How to prove $A=(A\setminus B)\cup (A\cap B)$ How to prove $A=(A\setminus B)\cup (A\cap B)$. I have seen this problem and the solution is clear to me. Initially I was satisfied by my prove but now I think it is wrong. 
How I have proved
$$(A\cup B)=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)\\
  (A\cup B)\cap A=[(A\setminus B)\cup(A\cap B)\cup(B\setminus A)]\cap A=[(A\setminus B)\cap A]\cup[(A\cap B)\cap A]\cup[(B\setminus A)\cap A]\\
  A=(A\setminus B)\cup(A\cap B)\cup\phi=(A\setminus B)\cup(A\cap B)$$
Have I proved it correctly. If yes then how? I mean that I have used $A=(A\setminus B)\cup (A\cap B)$ and $B=(B\setminus A)\cup (A\cap B)$ to get the $(A\cup B)=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)$ and then I'm using it to prove $A=(A\setminus B)\cup (A\cap B)$. This means that I'm using a statement to prove itself!
Kindly help me.
 A: Assume that if $x \in (A \setminus B) \cup (A \cap B)$
$\Leftrightarrow x \in (A \setminus B) \lor x \in (A \cap B)$
$\Leftrightarrow (x \in A \land x \in B^{\complement}) \lor (x \in A \land x \in B)$
$\Leftrightarrow (x \in A) \land (x \in B \lor x \in B^{\complement})$
$\Leftrightarrow (x \in A) \land (x \in U)$
$\Leftrightarrow x \in A$
A: Hint: $(A\setminus B)\cup (A\cap B)=(A\cap B')\cup (A\cap B)$
Take $(A\cap B)=X$, then we have,
$$(A\setminus B)\cup (A\cap B)=(A\cap B')\cup X$$
$$=(A\cup X)\cap (B'\cup X)$$
$$=A\cap (A \cup B')=A$$
To finish the proof show that: $(A\cup X)=A$ and $(B'\cup X)=(A\cup B')$
Can you finish from here?
A: Your proof is not correct.  If you start by assuming what you are trying to prove, then whatever follows doesn't prove anything.
The question you linked has several correct proofs, if you want to see the right way to go about proving this.  (I just added an answer to that question proving the claim directly using the algebra of sets, which might be the kind of proof you're looking for.)
