# First Midterm Exam

### Directions:

• Answer the problems on the exam pages.
• There are five problems on pages 2-6, for 100 total points plus 10 extra credit. Probable scale is somewhere around A=93, C=69.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.

```  Q1: 20 points
Q2: 20 points
Q3: 20 points
Q4: 20 points
Q5: 20 points plus 10 extra credit
Total: 100 points plus 10 extra credit
```

Questions 1, 2, and 3 deal with a set T of terriers including two specific terriers Angus (a) and Barney (b). If x is a terrier, the predicate C(x) means "x likes to chase squirrels" and D(x) means "x likes to dig". If x and y are two terriers, SS(x,y) means "x and y are the same size". You are given that SS is an equivalence relation.

• Question 1 (20): Translate the following statements as indicated:
• Statement I: (to English) (C(b) ∧ D(a)) → (D(b) ∧ ¬D(b))

If Barney likes to chase squirrels and Angus likes to dig, then Barney both does and does not like to dig.

• Statement II: (to symbols) If either Angus likes to chase squirrels or Barney likes to dig, or both, then Angus likes to dig.

(C(a) ∨ D(b)) → D(a)

• Statement III: (to English) ¬((C(a) → C(b)) ∧ (C(b) → C(a)))

It is not the case that both "if Angus likes to chase squirrels then so does Barney" and "if Barney likes to chase squirrels then so does Angus" are true. (Equivalently: Either Angus or Barney, but not both, likes to chase squirrels.)

• Statement IV: (to symbols) Given any terrier, there is a terrier of the same size that likes to dig.

∀x:∃y: SS(x,y) ∧ D(y)

• Statement V: (to English) ∀x: D(x) → ¬C(x)

If any terrier likes to dig, it does not like to chase squirrels.

• Statement VI: (to symbols) There are two terriers of the same size such that one likes to chase squirrels and the other doesn't.

∃x:∃y: SS(x,y) ∧ C(x) ∧ ¬C(y)

• Question 2 (20): Using Statements I, II, III, and V from Question 1, determine whether each of the propositions C(a), C(b), D(a), and D(b) are true or false. Check that given your answers, Statements I, II, and III are true, and that V is true in so far as it applies to Angus and Barney. (You may use truth tables or a deductive sequence proof.)

We can solve this with a sixteen-line truth table, with one line for each of the sixteen four settings of the four boolean variables C(a), C(b), D(a), and D(b). We would find that only one of these settings makes I, II, III, and the two specifications of V true.

But an equational sequence proof is easier. Since III tells us that either C(a) or C(b) is true, but not both, let's use Proof By Cases on C(a).

Assume C(a). By Specification from V, D(a) → ¬C(a), so by Modus Tollens (or Contrapositive followed by Modus Ponens) we get ¬D(a). Again by Modus Tollens or otherwise from II, we get ¬(C(a) ∨ D(b)), which by DeMorgan is ¬C(a) ∧ ¬D(b). But ¬C(a) forms a contradiction with the assumption C(a), so the assumption must be false. We have proved ¬C(a) by Contradiction.

From ¬C(a), C(b) follows from III (by Definition, DeMorgan, and Definition of Equivalence, III can be rewritten as C(a) ⊕ C(b)). By Specification on V, we have D(b) → ¬C(b), and C(b) therefore gives us ¬D(b) by Modus Tollens. Statement I says (C(b) ∧ D(a)) → 0 by Excluded Middle, and thus implies ¬(C(b) ∧ D(a)) which is equivalent to ¬C(b) ∨ ¬D(a) by DeMorgaon. Since C(b) is true, D(a) must be false.

We have found that C(b) is true and the other three booleans are false. We need to check that all our statements hold. I is true vacuously because C(b) ∧ D(a) is false. II is true vacuously because C(a) ∨ D(b) is false. III is true vacuously because C(b) → C(a) is false. The two Specifications of V are also both true vacuously because D(a) and D(b) are false.

• Question 3 (20): Using Statements I, II, III, IV, and V from Question 1, and any facts you were able to deduce from Question 2, prove Statement VI. Since both some of your premises and your conclusin involve quantifiers, you will need some or all of the quantifier rules of Existence, Instantiation, Specification, and Generalization.

We know from Question 2 that Angus and Barney are two terriers such that one (Barney) likes to chase squirrels and the other doesn't, but there is no reason to think that Angus and Barney are the same size. The only way to guarantee the existence of terriers of the same size is through IV. But how should we specialize IV?

If we specialize IV to some terrier x, we will get a terrier y of the same size as x, such that D(y) is true. Then from Specification of V we would get D(y) → ¬C(y) and thus ¬C(y). This will be enough to prove VI if our x satisfies C(x). So we want x to be Barney, and the proof goes:

1. Specify IV to b, getting ∃y:SS(b,y) ∧ D(y).
2. Let t be a terrier satisfying line 1, so that SS(b,t) ∧ D(t).
3. Specify V to t, getting D(t) → ¬C(t)
4. By Separation from line 2 get D(t).
5. By Modus Ponens from lines 4 and 3 get ¬C(t).
6. Recall that in Question 2 we proved C(b).
7. By Separation and Conjunction on lines 2, 6, and 5 get SS(b,t) ∧ C(b) ∧ ¬C(t).
8. By Existence, twice, get ∃x:∃y: SS(x,y) ∧ C(x) ∧ ¬C(y).

