The following is a proof that "For all sets A and B, P(An B) CP(A) P(B)". Fill in the blanks. Proof: Suppose that A and B are any sets, and suppose that S is a set such that definition of power sets, then SCAN B.. Since An BCA, then + and so, + and so, Similarly, since An BCB, then + S + P(A) P(B). +. By Hence, by definition of Therefore, P(An B) CP(A) P(B). Q.E.D. *Note here that in the choices, P(A) represents the power set of A.
The following is a proof that "For all sets A and B, P(An B) CP(A) P(B)". Fill in the blanks. Proof: Suppose that A and B are any sets, and suppose that S is a set such that definition of power sets, then SCAN B.. Since An BCA, then + and so, + and so, Similarly, since An BCB, then + S + P(A) P(B). +. By Hence, by definition of Therefore, P(An B) CP(A) P(B). Q.E.D. *Note here that in the choices, P(A) represents the power set of A.
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter14: Counting And Probability
Section14.CR: Chapter Review
Problem 2CC
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![The following is a proof that "For all sets A and B, P(An B) CP(A) P(B)". Fill in the blanks.
Proof:
Suppose that A and B are any sets, and suppose that S is a set such that
definition of power sets, then SCAN B..
Since An BCA, then
+ and so,
+ and so,
Similarly, since An BCB, then
+ S + P(A) P(B).
+. By
Hence, by definition of
Therefore, P(An B) CP(A) P(B).
Q.E.D.
*Note here that in the choices, P(A) represents the power set of A.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf4b66fc-dcb0-412d-aa28-d3c10dfc804e%2Fd4bb2a44-6d92-4d7d-bc69-7b1278afb8e7%2F8hozbq_processed.png&w=3840&q=75)
Transcribed Image Text:The following is a proof that "For all sets A and B, P(An B) CP(A) P(B)". Fill in the blanks.
Proof:
Suppose that A and B are any sets, and suppose that S is a set such that
definition of power sets, then SCAN B..
Since An BCA, then
+ and so,
+ and so,
Similarly, since An BCB, then
+ S + P(A) P(B).
+. By
Hence, by definition of
Therefore, P(An B) CP(A) P(B).
Q.E.D.
*Note here that in the choices, P(A) represents the power set of A.
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