Chp1 Logic
Which rule of inference is the basis of the following argument: "It is snowing. Therefore it is snowing or it is sunny"? Modus tollens Modus ponens Disjunctive syllogism Addition
Addition
Match the English expression on the left with the corresponding logical expression on the right. The domain of all variables is the integers.
Every positive number has an additive inverse. matches Choice ∀x∃y(x + y = 0) The product of two positive numbers is positive. matches Choice ∀x∀y((x > 0) ∧ (y > 0) → (xy > 0)) The product of two negative numbers is positive. matches Choice ∀x∀y((x < 0) ∧ (y < 0) → (xy > 0)) The product of a positive number and a negative number is negative. matches Choice ∀x∀y((x > 0) ∧ (y < 0) → (xy < 0))
Which of the following English sentences is a correct translation of ∀x ∃y(A(y) ∧ (S(x,y) ∨ ∃z(F(x, z)∧ S(z,y)))) where S(x,y) is "x speaks y," A(x) is "x is an Asian language," and F(x,y) is " x is a friend of y." The domain of x and z is all the students in a class and the domain of y is all languages.
Every student in the class either speaks an Asian language or has a friend who speaks an Asian language. All students in this class either speak an Asian language or have a friend who speaks an Asian language.
Which rule of inference is the basis for the following argument: "Felix is furry. Therefore, there is a furry cat"?
Existential generalization. Reason: Assuming the domain of x is all cats and F(x) is "x is furry." The argument is then concluding ∃x F(x) from F(Felix).
For which of these statements Q(x, y) is the truth value of ∀x∃y Q(x,y) different from ∃x∀y Q(x, y)? Assume the domain of the variables is the real numbers.
Q(x, y) is the statement "x + y = 0."
By doing which of these things do we show that p, q, and r are equivalent statements
Show that r → p, p → q, and q → r. Show that p → r, q → p, and r → q.
The premises "If the weather is good, people go to the beach," "If people go to the beach, they are happy," and "The weather is good" lead to the conclusion "People are happy." Let g be the proposition "the weather is good," b be the proposition "people go to the beach," and h be the proposition "people are happy." Each card is a step in a valid argument using the above premises to reach the stated conclusion. Drag each card to the stack that describes the reason that it is valid.
g → b matches Choice Premise b matches Choice Modus ponens using two premises h matches Choice Modus ponens using a premise and an intermediary result
Which of these are true statements? The domain of all variables is the integers. ∀x∀y∀z(x(yz) = (xy)z) ∃x∀y∀z(x = zy) ∀x∀y∃z(x÷y = z) ∀x∀y∃z(x + y = z)
∀x∀y∀z(x(yz) = (xy)z) Multiplication is associative for the integers. ∀x∀y∃z(x + y = z) The addition of two integers is always an integer.
Which of the following is a correct translation of "If someone is male and has the same parent as someone else, then he is a brother of that other person"? Assume that the domain of the quantifiers is all people. Use the predicates M(x) to represent "x is male," P(x,y) to represent "x is the parent of y," and B(x,y) to represent "x is the brother of y."
∀x∀y∃w ((M(x) ∧ P(w,x) ∧ P(w,y)) → B(x,y))
Match the logical expression on the left with the statement on the right which is logically equivalent to its negation, where no negation operator is to the left of a quantifier in its scope.
∀x∃y (P(x,y) ∧ Q(y,z)) matches Choice ∃x∀y (¬P(x,y) ∨ ¬Q(y,z)) ∃x∃y (P(x,y) ∨ Q(y,z)) matches Choice ∀x∀y (¬P(x,y) ∧ ¬Q(y,z)) ∃x∀y (P(x,y) ∧ Q(y,z)) matches Choice ∀x∃y (¬P(x,y) ∨ ¬Q(y,z)) ∃x∀y (P(x,y) → Q(y,z)) matches Choice ∀x∃y (P(x,y) ∧ ¬Q(y,z)) ∀x∃y ¬(P(x,y) ∧ Q(y,z)) matches Choice ∃x∀y (P(x,y) ∧ Q(y,z))
Translate the English sentence "All web pages owned by Ernest have a link to a page that is about chess" into predicate logic. Assume that the universe of discourse of all variables is all Web pages. For the predicates let L(x,y) be "x links to y," O(x, y) be "x owns y," and A(x, y) be "x is about y."
∀x∃y(O(Ernest,x) → (L(x, y) ∧ A(y, chess)))
Match the statement on the left with the domain for all variables that will make the statement true.
∀x∃y(xy = 1) matches Choice The positive real numbers ∃x∀y(x = 2y) matches Choice {0} ∀y∃x(x + y = 0) matches Choice The integers ∀x∀y(xy = xy + 1) matches Choice The empty set
Assume that the domain of x is {0, 1} and the domain of y is {2, 4, 7}. Which of the following is the correct representation of ∀x∃y Q(x, y) as a formula without quantification?
