Excuse me,how to do this type of question?
Using the laws of the algebra of sets show that
(a-b)∩(a∩b)'=a∩b'
$\displaystyle (A \backslash B) \cap (A \cap B)'$
$\displaystyle = (A \cap B') \cap (A' \cup B')$
$\displaystyle = A \cap B' \cap (B' \cup A')$
$\displaystyle = A \cap B'$
Where we used the following rules...
$\displaystyle (A \backslash B) = (A \cap B')$
$\displaystyle (A \cap B)' = (A' \cup B')$
$\displaystyle B' \cap (B' \cup A') = B'$
Hello everyone
All that Deadstar says is correct, but I think the statement$\displaystyle B' \cap(B'\cup A') = B'$(although obvious) requires further proof - it's not one of the usual Set Laws (although some people do include it as an Absorption Law).
You prove it like this:$\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$Grandad$\displaystyle =B'\cup(\oslash \cap A')$
$\displaystyle =B'\cup\oslash$
$\displaystyle =B'$
Hello mayThis uses the Distributive Law:$\displaystyle (P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$where I have replaced $\displaystyle P$ by $\displaystyle B'$, $\displaystyle Q$ by $\displaystyle \oslash$ and $\displaystyle R$ by $\displaystyle A'$.
Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!
Grandad
PS ... and to answer your extra question, following your edit:
$\displaystyle \oslash \cap A' = \oslash$ uses what I have always known as an Identity Law, but others sometimes refer to as a Domination Law.
See, for example, here.