Prove $(A \cap B) \cup (A \cap B')= A$ using Set Identities I recently started a Discrete Mathematics course in college and I am having some difficulties with one of the homework questions. I need to learn this, so please guide me through at least two steps to get the ball rolling. 
The question reads: Show that if $A$ and $B$ are sets, then: $(A \cap B) \cup (A \cap B')=A$
We are supposed to use set identities. I had a question prior, but it was simple: $(A \cap B \cap C)' = A'\cup B' \cup C'$  - Which would be one of De Morgan's laws.
I am at a loss. I have been reading the textbook and tried looking up some videos, but I am not sure exactly where to start. Any help you can provide, will be greatly appreciated! 
Thanks,
Kei
 A: The idea is to achieve get close $B$ and $B^c$. Then we use distributive property: $(A\cap B)\cup(A\cap B^c)=A\cap (B\cup B^c)=A\cap X=A $,
with $X$ the universe
A: Consider an element of $A$ - either it is in $B$, or it isn't, and thus is in the complement of $B$. Thus $A \subset (A \cap B)$ $\cup$ $(A \cap B')$. Now, try to argue on your own that the reverse "inclusion" holds: that we have $A \supset (A \cap B)$ $\cup$ $(A \cap B')$. 
A: You can use identities such as $(A\cap B) \cup (A \cap C) = A \cap (B \cup C)$ to get $(A \cap B) \cup (A \cap B') = A \cap (B \cup B') = A \cap U = A$.
But I prefer to think of what it is saying.  $A \cap B$ means "everything in A and in B" and $A \cap B'$ means "everything that is in A that is not in B" and $(A\cap B) \cup (A \cap B')$ means "every thing that is in A and B combined with everything that is not in B".  Is there a logical reason that "everything in A and B combined with everything in A and not in B" would be "A"?
Well, I hope it should be obvious.  Everything in A is either in B or not in B so combining the items of A that are not in B with those that are should give you all the items in A.
So the best way to express that idea directly would be:
$A = A \cap U = A \cap (B \cup B') = (A \cap B) \cup (A\cap B')$.  Or if that's a little too abstract, I rather like to do an element by element proof:
Let $x \in A$ either $x \in B$ or $x \in B'$.  If $x \in B$ then $x \in A \cap B$.  If $x \in B'$ then $x \in A \cap B'$.  Either way $x \in A \cap B$ or $x \in A\cap B'$ so $x \in (A \cap B) \cup (A\cap B)$.  So $A \subseteq (A\cap B) \cup (A \cap B)$.  Likewise if $y \in (A \cap B) \cup (A\cap B)$ then either $y \in (A \cap B) \subset A$ or $y \in (A\cap B') \subset A$.  Either way, $y \in A$ so $(A\cap B)\cup (A\cap B') \subseteq A$.
$A \subseteq (A\cap B) \cup (A \cap B)$ and $(A\cap B)\cup (A\cap B') \subseteq A$, so $$(A\cap B)\cup (A\cap B') = A$.
A fourth way is the big guns.
Let $x \in U$ now one of four things might happen:
1) $x \in A$ and $x \in B$.  Then $x \in A$ and $x \in A\cap B$ and $x \in (A \cap B) \cup (A \cap B')$.
2) $x \in A$ and $x \not \in B$. Then $x \in A$ and $x \in B'$ and $x \in A \cap B'$ and $x \in (A \cap B) \cup (A \cap B')$
3) $x \not \in A$ and $x \in B$.  Then $x \not \in A$ and $x \not \in A \cap B$ and $x \not \in A \cap B'$ so $x \not \in  (A \cap B) \cup (A \cap B')$.
4) $x \not \in A$ and $x \not \in B$. Then $x \not \in A$ and $x \not \in A \cap B$ and $x \not \in A \cap B'$ so $x \not \in  (A \cap B) \cup (A \cap B')$.
Looking at the four cases we see $x \in A \iff x  \in  (A \cap B) \cup (A \cap B')$.  Thus $A$ and $(A \cap B) \cup (A \cap B')$ have precisely the same elements and neither has any element the other doesn't.  In other words, $A = (A \cap B) \cup (A \cap B')$.