(Q(0, 2) ∨ Q(0, 4) ∨ Q(0, 7)) ∧ (Q(1, 2) ∨ Q(1, 4) ∨ Q(1, 7)) The universal quantifier means the statement must be true for all values of x. That is, we must have both ∃y Q(0, y) and ∃y Q(1, y). So we need a conjunction for all possible values from the domain of x. For ∃y Q(0, y) to be true, we only need one value of y that makes it true, so ∃y Q(0, y) is equivalent to a disjunction over all values of y. Likewise, ∃y Q(1, y) is the same as a disjunction.
Shown below are the steps of a proof by contraposition of the statement "If n is an integer and 5n + 3 is odd, then n is even." Match each step with its explanation.
1. Assume n is not even, that is, it is odd. matches Choice Negation of the conclusion 2.Then n = 2k + 1 for some integer k. matches Choice Definition of an odd integer 3. 5n + 3 = 5(2k + 1) + 3 = 10k + 8 = 2(5k + 4) matches Choice Arithmetic 4. Hence, 5n + 3 is even. matches Choice Definition of an even number
Prove by contradiction that if you have 367 people there are at least 2 that were born on the same day of the year. Match the step on the left with the justification on the right.
1. Let p be "At least 2 of the 367 people were born on the same day of the year." Then ¬p is "None of the 367 people were born on the same day of the year." matches Choice We let p be the statement we are trying to prove. 2. Then assume ¬p, the proposition that "None of the 367 people were born on the same day of the year." matches Choice The first step in a proof by contradiction is to assume the negation of what is to be proved. 3. ¬p contradicts the fact that there are not more than 366 days in any year. matches Choice If there are 367 people and 366 days in a year, at least two must have been born on the same day. 4. p must be true. At least 2 of the 367 people were born on the same day of the year. matches Choice Since assuming ¬p results in a contradiction, p must be true.
Show that the premises "Pegasus is a horse," "Pegasus has wings," "If Pegasus is a winged horse, then it is immortal," and "Pegasus is not immortal, if it is not magical" lead to the conclusion "Pegasus is magical." Let h be the proposition that "Pegasus is a horse, " w be the proposition that "Pegasus has wings," i be the proposition that "Pegasus is immortal," and m be the proposition that "Pegasus is magical." The premises are h, w, (h ∧ w) → i, ¬m →¬i, and the conclusion is m. Consider the following argument: 1.h 2. w 3.(h ∧ w) → i 4. ¬m → ¬i 5. h ∧ w 6. i 7. m Which of the steps is a result of the use of the rule of modus ponens?
6 Reason: From 5 and 3, 6 follows by modus ponens.
Given is a valid argument to show that the premises"The gun was bought from Chris or it was bought from Beth,""If the gun was bought from Chris, it was made in Chicago,""The gun was not made in Chicago and it is blue,""If the gun was bought from Beth then Beth sold an unregistered gun," and"If Beth sold an unregistered gun then she is a criminal"lead to the conclusion "Beth is a criminal." Let c be the proposition that "the gun was bought from Chris, " b be the proposition that "the gun was bought from Beth," m be the proposition that "the gun was made in Chicago," g be the proposition that "the gun is blue," u be the proposition that "Beth sold an unregistered gun," and r be the proposition that "Beth is a criminal." The premises are 1. c ∨ b, 2. c → m, 3.¬m ∧ g, 4. b → u, and 5. u → r. The conclusion is r. Match each premise on the left with the correct rule of inference on the right.
6. ¬m matches Choice Simplification from a premise 7. ¬c matches Choice Modus tollens from a premise and an intermediate step 8. b matches Choice Disjunctive syllogism from a premise and an intermediate step 9. u matches Choice Modus ponens using a premise and an intermediate step
In a proof by contradiction, which is the correct way to start a proof that 5⎯⎯√5 is irrational?
Assume that 5⎯⎯√5 can be expressed as ab�� for integers a, b where b ≠ 0, and a and b have no common factors.
Fill in the following, where each blank is to be filled with a single word. Moving a negation inward over a disjunction changes the disjunction into a _________ and each of the two statements is _______________. Moving a negation inward over an existential quantifier changes the quantifier to a _______________ quantifier and the rest of the expression is negated. Moving a negation inward over a universal quantifier changes the quantifier to an ___________________ quantifier and the rest of the expression is negated.
Blank 1: conjunction Blank 2: negated Blank 3: universal Blank 4: existential
Which of the following terms refer to true statements.
Corollary Theorem Result Lemma Fact
Match the English sentence on the left with the predicate calculus expression on the right. Let T(x,y) be "x has traveled to country y." The domain of x is all students in a class at your university and y is all countries in the world.
