For all sets A, B, and C, An (BU C) = (A N B) U (AN C). Fill in the blanks in the proof for the statement shown below. (In the proof, let n and U stand for the words "intersection" and "union," respectively.) Proof: Suppose A, B, and C are any sets. [To show that A N (B U C) = (A N B) U (AN C), we must show that A N (BU C) C (A N B) U (A N C) and that (A N B) U (A N C)CAN (BU C). (1) Proof that AN (BU C) S (AN B) U (A 0 C): Let x E AN (BU C). We must show that x E-Select-- By definition of n, x E-Select- v and x E BUC. Thus x E A and, by definition of U, x E B or x €-Select--- v . Case 1 (x € A and x E B): In this case, x E ANB by definition of n. Case 2 (x € A and x E C): In this case, x € ANC by definition of n. By cases 1 and 2, x E AN Blor x E ANC, and so, by definition of U, x E-Select--- [So AN (BUC) (A N B) U (AN C) by definition of subset.] (2) Proof that (AN B) U (A N C) S AN (BU C): Let x € (AN B) U (AN C). [We must show that x € A N (B U C).] By definition of U, x E ANB-Select- v XEANC. Case 1 (x EAN B): In this case, by definition of n, x EA and x E B. Since x E B, then XEBU C by definition of U. Case 2 (x € AN C): In this case, by definition of n, xEA-Select-v XE C. Since x E C, then x E BU C by definition of U. In both cases x E A and XEBUC, and so, by definition of n, xESelect--- So (AN B) U (AN C) CAN (BUC) by definition of -Select-- v (3) Conclusion: Since hoth cubcet relati
For all sets A, B, and C, An (BU C) = (A N B) U (AN C). Fill in the blanks in the proof for the statement shown below. (In the proof, let n and U stand for the words "intersection" and "union," respectively.) Proof: Suppose A, B, and C are any sets. [To show that A N (B U C) = (A N B) U (AN C), we must show that A N (BU C) C (A N B) U (A N C) and that (A N B) U (A N C)CAN (BU C). (1) Proof that AN (BU C) S (AN B) U (A 0 C): Let x E AN (BU C). We must show that x E-Select-- By definition of n, x E-Select- v and x E BUC. Thus x E A and, by definition of U, x E B or x €-Select--- v . Case 1 (x € A and x E B): In this case, x E ANB by definition of n. Case 2 (x € A and x E C): In this case, x € ANC by definition of n. By cases 1 and 2, x E AN Blor x E ANC, and so, by definition of U, x E-Select--- [So AN (BUC) (A N B) U (AN C) by definition of subset.] (2) Proof that (AN B) U (A N C) S AN (BU C): Let x € (AN B) U (AN C). [We must show that x € A N (B U C).] By definition of U, x E ANB-Select- v XEANC. Case 1 (x EAN B): In this case, by definition of n, x EA and x E B. Since x E B, then XEBU C by definition of U. Case 2 (x € AN C): In this case, by definition of n, xEA-Select-v XE C. Since x E C, then x E BU C by definition of U. In both cases x E A and XEBUC, and so, by definition of n, xESelect--- So (AN B) U (AN C) CAN (BUC) by definition of -Select-- v (3) Conclusion: Since hoth cubcet relati
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![For all sets A, B, and C, AN (BU C) = (A N B) U (AN C).
Fill in the blanks in the proof for the statement shown below. (In the proof, let n and U stand for the words "intersection" and "union," respectively.)
Proof: Suppose A, B, and C are any sets. [To show that A N (B U C) = (A N B) U (A N C), we must show that A N (B U C) C (A N B) U (A N C) and that (A N B) U (A N C) CAN (BU C).]
(1) Proof that AN (B U C) C (AN B) U (A N C):
Let x E AN (BU C). We
must show that x €--Select---
By definition of n, x E-Select-- v and x E BU C.
Thus x E A and, by definition of U, x E B or x e--Select--- v .
Case 1 (x € A and x E B): In this case, x E A N B by definition of n.
Case 2 (x € A and x E C): In this case, x € ANC by definition of n.
By cases 1 and 2, x E A N Blor x E AnC, and so, by definition of U, x E --Select---
[So AN (BU C) C (AN B) U (An C) by definition of subset.]
(2) Proof that (A N B) U (A N c) S A N (B U C):
Let x E (A N B) U (A N C). [We must show that x € A N (B U C).]
By definition of U, x E AN B--Select-- v xE ANC.
Case 1 (x E AN B): In this case, by definition of n, x E A and x E B.
Since x E B, then x E B U c by definition of U.
Case 2 (x E AN C): In this case, by definition of n, xEA -Select-- v x E C.
Since x E C, then x E BU C by definition of U.
In both cases x E A and XEBUC, and so, by definition of n, xE ---Select--
(AN B) U (AN C) CAN (BU C) by definition of --Select--- v
(3) Conclusion: Since both subset relations have been proved, it follows, by definition of set equality, that Select-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa28e667e-d0bd-42fc-8ff7-fbe8b481d237%2F4929639d-7271-4ad4-ab9e-0d30857f11b6%2Fd7b3kkbh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For all sets A, B, and C, AN (BU C) = (A N B) U (AN C).
Fill in the blanks in the proof for the statement shown below. (In the proof, let n and U stand for the words "intersection" and "union," respectively.)
Proof: Suppose A, B, and C are any sets. [To show that A N (B U C) = (A N B) U (A N C), we must show that A N (B U C) C (A N B) U (A N C) and that (A N B) U (A N C) CAN (BU C).]
(1) Proof that AN (B U C) C (AN B) U (A N C):
Let x E AN (BU C). We
must show that x €--Select---
By definition of n, x E-Select-- v and x E BU C.
Thus x E A and, by definition of U, x E B or x e--Select--- v .
Case 1 (x € A and x E B): In this case, x E A N B by definition of n.
Case 2 (x € A and x E C): In this case, x € ANC by definition of n.
By cases 1 and 2, x E A N Blor x E AnC, and so, by definition of U, x E --Select---
[So AN (BU C) C (AN B) U (An C) by definition of subset.]
(2) Proof that (A N B) U (A N c) S A N (B U C):
Let x E (A N B) U (A N C). [We must show that x € A N (B U C).]
By definition of U, x E AN B--Select-- v xE ANC.
Case 1 (x E AN B): In this case, by definition of n, x E A and x E B.
Since x E B, then x E B U c by definition of U.
Case 2 (x E AN C): In this case, by definition of n, xEA -Select-- v x E C.
Since x E C, then x E BU C by definition of U.
In both cases x E A and XEBUC, and so, by definition of n, xE ---Select--
(AN B) U (AN C) CAN (BU C) by definition of --Select--- v
(3) Conclusion: Since both subset relations have been proved, it follows, by definition of set equality, that Select-
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