$\big[(X \cap A^c)\cup A\big] \backslash \big[(C \cap A^c)\cup B\big] = \big[(X \cap A^c)\backslash (C \cap A^c)\big] \cup (A\backslash B)$?

Question

In J. Yeh's Real Analysis: Theory of Measure and Integration, he claims that

$$\big[(X \cap A^c)\cup A\big] \backslash \big[(C \cap A^c)\cup B\big] = \big[(X \cap A^c)\backslash (C \cap A^c)\big] \cup (A\backslash B)$$

if $X\cap A^c \supset C\cap A^c$ and $A \supset B$.

All I've done so far is to distribute the set difference over the union as shown below:

$ \begin{align*} \big[(X \cap A^c)\cup A\big] \backslash \big[(C \cap A^c)\cup B\big] &= \big[(X \cap A^c)\cup A\big] \cap \big[(C \cap A^c)\cup B\big]^c \\ &= \big[(X \cap A^c)\cup A\big] \cap \big[(C \cap A^c)^c \cap B^c \big]\\ &= \bigg[(X \cap A^c)\cap \big[(C \cap A^c)^c \cap B^c \big]\bigg] \cup \bigg[A \cap \big[(C \cap A^c)^c \cap B^c \big]\bigg]\\ &= \big[(X \cap A^c)\cap (C \cap A^c)^c \cap B^c \big] \cup \big[A \cap (C \cap A^c)^c \cap B^c \big] \end{align*}$

I can't seem to understand how he gets rid of $B^c$ from $\big[(X \cap A^c)\cap (C \cap A^c)^c \cap B^c \big]$ and $(C\cap A^c)^c$ from $ \big[A \cap (C \cap A^c)^c \cap B^c \big]$ with the relations given.

Original text included:

Any help given is highly appreciated!

Answer 1

To simplify your last expression:

$\big[(X \cap A^c)\cap (C \cap A^c)^c \cap B^c \big] \cup \big[A \cap (C \cap A^c)^c \cap B^c \big]$

You could take the following steps:

1. In the first term $\big[(X \cap A^c)\cap (C \cap A^c)^c \cap B^c \big] $ note that since $B^c \supset A^c$, then $X\cap A^c \cap B^c = X\cap A^c$ so you can remove the $B^c$ factor.

2. In the second term $\big[A \cap (C \cap A^c)^c \cap B^c \big]$ note that $(C\cap A^c)^c = A\cup C^c \supset A$ so that $A\cap (C\cap A^c)^c = A$. So you can remove this middle factor.

Now you should be left with $\big[(X \cap A^c)\cap (C \cap A^c)^c\big] \cup \big[A \cap B^c \big]$ which you know how to deal with.