Prove that if $B\subseteq A$, then $A\backslash (A\backslash B) = B$. So I tried this problem and I just wanted to make sure I did it right. Here it is:
Proof:
Given: $B \subseteq A$
I.) Let $x\in A\backslash(A\backslash B)$.
It follows that $x\in A$ AND $x\notin A\backslash B$, but $x\notin A\backslash B$ is logically equivalent to $x\notin A$ OR $x\in B$.
By the distributive law for logic that gives ($x\in A$ AND $x\notin A$) OR ($x\in A$ AND $x\in B$), which is just False OR ($x\in A$ AND $x\in B$).
That leaves $x\in A$ AND $x\in B$ which implies that $x\in B$, and thus $A\backslash(A\backslash B) \subseteq B.$
II.) Let $x\in B$.
Then its clear that $x\in A$ (because $B \subseteq A$) and $x\notin A\backslash B$ (because $x\in B$). However that's the same thing as saying that $x\in A\backslash(A\backslash B)$, so $B\subseteq A\backslash(A\backslash B)$.
Therefore $A\backslash (A\backslash B) = B$ if $B\subseteq A$.
Is my proof correct?
Your proof looks ok, but it can be much simpler if you just usebasic properties
$$
A \setminus (A \setminus B) = A \cap \overline{(A \cap \overline B)}=A\cap (\overline A \cup B) = (A \cap \overline A)\cup (A \cap B)=A \cap B = B
$$
let $$A=B\cup C$$ such that $B$ and $C$ are disjoint. Then, $$A-(A-B)=A-C=(B\cup C)-C=B\cup C-C=B$$ (Because $B$ and $C$ are disjoint)