Discrete Math Proof: $A \cup B$ I'm preparing ahead for a Discrete Math course coming up this year by doing some practice problems supplemented by online notes. 
The problem I'm having trouble proving is the following: 
$A \cup B = (A \cap B^C) \cup (A^C \cap B) \cup (A \cap B)$, where $^C$ denotes complement of a set
I know I should use some of the Laws of the algebra of sets, but this is all so new that it's difficult for me to conceptualize which law to begin with. 
UPDATE:
For any two finite sets A and B: $A \cup B = (A\B) \cup B$
$Using: (X\Y) = (X \cap Y^C)$
I found that:
$A \cup B = (A \cap B^C) \cup B$, substituting for A
$A \cup B = (A \cap B^C) \cup (B \cap A^C)$, substituting for A and B
SECOND UPDATE:
After gaining insight from everyone I've figured it out (or so I hope).
$A \cup B = (A \cap B^C) \cup (A^C \cap B) \cup (A \cap B)$
Rearrange using Commutative laws:
$(A \cap B^C) \cup (A \cap B) \cup (A^C \cap B)$
Applying the Distributive laws we get:
$(A \cap (B^C \cup B)) \cup (A^C \cap B)$
Then by Complement laws:
$A \cap U \cup (A^C \cap B)$
Using the Identity laws:
$A \cup (A^C \cap B)$
Using the Distributive laws:
$(A \cup A^C) \cap (A \cup B)$
By Complement laws:
$U \cap (A \cup B)$
Finally, using Identity laws we get:
$A \cup B$
 A: Intuitively the result says that if an element is in $A \cup B$, then either the element is in $A$ and not $B$, $B$ and not $A$, or in both $A$ and $B$. 
The arguement can then proceed as follows:
$\begin{array}{l l} x \in A \cup B & \Leftrightarrow x \in A \text{ and } x \not\in B \text{, or } x \not\in A \text{ and } x \in B \text{, or } x \in A \text{ and } x \in B\\  & \Leftrightarrow  x \in (A \cap B^c)\cup(A^c \cap B)\cup(A \cap B) \end{array}$
A: Two things.
First, you should not begin your proof by stating what you intend to prove. Instead of starting by writing $A \cup B = (A \cap B^C) \cup (A^C \cap B) \cup (A \cap B)$ and aiming to reach $A\cup B = A\cup B$, you should start with $(A \cap B^C) \cup (A^C \cap B) \cup (A \cap B)$ and show that it equals (through rewriting it in various forms that are the same set) as $A\cup B$.
(You can also show that two sets are equal by showing that each is a subset of the other; I don’t mean that my suggestion is the only approach, but it’s the appropriate way to write the proof using the particular mathematical argument you are presenting.)
This may seem like a silly question of style, but the fact is that you have only explained that if what you want to prove happens to be true, then it must follow that $A\cup B = A\cup B$. This is not a proof. It would also follows that $A\cup B = A\cup B$ even if what you began with is false. Any assumptions, even false ones, logically imply a true statement. A conjecture is not proven true by showing that you can use it to conclude a true statement. You have to show that you can proceed logically from true statements to prove it.
Second, you should add some parentheses. At one point (after you use the distributive law), you write the expression $ A \cap (B^C \cup B) \cup (A^C \cap B)$. You intend for this to mean $\left(A \cap (B^C \cup B)\right) \cup (A^C \cap B)$, and it’s safest to say so, because combinations of $\cup$ and $\cap$ may give different results depending on the order in which the operators are evaluated.
