth dog, Ace (who is not equal to Biscuit, Cardie, or Duncan), perhaps along with other dogs. From this assumption and Statements V, VI, and VII, prove that SC is not a partial order.

    We need to prove that SC is not antisymmetric, since we know that it is reflexive and symmetric from Statement VII. Specify Statement VI to Ace, getting SC(a, b) (since a ≠ d) and SC(b, a) (since a ≠ c). Now we have SC(a, b) ∧ SC(b, a) ∧ (a ≠ b). By Existence, twice, we get ∃x: ∃y: SC(x, y) ∧ SC(y, x) ∧ (x ≠ y), which is the negation of the definition for "SC is antisymmetric".

  • (d,10) For this part only, assume that D contains only the three dogs Biscuit, Cardie, and Duncan. It turns out that in this case, Statements V, VI, and VII determine SC entirely in that they imply truth values for SC(x, y) for all nine possible pairs of dogs (x, y).

    Determine these nine values. (Hint: You may use Statement VIII even if you didn't get part (b) of this question.)

    By Specification from the Reflexive part of VII, we know that SC(b, b), SC(c, c), and SC(d, d) are all true. By Specification from VIII, we get that SC(b, d), SC(c, b), and SC(c, d) are all true. By Specification from VI we find that SC(d, b) and SC(b, c) are false. This leaves only the truth value of SC(d, c) to determine, and this is false because if it were true, specializing from transitivity would give us SC(d, b) (known to be false) from SC(d, c) and SC(c, b).

    Show that in this case SC is a partial order, and give its Hasse diagram.

    By Statement VII we have only to check antisymmetry, and we can do this by noting that exactly one of SC(b, c) and SC(c, b), exactly one of SC(b, d) and SC(d, b), and exactly one of SC(c, d) and SC(d, c) are true.

    The Hasse diagram has c at the top, b below it connected by an edge, and d below b connected by an edge. By inspection we can see that SC(x, y) is true if and only if x has a downward path to y in this diagram.

Last modified 21 October 2010