Download CMSC250, Fall 2003 Homework 5 Answers

CMSC250, Fall 2003
Homework 5 Answers
Due Wednesday, October 8 at the beginning of your discussion section.
You must write the solutions to the problems single-sided on your own lined paper,
with all sheets stapled together, and with all answers written in sequential order or
you will lose points.
1. Complete the following proofs using the method described in class.
(a)
P1 ∀x ∈ D R(x) ∨ [Q(x) → T (x)]
P2 ∃y ∈ D P (y) ∨ [∼ R(y) ∧ Q(y)]
P3 ∀z ∈ D ∼ T (z) →∼ P (z)
∴ ∃w ∈ D T (w)
Answer:
#
1
2
3
4
5
6
7
8
9
10
11
12
13
Statement
P (a) ∨ [∼ R(a) ∧ Q(a)]
P (a)
T (a)
P (a) → T (a)
∼ R(a) ∧ Q(a)
∼ R(a)
Q(a)
R(a) ∨ [Q(a) → T (a)]
Q(a) → T (a)
T (a)
[∼ R(a) ∧ Q(a)] → T (a)
T (a)
∃w ∈ D T (w)
Rule
∃ instantiation
Assume
∀ modus tollens
Closing cond world w/o contra
Assume
Conjunctive simplification
Conjunctive simplification
∀ instantiation
Disjunctive syllogism
Modus ponens
Closing cond world w/out contra
Dilemma
∃ generalization
Lines Used
P2
—
2, P3
2–3
—
5
5
P1
8, 6
9, 7
5–10
1, 4, 7
12
Another way using a conditional world with contradiction:
#
1
2
3
4
5
6
7
8
9
10
11
12
13
Statement
P (m) ∨ [∼ R(m) ∧ Q(m)]
R(m) ∨ [Q(m) → T (m)]
∼ T (m) → ∼ P (m)
∼ T (m)
∼ P (m)
∼ R(m) ∧ Q(m)
∼ R(m)
Q(m) → T (m)
Q(m)
∼ Q(m)
∼ Q(m) ∧ Q(m)
T (m)
∃w ∈ D T (w)
Rule
∃ instantiation
∀ instantiation
∀ instantiation
Assume
Modus ponens
Disjunctive syllogism
Conjunctive simplification
Disjunctive syllogism
Conjunctive simplification
Modus tollens
Conjunctive addition
Closing cond world w/ contra
∃ generalization
1
Lines Used
P2
P1
P3
—
3, 4
5, 1
6
2, 7
6
8, 4
9, 10
4–11
12
(b)
P1 ∀t ∈ D
P2 ∀u ∈ D
P3 ∀v ∈ D
∴ ∀h ∈ D
Answer:
∼ L(t)
S(u) → (R(u) ∧ T (u))
[L(v) → ∼ S(v)] → [R(v) → L(v)]
∼ S(h)
#
1
2
3
4
5
Statement
∼ L(a)
[L(a) →∼ S(a)] → [R(a) → L(a)]
[∼ L(a)∨ ∼ S(a)] → [∼ R(a) ∨ L(a)]
∼ [∼ L(a)∨ ∼ S(a)] ∨ [∼ R(a) ∨ L(a)]
((L(a) ∧ S(a)) ∨ ∼ R(a)) ∨ L(a)
6
7
8
9
10
11
12
13
(L(a) ∧ S(a)) ∨ ∼ R(a)
∼ L(a)∨ ∼ S(a)
∼ (L(a) ∧ S(a))
∼ R(a)
∼ R(a)∨ ∼ T (a)
∼ (R(a)∧ ∼ T (a))
∼ S(a)
∀h ∈ D ∼ S(h)
Rule
∀ instantiation
∀ instantiation
Definition of →
Definition of →
DeMorgan’s law
and associative law
Disjunctive syllogism
Disjunctive addition
DeMorgan’s law
Disjunctive syllogism
Disjunctive addition
DeMorgan’s law
∀ modus tollens
∀ generalization
Lines Used
P1
P3
2
3
4
5, 1
1
7
6, 8
9
10
11, P2
12
Another way using a conditional world with contradiction:
#
1
2
3
4
5
6
7
8
9
10
11
12
13
Statement
∼ L(m)
S(m) → (R(m) ∧ T (m))
[L(m) → ∼ S(m)] → [R(m) → L(m)]
S(m)
R(m) ∧ T (m)
∼ L(m)∨ ∼ S(m)
L(m) → ∼ S(m)
R(m) → L(m)
∼ R(m)
R(m)
R(m)∧ ∼ R(m)
∼ S(m)
∀h ∈ D ∼ S(h)
Rule
∀ instantiation
∀ instantiation
∀ instantiation
Assume
Modus ponens
Disjunctive addition
Definition of →
Modus ponens
Modus tollens
Conjunctive simplification
Conjunctive addition
Closing cond world w/ contra
∀ generalization
Lines Used
P1
P2
P3
—
2, 4
1
6
3, 7
8, 1
5
9, 10
4–11
12
2. Assume you know that ∃x ∈ S ∀y ∈ T P (x, y). Do you necessarily know that
∀y ∈ T ∃x ∈ S P (x, y)? Explain your answer in English. You may wish to provide an example
to support your reasoning.
