for this problem (A-B) U (A (upside U) B). simplify using laws of set theory.

where do i even begin. my first time using this. not even sure how to put in correct symbols.
any help would be awesome.

Originally Posted by nvm0019

for this problem (A-B) U (A (upside U) B). simplify using laws of set theory.

Consider $\displaystyle U$ as universal set (for example $\displaystyle U=A\cup B$) then:

$\displaystyle (A-B)\cup (A\cap B)=(A\cap B^c)\cup (A\cap B)=\ldots=A$

(Use one distributive law, $\displaystyle M^c\cup M=U$ and $\displaystyle N\cap U=N$ )

Hello, nvm0019!

$\displaystyle \text{Simplify: }\:(A - B) \cup (A \cap B)$

$\displaystyle \begin{array}{cccccccc}1. & (A-B) \cap (A\cap B) && 1. & \text{Given} \\ \\[-3mm] 2. & (A\;\cap \sim\!B) \cup (A \cap B) && 2. & \text{def. Subtr'n} \\ \\[-3mm] 3. & (A \cup A) \cap (A \cup B) \cap (\sim\!B \cup A) \cap (\sim\!B \cup B) && 3. & \text{Distributive} \\ \\[-3mm] 4. & A \cap (A \cup B) \cap (A\;\cup \sim\!B) \cap U && 4. & s \cup s \:=\:s \\ &&&& \sim\!s \cup s \:=\:U \\ \\[-3mm] 5. & A \cap (A \cup B) \cap (A\; \cup \sim\!B) && 5. & s \cap U \:=\:s \\ \\[-3mm] 6. & A \cap \big[A \cup (B \;\cap \sim\!B)\big] && 6. & \text{Distributive} \\ \\[-3mm] 7. & A \cap \big[A \cup \emptyset\big] && 7. & s\;\cap\sim\!s \:=\:\emptyset \\ \\[-3mm] 8. & A \cap A && 8. & s \cup \emptyset \:=\:s \\ \\[-3mm] 9. & A && 9. & s \cap s \:=\:s \end{array}$