Logic and Reasoning Final Exam

Practice – Solutions

Name _________________________________ Section Number ________

The final examination is worth 100 points.

1. (10 points) What is an argument? Explain what is meant when one says that logic is the

normative study of arguments. Compare and contrast the deductive and inductive

standards for assessing the goodness of an argument.

An argument is a collection of sentences. Some of the sentences, called the premisses, are supposed to evidentially support one of the other sentences, called the conclusion. When one says that logic is the normative study of arguments (as opposed to a descriptive study), one means that logic is the study of what makes arguments good or bad. Logic is not the study of what arguments people find persuasive or of how people actually argue. On a deductive standard, an argument is good (valid) just in case whenever all of the premisses are true, the conclusion is guaranteed to be true. On an inductive standard, an argument might be good even though it is possible for the premisses to be true while the conclusion is false. For example, an argument might be inductively good if the conclusion is very likely (but not certain) to be true when the premisses are all true.

2. (5 points) Fill in the truth tables below, where (P % Q ) = ((P ∧ ~Q ) → (~P ∨ Q )):

P Q (P ∧ Q) (P ∨ Q) (P → Q) (P % Q)

T T T T T T

T F F T F F

F T F T T T

F F F F T T

3. (5 points) Use truth tables to decide whether the argument below is valid or invalid.

~(~P ∧ ~Q ) (R → ~P ) (Q → ~R )

~R

P Q R ~(~P ∧ ~Q ) (R → ~P ) (Q → ~R ) ~R

T T T T F F F

T T F T T T T

T F T T F T F

T F F T T T T

F T T T T F F

F T F T T T T

F F T F T T F

F F F F T T T

Since there is no row in which all of the premisses are true and the conclusion is false, the argument is valid. That is, every row in which all of the premisses are true is a row in which the conclusion is true.

4. In the following problems, you are asked to prove some alternative rules of inference.

a. (3 points) Double Negation Introduction { P } ⊢ ~~P [5 lines] 1 (1) P A 2 (2) ~P A* 1 (3) ~P → P 1 →I (4) ~P → ~P 2 CP 1 (5) ~~P 3,4 ~I

b. (3 points) Absorption { (P → Q ) } ⊢ (P → (P ∧ Q ))

[5 lines] 1 (1) P → Q A 2 (2) P ACP 1,2 (3) Q 1,2 →E 1,2 (4) P ∧ Q 2,3 ∧I 1 (5) P → (P ∧ Q ) 2,4 CP c. (4 points) Modus Tollens { (P → Q ), ~Q } ⊢ ~P [4 lines] 1 (1) P → Q A 2 (2) ~Q A 2 (3) P → ~Q 2 →I 1,2 (4) ~P 1,3 ~I

5. (5 points) Suppose we think of the turnstile, ⊢, as a relation defined with respect to a

universe of sentences, such that we may translate “X ⊢ Y” as “Y is provable from X.”

Give informal arguments showing that the turnstile is reflexive and transitive. Then give

an example showing that it is not symmetric.

A relation R is reflexive if and only if everything is R-related to itself. If the turnstile is reflexive, then everything is provable from itself. This is obviously true. Pick an arbitrary sentence, P. Since P is given as a premiss, we are allowed to write down P by the rule of assumption. Having written down P, we are finished: we have proved that P. A relation R is transitive if and only if for any three things X, Y, and Z you pick, if X is R-related to Y and Y is R-related to Z, then X is R-related to Z. Pick arbitrary sentences P, Q, and S. Suppose that P ⊢ Q and Q ⊢ S. Then there is a proof of Q from P and a proof of S from Q. If there is a proof of S from P, then the turnstile is transitive. To get such a proof, just paste together the two proofs that we already have. That is, carry out the proof from P to Q. Then start the proof from Q to S. The result is a proof of S from P. From (P ∧ Q ) we can prove P by conjunction elimination. But we cannot prove (P ∧ Q ) from P. Hence, the turnstile is not symmetric.

6. Address the problems below about small worlds, models, and validity.

a. (2 points) Define validity for first-order logic. Then explain the definition.

An argument in first-order logic is valid if and only if every small world that is a model for the premisses of the argument is also a model for the conclusion. In other words, every world that makes the premisses true also makes the conclusion true. It is impossible for the conclusion to be false if the premisses are all true.

b. (2 points) Produce one small world that is a model for (∀x)(Fx → Gx) AND

one small world that is NOT a model for (∀x)(Fx → Gx).

Consider a world with one constant, a. Suppose that Fa and Ga. Such a world is a model for the sentence (∀x)(Fx → Gx). A world in which Fa and ~Ga is not a model for the sentence (∀x)(Fx → Gx).

c. (1 point) Compare and contrast small worlds with truth tables.

A small world and a row in a truth table both represent a possible world: a way things might be. Truth tables and small worlds are tools for evaluating arguments in zeroth-order logic and in first-order logic, respectively. But whereas truth tables may be used to show that arguments are valid, small worlds cannot always be used to show that arguments are valid.

