How to prove $A = (A \cap B) \cup (A - B)$ and show that $(A \cap B)$ and $(A - B)$ are non-intersecting. I thought about starting with $A = (A \cap B) \cup (A)$ since $(A-B)$ is essentially saying only consider $A$. This would then lead me to proving $A = (A \cap B)$. However I'm completely stuck on how to prove they are non-intersecting.
 A: HINT
You can use $A - B = A \cap B^C$ 
A: By definition:
$A\cap B=\{x|x \in A \text { and }x\in B\}$
$A-B=\{x|x \in A \text { and } x \not \in B\} $
$C\cup D=\{x| x \in C \text { OR } x\in D\} $
So $(A\cap B)\cup (A-B)=$
$\{x|(x\in A \text { and }x\in B)\text { OR } (x\in A \text { and }x \not \in B)\}= $
$\{x|x \in A \text { and } x \text{ either is or isn't in } B\}=$
$\{x|x \in A \text { and nothing can be said about whther} x \text { is in }B\}= $
$\{x|x\in A\} =$
$A $.
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Likewise
$(A\cap B)\cap (A-B)=$
$\{x|(x\in A\text{ and } x\in B)\text{ and } (x\in A \text { and }x\not \in B\} =$
$\{x|x \in A \text { and } x \text { is both in and not in }B\} $
There can't be any elements that are both in and not in a set.  So these do not intersect.
A: Let $A,B \subset X$.
Using Bram28's hint:  $A - B = A \cap B^C$.
1) Distributive Law:
$(A \cap B) \cup (A \cap B^C) = $
$A \cap (B \cup B^C) = A \cap X = A$ .
2) Associative and Commutative Law:
$(A \cap B) \cap (A \cap B^C) = $
$A \cap (B \cap B^C) = A \cap \emptyset = \emptyset$.