• Question 4 (20): In this problem, Amy, Bob, and Claire are three students who each have zero or more majors taken from the set {Dance, English, French}. We write M(x,y) to mean "student x has major y" (possibly with other majors as well). You are given that M(a,d) and M(a,e) are false, and that M(b,d) and M(c,e) are true. Recall that a relation R from one set to another is a function if and only if it is both total (∀x:∃y: R(x,y)) and well-defined (∀x:∀y:∀z:[R(x,y) ∧ R(x,z)] → (y=z)).

• (a,5) If we are given, in addition to the four truth values above, that M is a function, deterine the truth values of the five propositions M(x,y) that are not given above.

M(a,f) must be true for M to be total, as a does not map to d or e. The other four values must be false for M to be well-defined: M(b,e) and M(b,f) must be false because M(b,d) is true, and M(c,d) and M(c,f) must be false because M(c,e) is true.

• (b,5) If M is a function as in (a), is it a surjection (and onto function)? Is it an injection (a one-to-one function)? Explain your answer indicating that you understand the relevant definitions.

Yes, it is both one-to-one and onto. It is one-to-one because each domain value maps to only one range value, and it is onto because each of the three range values is mapped to by at least one domain value.

• (c,5) Give truth values for the five propositions not given above that make M a total relation that is not well-defined. Explain why your values do this.

The simplest thing is to make all five values true. Then we are total because we have M(a,f), M(b,d), and M(c,e), and we are not well-defined because we have both M(b,d) and M(b,e), for example.

• (d,5) Give truth values for the five propositions not given above that make M a well-defined relation that is not total. Explain why your values do this.

The simplest thing is to make all five values false. Then we are not total because there is no y such that M(a,y). But we are well-defined because we never have any domain element mapped to more than one range element -- a, b, and c are mapped to zero, one, and one respectively.

• Question 5 (20+10): Let X be an arbitrary set and let R and S be two arbitrary partial orders on X. That is, R is:

• Reflexive: ∀x: R(x,x)
• Antisymmetric: ∀x:∀y: [R(x,y) ∧ R(y,x)] → (x=y)
• Transitive: ∀x:∀y:∀z: [R(x,y) ∧ R(y,z)] → R(x,z)

• (a,20) Let T be the binary relation defined by T(x,y) ↔ (R(x,y) ∧ S(x,y)) (that is, T is the intersection of R and S). Prove that T is a partial order by carefully verifying each of the three properties above using quantifier rules and the fact that R and S are partial orders.

T is reflexive: Let x be arbitrary. R(x,x) and S(x,x) are true by Specification on the reflexivity of R and S respectively. So T(x,x) is true by the definition of T. Since x was arbitrary, we have proved ∀x:T(x,x).

T is antisymmetric: Let x and y be arbitrary. Assume T(x,y) ∧ T(y,x). By definition of T and Separation, we have R(x,y) and R(y,x). By Specification on the antisymmetry of R and Modus Ponens, we have that x=y. We have proved (T(x,y) & T(y,x)) → (x=y) by direct proof. Since x and y were arbitrary, we have proved ∀x:∀y:(T(x,y) ∧ T(y,x)) → (x=y).

T is transitive: Let x, y, and z be arbitrary. Assume T(x,y) ∧ T(x,z). By the definition of T and Separation we get R(x,y), S(x,y), R(y,z), and S(y,z). By Specification on the transitivity of R to x, y, and z, and Modus Ponens, we get R(x,z). Similarly from the transitivity of S we get S(x,z). Then we get R(x,z) ∧ S(x,z) by Conjunction, and this is the definition of T(x,z). We have shown (T(x,y) ∧ T(y,z)) → T(x,z) by direct proof. Since x, y, and z were arbitrary, we have proved the transitivity statement by Generalization.

• (b,10 XC) Let U be the union of R and S, so that U(x,y) ↔ (R(x,y) ∨ S(x,y)). Is U necessarily a partial order? For each of the three properties, either prove that the property holds for U given any partial orders R and S, or indicate how it could fail for U on some particular partial orders R and S.

U is reflexive: Let x be arbitrary. Since we have R(x,x) by Specification on the relexivity of R, we can get R(x,x) ∨ S(x,x) by Joining, and this is U(x,x). Since x was arbitrary, we have proved ∀x:U(x,x) by Generalization.

U need not be antisymmetric: Consider a universe with only two elements a and b, where only the reflexive statements, R(a,b), and S(b,a) are true. (Here R and S are each still antisymmetric and transitive.) Now U(a,b) and U(b,a) are both true but a is not equal to b.

U need not be transitive: Consider a universe with only three elements a, b, and c, where only the reflexive statements, R(a,b), and S(b,c) are true. Then U(a,b) and U(b,c) are both true, but U(a,c) is false.