Quiz

Note: It is recommended that you save your response as you complete each question.

Question 1 (1 point)

What is meant by the term discrete when it is used in the context of this course, as in discrete structures, or discrete mathematics?

Question 1 options:

	Private
	Countable
	Binary (1 or 0)
	Small

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Question 2 (1 point)

Give at least two examples of why logic is relevant in Computer Science. (Pick the best 2)

Question 2 options:

	Programming
	Getting a good grade
	Problem Solving
	Writing Reports

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Question 3 (1 point)

Create a truth table to determine whether the following proposition is valid:

(p & q) (~p v q)

Question 3 options:

	p q -p p^q -pq (p^q)(-pq) T T F T T T T F F F F F F T T F T F F F T F T F The statement is not valid
	p q -p p^q -pq (p^q)(-pq) T T F T T T T F F F F T F T T F T T F F T F T T The statement is not valid
	p q -p p^q -pq (p^q)(-pq) T T F T T T T F F F F T F T T F T T F F T F T T The statement is valid
	p q -p p^q -pq (p^q)(-pq) T T F T T T T F F F F T F T T F T T F F F F F F The statement is not valid

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Question 4 (1 point)

Create a truth table to determine whether the following proposition is true

(p v q) ~(~p & ~q)

Question 4 options:

	p q ~p ~q ~p^~q pvq ~(~p^~q) (pvq)~(~p^~q) T T F F F T T T T F F T F T T T F T T F F T T T F F T T T F F F The Statement is not valid
	p q ~p ~q ~p^~q pvq ~(~p^~q) (pvq)~(~p^~q) T T F F F T T T T F F T T T F T F T T F T T F T F F T T T F F T The Statement is valid
	p q ~p ~q ~p^~q pvq ~(~p^~q) (pvq)~(~p^~q) T T F F F T T T T F F T F T T T F T T F F T T T F F T T T F F T The Statement is not valid
	p q ~p ~q ~p^~q pvq ~(~p^~q) (pvq)~(~p^~q) T T F F F T T T T F F T F T T T F T T F F T T T F F T T T F F T The Statement is valid

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Question 5 (1 point)

Create a truth table to determine whether the following two propositions are equivalent, i.e. are true under the same circumstances

(p v q) and (~p q)

Question 5 options:

	p q ~p pvq ~pq T T F T T T F F T T F T T T T F F T F F They are the same
	p q ~p pvq ~pq T T F T T T F F T T F T F T F F F T F F They are not the same
	p q ~p pvq ~pq T T F T T T F F F T F T T T T F F T F F They are not the same
	p q ~p pvq ~pq T T F T T T F F F F F T T T T F F T F F They are the same

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Question 6 (1 point)

Consider the following argument:

If Han obeys the rules, he keeps his credit card.

Han does not obey the rules.

Therefore, he does not keep his credit card.

Create a truth table to determine whether the argument is true or false

Question 6 options:

	Let O = Han obeys the rules Let K = Han keeps his credit card O K ~O OK T T F T T F F F F T T T F F T T The Argument is true
	Let O = Han obeys the rules Let K = Han keeps his credit card O K ~O OK T T F T T F T F F T F T F F T T The Argument is true
	Let O = Han obeys the rules Let K = Han keeps his credit card O K ~O OK T T F T T F T F F T F T F F T T The Argument is false
	Let O = Han obeys the rules Let K = Han keeps his credit card O K ~O OK T T F T T F F F F T T T F F T T The Argument is false

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Question 7 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

what is the best way to render the following predicate logic statement in English?

(x)[(C(x) & S(x)) (y)[M(y) & O(x,y)]]

Question 7 options:

	For all cars that shine the exists a man who owns it
	For each car that shines it implies that there exists a man who owns the car.
	All shiney cars own a man.
	All cars that shine imply that there exists a man who owns it.

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Question 8 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

what is the best way to render the following predicate logic statement in English?

(x)[(M(x) & (y)[C(y) & O(x,y)]) P(x)]

Question 8 options:

	Each man that owns a shiney car is pleased
	Every man who owns a car is pleased
	There exists a car that all men own and they are pleased.
	For all x that are men, there exists a y that is a car and the man owns the car which implies that the man is pleased.

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Question 9 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

what is the best way to render the following predicate logic statement in English?

(x)[(C(x) & ~(y)[M(y) & O(y,x)]]

Question 9 options:

	There exists a car and not exists a man and the man owns the car.
	No men own cars.
	Cars do not own men.
	There is a car that no-one owns.

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Question 10 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

what is the best way to render the following predicate logic statement in English?

(x)[(C(x) ~(y)[M(y) & O(x,y)]]

Question 10 options:

	No man owns every car.
	No car owns every man
	No car owns a man
	There exists a man that no car owns.

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Question 11 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

translate the following English statement into predicate logic:

All men who own cars wash them

Question 11 options:

	x[M(x)^y(C(y)^O(x,y)]W(x,y)
	x[M(x)^y(C(y)^O(x,y)]W(y,x)
	x[M(x)^x(C(x)^O(x,y)]W(x,y)
	x[M(x)^y(C(y)^O(y,x)]W(y,x)

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Question 12 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

translate the following English statement into predicate logic:

If a man washes a car, the car shines and the man is pleased

Question 12 options:

	x[M(x)^C(y)^W(x,y)][S(x)^P(x)]
	xy[M(x)^C(y)^W(x,y)][S(y)^P(x)]
	xy[M(x)^C(y)^W(x,y)][S(x)^P(y)]
	[M(x)^C(y)^W(x,y)][S(y)^P(x)]

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Question 13 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

translate the following English statement into predicate logic:

Every man owns a car that shines.

Question 13 options:

	x(M(x))y[(C(y)^O(x,y)^S(y)]
	xy(M(x)^(C(y)^O(x,y)^S(y))
	x(M(x))y[(C(y)^O(x,y)^S(y)]
	xy(M(x))[(C(y)^O(x,y)^S(y)]

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Question 14 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

translate the following English statement into predicate logic:

There is a car that does not shine and there is a man who owns it and who is not pleased.

Question 14 options:

	x[C(x)^~S(x)]^y[M(y)^O(y,x)^~P(y)]
	[C(x)^~S(x)]^[M(y)^O(y,x)^~P(y)]
	x[C(x)^~S(x)]^y[M(y)^O(y,x)^~P(y)]
	x[C(x)^~S(x)]^y[M(y)^O(x,y)^~P(y)]

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Question 15 (1 point)

Using the following predicates

C(x)	x is a car
M(x)	x is a man
O(x,y)	x owns y
W(x,y)	x washes y
S(x)	x shines
P(x)	x is pleased

translate the following English statement into predicate logic:

If a man is pleased, he owns a car and washes it.

Question 15 options:

	x[M(x)^P(x)]y[C(y)^O(x,y)^W(x,y)]
	x[M(x)^P(x)]y[C(y)^O(x,y)^W(x,y)]
	x[M(x)^P(x)]y[C(y)^O(y,x)^W(y,x)]
	x[M(x)^P(x)]^y[C(y)^O(x,y)^W(x,y)]

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