7. (5 points) Translate the following argument into one of our formal languages and then

either show that the argument is invalid or give a natural deduction proof of the

conclusion from the premisses. Given that the argument is valid, are you rationally

compelled to believe that the universe has a cause? Explain your answer.

Everything that began to exist has a cause.

If so, then either the universe has a cause or the universe did not begin to exist.

The universe began to exist.

The universe has a cause.

Let P = “Everything that began to exist has a cause.” Let Q = “The universe has a cause.” Let R = “The universe began to exist.” Then the argument may be translated as … P P → (Q ∨ ~R) R

Q 1 (1) P A 2 (2) P → (Q ∨ ~R) A 3 (3) R A 1,2 (4) Q ∨ ~R 1,2 →E 5 (5) Q ACP (6) Q → Q 5 CP 7 (7) ~R ACP 7 (8) ~Q → ~R 7 →I 3 (9) ~Q → R 3 →I 3,7 (10) ~~Q 8,9 ~I 3,7 (11) Q 10 ~E 3 (12) ~R → Q 7,11 CP 1,2,3 (13) Q 4,6,12 ∨E You are not rationally compelled to believe that the universe has a cause, since you may reject one or more of the premisses in the argument.

8. Suppose we construct the natural numbers using sets as follows, where “x =df y” means

that x is defined to be identical to y:

0 =df { } That is, zero is defined to be identical to the empty set.

1 =df {0}

2 =df {0, 1}

3 =df {0, 1, 2}

:

And in general, the natural number n is defined to be the set of all natural numbers

smaller than n.

a. (3 points) How many elements are in (i) the natural number 3, (ii) the natural

number 36, and (iii) the natural number n?

3 36 n (i) (ii) (iii) b. (3 points) If B is the set {1, 3, {5}, 7}, what is the intersection of 6 and B?

The intersection of B and 6 is {1, 3}.

c. (2 points) List all of the subsets of the natural number 3.

{} {0} {1} {2} {0, 1} {0, 2} {1, 2} {0, 1, 2}

d. (2 points) Would it work to define “x + y” to be the same as x ∪ y? Explain

your answer.

No. Given that definition, the sum of 1 and 2 would be 2.

9. Answer the questions about probability below by referring to the following diagram:

a. (3 points) What is the probability that a randomly-chosen object is unshaded?

“Randomly-chosen” implies that all objects are equally likely to be drawn. Hence, the probability is the number of unshaded out of the number of objects: 13 / 25. b. (3 points) What is the probability that a randomly-chosen object is a

shaded triangle?

Probability of shaded and triangle is 5 / 25.

c. (2 points) What is the probability that a randomly-chosen object is a triangle

given that it is shaded?

Conditional probability of triangle given shaded is the probability of shaded and triangle divided by the probability of shaded: (5 / 25) / (12 / 25) = 5 / 12

d. (2 points) Is being a square independent of being a circle?

No. If an object is a square, then it cannot be a circle. Since there are some squares and circles, Pr(square | circle) ≠ Pr(square).

10. Answer the following conceptual questions about probability.

a. (4 points) Suppose your friend Rudy says he thinks that there is a 25 percent

chance that it will rain tomorrow and that it is twice as likely that it

will not rain tomorrow. Is Rudy’s statement coherent? Explain

your answer.

No. Since “It will rain tomorrow,” and “It will not rain tomorrow” are mutually exclusive and exhaustive, the sum of their probabilities must be 1. But Rudy thinks the sum of their probabilities is only 0.75. b. (3 points) Rudy assigns probability 0.8 to the sentence (∀x)Fx. What

probability should Rudy assign to the sentence (∃x)~Fx?

Explain your answer.

Since (∃x)~Fx is logically equivalent to the negation of (∀x)Fx, and Rudy assigns probability 0.8 to the sentence (∀x)Fx, Rudy should assign probability 0.2 to the sentence (∃x)~Fx.

c. (3 points) One day, Rudy is talking to an epidemiologist. Rudy asserts that

the probability of a world-wide pandemic this year is ½, since

either there will be a pandemic or not. What should the

epidemiologist say in reply? Explain your answer.

The epidemiologist should first remind Rudy that it is an idealization (like frictionless surfaces in physics) to treat all possibilities as equally likely. In many cases, all possibilities are not equally likely, and we should adjust the probabilities we assign in light of our evidence. The epidemiologist should then cite relevant evidence that makes the probability of a world-wide pandemic different from ½, assuming that it is, in fact different from that value. Based on what I have read, I would expect the probability of a world-wide pandemic to be much lower than ½.

11. (10 points) [Modified Student Question] Suppose that 68% of Urban Outfitters’

customers are politically liberal. Suppose that among Urban Outfitters’ liberal customers,

71% are women and 29% are men, and among their conservative customers, 37.5% are

women and 62.5% are men. Urban Outfitters is developing a new Politics line of

women’s clothing, which consists of clothing emblazoned with liberal or conservative

political slogans. They want to stock their new clothing line in such a way that the

percentage of clothing items that have a liberal (or conservative) slogan matches the

percentages of liberal (or conservative) women who shop at their stores. What percentage

of their Politics line of women’s clothing should have a liberal slogan, and what

percentage should have a conservative slogan?

