Chapter 5: Problem 5
$$ \text { Prove that for all sets } A \text { and } B, B-A=B \cap A^{c} \text {. } $$
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Chapter 5: Problem 5
$$ \text { Prove that for all sets } A \text { and } B, B-A=B \cap A^{c} \text {. } $$
These are the key concepts you need to understand to accurately answer the question.
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a. Write a negation for the following statement: \(\forall\) sets \(A\), if \(A \subseteq \mathbf{R}\) then \(A \subseteq \mathbf{Z}\). Which is true, the statement or its negation? Explain. b. Write a negation for the following statement: \(\forall\) sets \(S\), if \(S \subseteq Q^{+}\)then \(S \subseteq Q^{-} .\)Which is true, the statement or its negation? Explain.
For all sets \(A, B\), and \(C\), $$ (A-B) \cup(C-B)=(A \cup C)-B $$
Assume that \(B\) is a Boolean algebra with operations \(+\) and . Prove each statement without using any parts of Theorem \(5.3 .2\) unless they have already been proved. You may use any part of the definition of a Boolean algebra and the results of previous exercises, however. For all \(x, y\), and \(z\) in \(B\), if \(x+y=x+z\) and \(x \cdot y=x \cdot z\), then \(y=z\).
$$ \text { For all sets } A \text { and } B, A \cup(A \cap B)=A \text {. } $$
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) \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C):\) Let \(x \in A \cap(B \cup C)\). [We must show that \(x \in \underline{\text { (a) }}\).] By definition of intersection, \(x \in \underline{\text { (b) }}\) and \(x \in \frac{\text { (c) }}{\text { Thus }}\) \(x \in A\) and, by definition of union, \(x \in B\) or \((\mathrm{d})\) Case \(1(x \in A\) and \(x \in B):\) In this case, by definition of intersection, \(x \in \underline{(e)}\), 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 \(\underline{\text { (a) }}\).] By definition of union, \(x \in \stackrel{\text { (b) }}{\text { or } x \in \stackrel{\text { (c) }}{\text { Let } x \in} \text {. }}\) Case \(1(x \in A \cap B):\) In this case, by definition of intersection, \(\frac{(\mathrm{d})}{}\) and \((\mathrm{e})\). 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, \(x \in \underline{\text { (f) }}\). Case \(2(x \in A \cap C):\) In this case, \((g) .\) 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. [Since both subset relations have been proved, it follows, by definition of set equality, that \(\stackrel{(\mathrm{h})}{-}\).]
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