Chapter 2.2, Problem 8E

To determine

Prove that $\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)-\left(B\cap A\right)$ from Exercise 4.

Prove that ${\left(A\cap B\right)}^c=A^c\cup B^c$ from Exercise 6.

Prove that $\left(A\cap B\right)\cup C=\left(A\cup C\right)\cap \left(B\cup C\right)$ from Exercise 7.

Explanation of Solution

Calculation:

Define the events A and B. The events A and B are considered as equal when $A\subseteq B$ and $B\subseteq A$. The event A would be subset of B for an element x, if $x\in A$ then $x\in B$ is true.

Pair of events $\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)-\left(B\cap A\right)$:

Let x be an element.

$\begin{array}{l}x\in \left(A-B\right)\cup \left(B-A\right)\iff x\in \left(A-B\right)\text{or}x\in \left(B-A\right)\\ \iff \left(x\in A\text{but}x\notin B\right)\text{or}\left(x\in B\text{but}x\notin A\right)\\ \iff x\in A\text{or}x\in B\\ \iff x\in \left(A\cup B\right)\text{and}x\notin \left(A\cap B\right)\\ \iff x\in \left(A\cup B\right)-\left(A\cap B\right)\end{array}$

The event $\left(A-B\right)\cup \left(B-A\right)\subseteq \left(A\cup B\right)-\left(B\cap A\right)$ because $x\in \left(A-B\right)\cup \left(B-A\right)$ then $x\in \left(A\cup B\right)-\left(A\cap B\right)$ is also true.

$\begin{array}{l}x\in \left(A\cup B\right)-\left(A\cap B\right)\iff x\in \left(A\cup B\right)\text{and}x\notin \left(A\cap B\right)\\ \iff x\in A\text{or}x\in B\\ \iff \left(x\in A\text{but}x\notin B\right)\text{or}\left(x\in B\text{but}x\notin A\right)\\ \iff x\in \left(A-B\right)\text{or}x\in \left(B-A\right)\\ \iff x\in \left(A-B\right)\cup \left(B-A\right)\end{array}$

The event $\left(A\cup B\right)-\left(B\cap A\right)\subseteq \left(A-B\right)\cup \left(B-A\right)$ because $x\in \left(A\cup B\right)-\left(A\cap B\right)$ then $x\in \left(A-B\right)\cup \left(B-A\right)$ is also true.

Hence, it is proved that $\left(A-B\right)\cup \left(B-A\right)=\left(A\cup B\right)-\left(B\cap A\right)$.

Pair of events ${\left(A\cap B\right)}^c=A^c\cup B^c$:

$\begin{array}{l}x\in {\left(A\cap B\right)}^c\iff x\notin A\cap B\\ \iff x\in \left(A-B\right)\text{or}x\in \left(B-A\right)\text{or}x\in {\left(A\cup B\right)}^c\\ \iff \left[x\in \left(A-B\right)\text{or}x\in {\left(A\cup B\right)}^c\right]\text{or}\left[x\in \left(B-A\right)\text{or}x\in {\left(A\cup B\right)}^c\right]\\ \iff x\in B^c\text{or}x\in A^c\\ \iff x\in \left(A^c\cup B^c\right)\end{array}$

The event ${\left(A\cap B\right)}^c\subseteq A^c\cup B^c$ because $x\in {\left(A\cap B\right)}^c$ then $x\in \left(A^c\cup B^c\right)$ is also true.

$\begin{array}{l}x\in \left(A^c\cup B^c\right)\iff x\in B^c\text{or}x\in A^c\\ \iff \left[x\in \left(A-B\right)\text{or}x\in {\left(A\cup B\right)}^c\right]\text{or}\left[x\in \left(B-A\right)\text{or}x\in {\left(A\cup B\right)}^c\right]\\ \iff x\in \left(A-B\right)\text{or}x\in \left(B-A\right)\text{or}x\in {\left(A\cup B\right)}^c\\ \iff x\notin A\cap B\\ \iff x\in {\left(A\cap B\right)}^c\end{array}$

The event $A^c\cup B^c\subseteq {\left(A\cap B\right)}^c$ because $x\in \left(A^c\cup B^c\right)$ then $x\in {\left(A\cap B\right)}^c$ is also true.

Hence, it is proved that ${\left(A\cap B\right)}^c=A^c\cup B^c$.

Pair of events $\left(A\cap B\right)\cup C=\left(A\cup C\right)\cap \left(B\cup C\right)$:

$\begin{array}{l}x\in \left(A\cap B\right)\cup C\iff x\in \left(A\cap B\right)\text{or}x\in C\\ \iff \left[x\in A\text{and}x\in B\right]\text{or}x\in C\\ \iff \left[x\in A\text{or}x\in C\right]\text{and}\left[x\in B\text{or}x\in C\right]\\ \iff x\in A\cup C\text{and}x\in B\cup C\\ \iff x\in \left(A\cup C\right)\cap \left(B\cup C\right)\end{array}$

The event $\left(A\cap B\right)\cup C\subseteq \left(A\cup C\right)\cap \left(B\cup C\right)$ because $x\in \left(A\cap B\right)\cup C$ then $x\in \left(A\cup C\right)\cap \left(B\cup C\right)$ is also true.

$\begin{array}{l}x\in \left(A\cup C\right)\cap \left(B\cup C\right)\iff x\in A\cup C\text{and}x\in B\cup C\\ \iff \left[x\in A\text{or}x\in C\right]\text{and}\left[x\in B\text{or}x\in C\right]\\ \iff \left[x\in A\text{and}x\in B\right]\text{or}x\in C\\ \iff x\in \left(A\cap B\right)\text{or}x\in C\\ \iff x\in \left(A\cap B\right)\cup C\end{array}$

The event $\left(A\cup C\right)\cap \left(B\cup C\right)\subseteq \left(A\cap B\right)\cup C$ because $x\in \left(A\cup C\right)\cap \left(B\cup C\right)$ then $x\in \left(A\cap B\right)\cup C$ is also true.

Hence, it is proved that $\left(A\cap B\right)\cup C=\left(A\cup C\right)\cap \left(B\cup C\right)$.