91Ó°ÊÓ

$$ \text { For all sets } A, A \times \emptyset=\emptyset \text {. } $$

Let \(A=\\{1,2,3\\}, B=\\{u, v\\}\), and \(C=\\{m, n\\} .\) List the elements of each of the following sets: a. \(A \times(B \times C)\) b. \((A \times B) \times C\) c. \(A \times B \times C\)

For all sets \(A\) and \(B, A-(A \cap B)=A-B\).

For all sets \(A\) and \(B\), $$ (A-B) \cup(B-A)=(A \cup B)-(A \cap B) $$

For all sets \(A, B, C\), and \(D\), if \(A \cap C=\emptyset\) then \((A \times B) \cap(C \times D)=\emptyset\).

Use mathematical induction and the following definitions to prove each statement in 35-37. If \(n\) is an integer with \(n \geq 3\) and if \(C_{1}, C_{2}, C_{3}, \ldots, C_{n}\) are any sets, \(C_{1} \cup C_{2} \cup C_{3} \cup \cdots \cup C_{n}=\left(C_{1} \cup C_{2} \cup C_{3} \cup \cdots \cup C_{n-1}\right) \cup C_{n}\), and \(C_{1} \cap C_{2} \cap C_{3} \cap \cdots \cap C_{n}=\left(C_{1} \cap C_{2} \cap C_{3} \cap \cdots \cap C_{n-1}\right) \cap C_{n} .\) (More rigorous versions of the definitions are given in Section 8.4.) Generalized Distributive Law for Sets: For any integer \(n \geq 1\), if \(A\) and \(B_{1}, B_{2}, B_{3}, \ldots, B_{n}\) are any sets, then $$ \begin{aligned} \left(A \cap B_{1}\right) \cup\left(A \cap B_{2}\right) \cup & \cdots \cup\left(A \cap B_{n}\right) \\ &=A \cap\left(B_{1} \cup B_{2} \cup B_{3} \cup \cdots \cup B_{n}\right) \end{aligned} $$

Consider the following set property: For all sets \(A, B\), and \(C,(A-B) \cup(B-C)=(A \cup B)-(B \cap C)\). a. Use an element argument to derive the property. b. Use an algebraic argument to derive the property (by applying properties from Theorem \(5.2 .2\) ). c. Comment on which method you found easier.

Refer to the definition of symmetric difference given above. Prove each of \(40-45\), assuming that \(A, B\), and \(C\) are all subsets of a universal set \(U\). $$A \Delta B=B \Delta A$$

Assume that \(B\) is a Boolean algebra with operations \(+\) and . Give the reasons needed to fill in the blanks in the proofs, but do not use 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 \(a\) in \(B, a \cdot a=a\). Proof: Let \(a\) be any element of \(B\). Then $$ \begin{array}{rlr} a & =a \cdot 1 & \frac{(a)}{(b)} \\ & =a \cdot(a+\bar{a}) & \frac{(b)}{(a)} \\ & =(a \cdot a)+(a \cdot \bar{a}) & \frac{(d)}{(d)} \\ & =(a)+0 & \\ & =a \cdot a & \text { (e) } \end{array} $$

Assume that \(B\) is a Boolean algebra with operations \(+\) and . Give the reasons needed to fill in the blanks in the proofs, but do not use 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 \(a\) and \(b\) in \(B,(a+b) \cdot a=a\). Proof: Let \(a\) and \(b\) be any elements of \(B\). Then $$ \begin{aligned} (a+b) \cdot a &=a \cdot(a+b) & & \frac{(\mathrm{a})}{(\mathrm{b})} \\ &=a \cdot a+a \cdot b & & \frac{(\mathrm{d})}{(\mathrm{c})} \\ &=a+a \cdot b & & \frac{(\mathrm{d})}{(\mathrm{d})} \\ &=a \cdot 1+a \cdot b & & \\ &=a \cdot(1+b) & & \frac{(\mathrm{e})}{\mathrm{by} \text { exercise } 49} \\ &=a \cdot(b+1) & & \text { (f) } \\ &=a \cdot 1 & & \end{aligned} $$