Let h = Customer is politically liberal. Let e = Customer is a woman. The base rate for political liberalism among Urban Outfitters’ customers is 0.68. Hence, Pr(h) = 0.68 and by the negation rule, Pr(~h) = 0.32. The probability that a customer is a woman given that she is liberal is 0.71, and the probability that a customer is a woman given that she is not a liberal is 0.375. Hence, Pr(e | h) = 0.71 and Pr(e | ~h) = 0.375. By Bayes’ Theorem, Pr(h | e) = Pr(e | h) · Pr(h) / Pr(e), and by the law of total probability, Pr(e) = Pr(e | h) · Pr(h) + Pr(e | ~h) · Pr(~h). Hence, Pr(e) = 0.71 · 0.68 + 0.375 · 0.32 ≅ 0.603, and Pr(h | e) = 0.71 · 0.68 / 0.603 ≅ 0.80. Therefore, about 80% of their Politics line of women’s clothing should have a liberal slogan and about 20% should have a conservative slogan.

12. (10 points) State the definition of conditional probability and then use it together with

derived rules as needed to prove Bayes’ Theorem.

According to our definition, Pr(A | B) = Pr(A ∧ B) / Pr(B). Bayes’ Theorem states that Pr(A | B) = Pr(B | A) · Pr(A) / Pr(B). “Proof.” By definition, Pr(B | A) = Pr(B ∧ A) / Pr(A). Multiplying both sides by Pr(A), we have Pr(B | A) · Pr(A) = Pr(B ∧ A). According to one of our derived rules, if two sentences are logically equivalent, then they have the same probability. The sentence (A ∧ B) is logically equivalent to the sentence (B ∧ A). So, Pr(B | A) · Pr(A) = Pr(A ∧ B). By definition, Pr(A | B) = Pr(A ∧ B) / Pr(B). Substituting Pr(B | A) · Pr(A) for Pr(A ∧ B) in the definition yields Bayes’ Theorem. []

13. Answer the following questions about independence.

a. (3 points) Give a formal statement of what it means for two variables to be

independent. Then explain what it means for two variables to be independent, using

ordinary language.

Two variables X and Y are independent if and only if for all values x and y, Pr(X = x | Y = y ) = Pr(X = x ). In ordinary language, if two variables are independent, then they are informationally disconnected. Knowing the value that one of them takes tells you nothing about what value the other might take.

b. (2 points) Suppose the table below represents a population for which every unit is

equally likely to be drawn. Are Studies and Grade independent? Explain your answer.

Unit Studies Grade

1 never B

2 sometimes D

3 never D

4 often B

5 sometimes F

6 often C

7 often A

8 sometimes B

9 never F

Pr(Grade = A | Studies = often) = 1/3 Pr(Grade = A ) = 1/9 Since 1/3 ≠ 1/9, the variables Grade and Studies are not independent.

14. (5 points) Suppose you have two trick coins. One has a bias of 1/3 for heads, and the

other has a bias of 2/3 for heads. You cannot tell them apart just by looking, and you have

forgotten which one is which. So, you decide to flip one of them several times and make

a guess at which it is. You flip the coin ten times and observe heads come up six times.

Set up the formula that you need in order to calculate the probability that you have been

flipping the coin that has a 1/3 probability of heads on any given flip. Do not calculate a

numerical answer. How would your answer change if you had picked your coin out of a

bag of 20 coins in which 19 had a bias of 1/3 for heads and one had a bias of 2/3 for

heads?

We have two hypotheses about the coin: h = Coin has 1/3 bias for Heads, and ~h = Coin has 2/3 bias for Heads. We flip one coin and see six Heads and four Tails. That is our evidence, e. We have Pr(h) = Pr(~h) = ½. We need to find Pr(e | h) and Pr(e | ~h) in order to apply Bayes Theorem to get an answer. Sequences of coin flips are well-modeled by the binomial distribution. So, we compute Pr(e | h) and Pr(e | ~h) using the binomial distribution.

6 4

10 1 2Pr( | )

6 3 3e h

6 4

10 2 1Pr( | ~ )

6 3 3e h

Plugging into Bayes Theorem is a complete answer. Although we didn’t need to compute a numerical answer, we can get one without too much difficulty:

6 4 2

6 4 4 6 2 2

1 2 1

13 3 3Pr( | )

51 2 1 2 1 2

3 3 3 3 3 3

h e

If we had picked our coin out of a bag of 20 coins in which 19 had a bias of 1/3 for heads and one had a bias of 2/3 for heads, the priors for h and ~h would have been 19/20 and 1/20, respectively. Consequently, we wouldn’t be able to cancel the ½ in the numerator and denominator of Bayes Theorem. Plugging into Bayes Theorem, we get:

6 4 2

6 4 4 6 2 2

19 1 2 19 1

1920 3 3 20 3Pr( | )

2319 1 2 1 1 2 19 1 1 2

20 3 3 20 3 3 20 3 20 3

h e