Let A,B and C be sets. Using set identities, show that: Let A,B and C be sets. Using set identities, show that:
(a) $A\cup( B-A ) = A\cup B$
(b) $(A-B)-C = (A-C)-(B-C)$
How should I solve this problems?
Thank you so much.
 A: $$A\cup(B\setminus A)=A\cup(B\cap A^{c})=(A\cup B)\cap(A\cup A^{c})=A\cup B$$
using the distributive law. Show the other one similarly.
A: (b) $(A-B)-C = (A-C)-(B-C)$
I will use other approach proving the above:
First I'll show that $(A\setminus B)\setminus C \subseteq (A\setminus C)\setminus(B\setminus C)$. 
Let $x\in (A\setminus B)\setminus C$. Thus $x\in (A\setminus B) $ and $x\notin C$, or  $x\in A $ and $x\notin B$ and $x\notin C$. Therefore $x\in (A\setminus C) $ and $x\notin B\setminus C$, thus $x\in (A\setminus C)\setminus( B\setminus C)$. Hence, $(A\setminus B)\setminus C \subseteq (A\setminus C)\setminus(B\setminus C)$.
Now, I will show that  $(A\setminus B)\setminus C \supseteq (A\setminus C)\setminus(B\setminus C)$
Let $x\in (A\setminus C)\setminus(B\setminus C)$. Then $x\in (A\setminus C)$ and $x\notin (B\setminus C)$. Thus, $x\in A$ and $x\notin C$, and $x\notin B$. I other words, $x\in (A\setminus B)\setminus C$, Hence, $(A\setminus B)\setminus C \supseteq (A\setminus C)\setminus(B\setminus C)$.
From these two above inclusions, we deduce $(A\setminus B)\setminus C =(A\setminus C)\setminus(B\setminus C)$