Answer: Yes. Let’s say S = T = {all people} and P (x, y) = x likes y. Since you know
that “∃x ∈ S ∀y ∈ T P (x, y)”, you know that it is true that there is a person (lets call him
Bob) who does indeed like every person. “∀y ∈ T ∃x ∈ S P (x, y)” means that if we go
down the list of all people one at a time, for each of them we can find a person who likes
them. Since Bob is known to like everyone from the first statement, we could (if there were
nobody else) pick Bob for each person in the list.
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3. Assume you are given the following paragraph:
1
Tony, Mike, and John belong to the Alpine Club. Every member of the Alpine
Club is either a skier, a mountain climber, or both. No mountain climber likes rain,
and all skiers like snow. Mike dislikes whatever types of weather Tony likes and
likes whatever types of weather Tony dislikes. Tony likes rain and snow.
For this problem, you may only use the use the two sets P = {all people} and W =
{all types of weather}, and the following predicates:
• A(x) = “x is a member of the Alpine Club,”
• M (x) = “x is a mountain climber,”
• S(x) = “x is a skier,” and
• L(x, y) = “x likes y.”
You may also use the following constants: tony, mike, john, rain, snow.
(a) Represent the information in the paragraph as predicate calculus statements. Label each
statement as P1, P2, etc. Hints: You may wish to represent some of the English sentences
as more than one predicate calculus statement. You will not need to use any existential
quantifiers in your translations, and some translations will not need any quantifiers at
all.
Answer:
P1 A(tony)
P2 A(mike)
P3 A(john)
P4 ∀x ∈ P A(x) → (M (x) ∨ S(x))
P5 ∀x ∈ P M (x) → ∼ L(x, rain)
P6 ∀x ∈ P S(x) → L(x, snow)
P7 ∀w ∈ W L(tony, w) → ∼ L(mike, w)
P8 ∀w ∈ W ∼ L(tony, w) → L(mike, w)
P9 L(tony, rain)
P10 L(tony, snow)
(b) Complete a proof using the premises you defined to answer the question, “Of the members of the Alpine Club, who is a mountain climber but not a skier?” Follow the same
format described in class for predicate calculus proofs.
Answer:
# Statement
Rule
Lines Used
1 ∼ L(mike, rain)
∀ modus ponens
P7, P9
2 ∼ L(mike, snow)
∀ modus ponens
P7, P10
3 ∼ S(mike)
∀ modus tollens
P6, 2
4 M (mike) ∨ S(mike)
∀ modus ponens
P2, P4
5 M (mike)
Disjunctive syllogism
3, 4
6 M (mike)∧ ∼ S(mike)
Conjunctive addition
3, 5
7 A(mike) ∧ M (mike)∧ ∼ S(mike) Conjunctive addition
P2, 6
1
Adapted from a problem in Artificial Intelligence: A New Synthesis, by Nils J. Nilsson.
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(c) Of the members of the Alpine Club, who is a mountain climber but not a skier?
Answer: Mike.
4. For each of the following statements, give a proof to show it is true, or give and verify a
counterexample to show it is false.
(a) Given any even integer x, the sum of x − 1 and x + 1 is divisible by 4.
Answer: TRUE.
Proof: Let x be an arbitrary even integer.
So ∃k ∈ Z such that x = 2k by the definition of even.
Therefore, x − 1 = 2k − 1 and x + 1 = 2k + 1 by substitution.
So (x − 1) + (x + 1) = (2k − 1) + (2k + 1) by substition,
and (2k − 1) + (2k + 1) = 2k + 2k − 1 + 1 = 4k by algebra.
Since k ∈ Z and (x − 1) + (x + 1) = 4k, 4|((x − 1) + (x + 1)) by the definition of divides.
(b) If x and y are rational, but x/y is not rational, then y = 0.
Answer: TRUE.
Proof: Let x and y be arbitrary in Q.
So ∃a, b, c, d ∈ Z such that x = a/b and b 6= 0, and y = c/d and d 6= 0 by the definition
of rational.
Therefore x/y = (a/b)/(c/d) by substitution,
and (a/b)/(c/d) = (ad)/(bc) by algebra.
Since we know that x/y = (ad)/(bc) is not rational, we know that either ad ∈
/ Z, or
bc ∈
/ Z, or bc = 0.
By closure of the integers under multiplication, we know that ad and bc both are in Z.
So (by disjunctive syllogism) we know that bc = 0.
Since bc = 0, either b or c is zero. We already know that b 6= 0, so therefore c = 0, by
disjunctive syllogism.
Therefore, y = 0/d = 0 by algebra.
(c) If x and y are both even integers, then their sum is divisible by 4.
Answer: FALSE.
Counterexample: Let x = 2 and y = 4. Both 2 and 4 are even integers, but their sum,
6, is not divisible by 4.
(d) x2 − 1 is odd for all integers x.
Answer: FALSE.
Counterexample: Let x = 3. So x2 − 1 = 32 − 1 = 9 − 1 = 8, which is not odd.
(e) The sum of any two perfect squares is a perfect square.
Answer: FALSE.
Counterexample: Let x = y = 1. So x2 = y 2 = 1, so both x2 and y 2 are perfect
squares. But x2 + y 2 = 1 + 1 = 2, which is not a perfect square.
(f) There are two perfect squares such that their sum is a perfect square.
Answer: TRUE.
Proof: Let x = 3 and y = 4. So x2 = 9 and y 2 = 16, so both x2 and y 2 are perfect
squares. Therefore x2 + y 2 = 9 + 16 = 25, which is a perfect square (25 = 52 ).
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