Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true". I don't know how that makes sense. So, please explain this part. Next, (2) Suppose that the second proposition is true. Let x be an element in the domain. There is a y such that Q(y) is true, so ∃xQ(x) is true. Because ∀xP(x) is also true, it follows that the first proposition is true. Ok, I get this part. . . . b) ∀xP(x) ∨ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∨ Q(y)) (1) Suppose that ∀xP(x) ∨ ∃xQ(x) is true. Then either P(x) is true for all x, or there exists a y for which Q(y) is true. Shouldn't it be "there exists a x for which Q(x) is true." instead of "there exists a y for which Q(y) is true"? Even though their domains are the same, I think what I'm thinking is right. In the former case, P(x) ∨ Q(y) is true for all x, so ∀x∃y(P(x) ∨ Q(y)) is true. Ok, I understand. In the latter case, Q(y) is true for a particular y, so P(x) ∨ Q(y) is true for all x. ??? What I think is that since Q(y) is true for a particular y, P(x) ∨ Q(y) is true for a particular y. I don't understand why this makes sense...Please explain Consequently, ∀x∃y(P(x) ∨ Q(y)) is true. (2) Suppose that the second proposition is true. [∀x∃y(P(x) ∨ Q(y))] If P(x) is true for all x, then the first proposition is true. If not, P(x) is false for some x, and for this x there must be a y such that P(x) ∨ Q(y) is true. "for this x there must be a y such that P(x) ∨ Q(y) is true." I don't understand this part. Please explain. I mean, what's the connection between some x and y that makes P(x) ∨ Q(y) true??? Hence, Q(y) must be true, so ∃yQ(y) is true.

Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true". I don't know how that makes sense. So, please explain this part. Next, (2) Suppose that the second proposition is true. Let x be an element in the domain. There is a y such that Q(y) is true, so ∃xQ(x) is true. Because ∀xP(x) is also true, it follows that the first proposition is true. Ok, I get this part. . . . b) ∀xP(x) ∨ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∨ Q(y)) (1) Suppose that ∀xP(x) ∨ ∃xQ(x) is true. Then either P(x) is true for all x, or there exists a y for which Q(y) is true. Shouldn't it be "there exists a x for which Q(x) is true." instead of "there exists a y for which Q(y) is true"? Even though their domains are the same, I think what I'm thinking is right. In the former case, P(x) ∨ Q(y) is true for all x, so ∀x∃y(P(x) ∨ Q(y)) is true. Ok, I understand. In the latter case, Q(y) is true for a particular y, so P(x) ∨ Q(y) is true for all x. ??? What I think is that since Q(y) is true for a particular y, P(x) ∨ Q(y) is true for a particular y. I don't understand why this makes sense...Please explain Consequently, ∀x∃y(P(x) ∨ Q(y)) is true. (2) Suppose that the second proposition is true. [∀x∃y(P(x) ∨ Q(y))] If P(x) is true for all x, then the first proposition is true. If not, P(x) is false for some x, and for this x there must be a y such that P(x) ∨ Q(y) is true. "for this x there must be a y such that P(x) ∨ Q(y) is true." I don't understand this part. Please explain. I mean, what's the connection between some x and y that makes P(x) ∨ Q(y) true??? Hence, Q(y) must be true, so ∃yQ(y) is true.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Related questions

Question

) Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true". I don't know how that makes sense. So, please explain this part. Next, (2) Suppose that the second proposition is true. Let x be an element in the domain. There is a y such that Q(y) is true, so ∃xQ(x) is true. Because ∀xP(x) is also true, it follows that the first proposition is true. Ok, I get this part. . . . b) ∀xP(x) ∨ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∨ Q(y)) (1) Suppose that ∀xP(x) ∨ ∃xQ(x) is true. Then either P(x) is true for all x, or there exists a y for which Q(y) is true. Shouldn't it be "there exists a x for which Q(x) is true." instead of "there exists a y for which Q(y) is true"? Even though their domains are the same, I think what I'm thinking is right. In the former case, P(x) ∨ Q(y) is true for all x, so ∀x∃y(P(x) ∨ Q(y)) is true. Ok, I understand. In the latter case, Q(y) is true for a particular y, so P(x) ∨ Q(y) is true for all x. ??? What I think is that since Q(y) is true for a particular y, P(x) ∨ Q(y) is true for a particular y. I don't understand why this makes sense...Please explain Consequently, ∀x∃y(P(x) ∨ Q(y)) is true. (2) Suppose that the second proposition is true. [∀x∃y(P(x) ∨ Q(y))] If P(x) is true for all x, then the first proposition is true. If not, P(x) is false for some x, and for this x there must be a y such that P(x) ∨ Q(y) is true. "for this x there must be a y such that P(x) ∨ Q(y) is true." I don't understand this part. Please explain. I mean, what's the connection between some x and y that makes P(x) ∨ Q(y) true??? Hence, Q(y) must be true, so ∃yQ(y) is true.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.