The following is a proof that for any sets A, B, and C, A (B C) = (A B) (A C). Fill in

Chapter 6, Problem 6

(choose chapter or problem)

The following is a proof that for any sets A, B, and C, \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\). Fill in the blanks.

Proof: Suppose A, B, and C are any sets.

(1) Proof that \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C)\):

Let \(x \in A \cap(B \cup C)\). [We must show that \(x \in\)   (a)  .] By definition of intersection, \(x \in\)   (b)   and \(x \in\)   (c)  . Thus \(x \in A\) and, by definition of union, \(x \in B\) or   (d)  .

Case 1 (\(x \in A\) and \(x \in B\)): In this case, by definition of intersection, \(x \in\)   (e)  , and so, by definition of union, \(x \in(A \cap B) \cup(A \cap C)\).

Case 2 (\(\(x \in A\)\) and \(\(x \in C\)\)): In this case,   (f)  . Hence in either case, \(x \in(A \cap B) \cup(A \cap C)\) [as was to be shown].

[So \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C)\) by definition of subset.]

(2) \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C)\):

Let \(x \in(A \cap B) \cup(A \cap C)\). [We must show that   (a)  .] By

definition of union, \(x \in A \cap B\)   (b)   \(x \in A \cap C\).

Case 1 \((x \in A \cap B)\): In this case, by definition of intersection, \(x \in A\)   (c)   \(x \in B\). Since \(x \in B\), then by definition of union, \(x \in B \cup C\). Hence \(x \in A\) and \(x \in B \cup C\), and so, by definition of intersection, \(x \in\)   (d)  .

Case 2 \((x \in A \cap C)\): In this case,   (e)  . In either case, \(x \in A \cap(B \cup C)\) [as was to be shown]. [Thus \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C)\) by definition of subset.]

(3) Conclusion: [Since both subset relations have been proved, it follows, by definition of set equality, that (a) .]

Text Transcription:

A cap(B cup C)=(A cap B) cup(A cap C)

A cap(B cup C) subseteq(A cap B) cup(A cap C)

x in A cap(B cup C)

x in

x in

x in

x in A

x in B

x in A

x in B

x in

x in(A cap B) cup(A cap C)

x in A

x in C

x in(A cap B) cup(A cap C)

A cap(B cup C) subseteq(A cap B) cup(A cap C)

(A cap B) cup(A cap C) subseteq A cap(B cup C)

x in(A cap B) cup(A cap C)

x in A cap B

x in A cap C

(x in A cap B)

x in A

x in B

x in B

x in B cup C

x in A

x in B cup C

x in

(x in A cap C)

x in A cap(B cup C)

(A cap B) cup(A cap C) subseteq A cap(B cup C)