Everyone in the class has traveled to every country. matches Choice ∀x∀y T(x,y) Someone in the class has traveled to every country. matches Choice ∃x∀y T(x,y) Everyone in the class has traveled to some country. matches Choice ∀x∃y T(x,y) Someone in the class has traveled to some country. matches Choice ∃x∃y T(x,y) There is some country to which everyone in the class has traveled. matches Choice ∃y∀x T(x,y) Everyone in the class has traveled to France. matches Choice ∀x T(x, France) Sundari has traveled to France. matches Choice T(Sundari, France) Brigitte has traveled to every country. matches Choice ∀y T(Brigitte, y) Everyone in the class has not traveled to some country. matches Choice ∀x∃y ¬T(x, y)
Which of the following is the correct translation of ∀x∀y((x < 0) ∧ (y < 0) → (xy > 0)) into English? You can assume the domain is all real numbers.
For any two negative numbers, the product is positive.
Which statements can easily be proved by a direct proof?
If n is odd, then n3 + 1 is even. If n = (ab)2 where a and b are positive integers, then n ≥ a2 and n ≥ b2. If n is a perfect square where n is a positive integer, then n + 1 is not a perfect square.
For which of the following predicates is ∀x∀y∃z Q(x, y, z) true? Assume the domain of quantification for all variables is the real numbers.
Let Q(x, y, z) be the statement "(xy)z = x(yz)." Let Q(x, y, z) be the statement "xy = z."
Which rule of inference is the basis of the following argument: "If it rains, then the grass grows. It is raining. Therefore, the grass grows"?
Modus ponens
A __________ is a valid argument that establishes the truth of a theorem. The statements used in a proof can include _______________, which are statements we assume to be true, of the theorem, and previously proven theorems.
Proof Axiom
For which of these would a proof by contraposition be a better approach than a direct approach?
Prove that if 1/x is irrational then x is irrational. If 3n + 2 is odd, then n is odd.
Match the terms on the left with their definitions on the right.
Theorem matches Choice A statement that can be shown to be true Proof matches Choice A demonstration that a theorem is true Lemma matches Choice A less important theorem helpful in the proof of other results Corollary matches Choice A theorem that can be proved directly from some other theorem Conjecture matches Choice A statement proposed to be true Axiom matches Choice A statement assumed to be true
Which rule of inference is the basis for the following argument: "All cats are furry. Felix is a cat. Therefore Felix is furry"?
Universal instantiation.
How should we go about proving that 9n - 2 is even if and only if n is even.
Use a direct proof for one direction and contraposition for the other. .
Drag and drop the English translation on the right against the corresponding predicate calculus expression on the left. Let L(x, y) be "x links to y" and O(x, y) be "x owns y." The domain of the variables consists of all web pages.
∀x∀y ((O(Bob, x) ∧ O(Fareed, y)) → L(x,y)) matches Choice All Bob's web pages link to all of Fareed's web pages. ∀x∃y O(Bob, x) → (O(Fareed, y) ∧ L(x,y)) matches Choice All Bob's Web pages link to some Web page owned by Fareed. ∃y∀x ((O(Fareed, y) ∧ O(Bob, x)) → L(x, y)) matches Choice There is a Web page owned by Fareed that is linked to by all of Bob's Web pages. ∃x∃y (O(Bob, x) ∧ O(Fareed, y) ∧ L(x, y)) matches Choice There is a Web page owned by Bob that links to a Web page owned by Fareed. ∃x∃y O(Bo, x) → (O(Fareed, y) ∧ L(x, y)) matches Choice There is a Web page not owned by Bob or there is a page owned by Fareed that has some page linking to it.
Match the expression on the left with the explanation on the right of the nested quantifications.
∀x∀y P(x, y) matches Choice P(x, y) is true for every pair x, y ∀x∃y P(x, y) matches Choice For every x, there is a y for which P(x, y) is true. ∃x∀y P(x, y) matches Choice There is an x for which P(x, y) is true for every y. ∃x∃y P(x, y) matches Choice There is a pair x, y for which P(x, y) is true.
Match the expression on the left with the conditions on the right that cause the expression to be false. Drag and drop the conditions that cause the expression to be false against the corresponding expression on the left.
∀x∀y P(x, y) matches Choice There is a pair x, y for which P(x, y) is false. ∀x∃y P(x, y) matches Choice There is an x such that P(x, y) is false for every y. ∃x∀y P(x, y) matches Choice For every x there is a y for which P(x, y) is false. ∃x∃y P(x, y) matches Choice P(x, y) is false for every pair x, y.
Match the statement on the left with the domain for all variables that will make the statement true.
∀x∃y(xy = 1) matches Choice The positive real numbers ∃x∀y(x = 2y) matches Choice {0} ∀y∃x(x + y = 0) matches Choice The integers ∀x∀y(xy = xy + 1) matches Choice The empty set
Which of the following are true statements? The domain of all variables is the integers.
∃x∀y (xy = y) ∀x∀y (xy = yx)
