Simplifying number of sets in a relationship Got this monster set
((A∩B) ∪C ) ∪ (A∪(B∩C))
I'm trying to reduce the number of sets to be as small as possible using set identities
Set Rules 
All I can think of is to apply distribution law
((C∪A) ∩ (C∪B) ∪ (A∪B) ∩ (A∪C))
Maybe I could play around with the middle terms (C∪B) ∪ (A∪B) but is that allowed?
 A: You have a union, ultimately; this can be "simplified" to:
$$(A \cap B) \cup C \cup A \cup (B \cap C).$$
But $(A \cap B) \cup A = A$, and $C \cup (B \cap C) = C$, so this simplifies further to just $A \cup C$.
That is,
$$(A \cap B) \cup C \cup A \cup (B \cap C) = A \cup C.$$
Try drawing a Venn diagram for small problems like this, to get an idea for what's going on.
A: Here is a hint.  
What is a simpler way to describe $A \cup (A \cap B)$? 
What is a simpler way to describe $C \cup (B \cap C)$?
A: Formally, set union is associative and commutative, so you can unbracket and reshuffle the set union operations there. Then the intersection of a set with another set is no bigger than either of those sets, and making a union back with one of those sets just gives you that set - nothing extra, $ (S\cap T) \cup S = S$.
\begin{align}((A \cap B) \cup C) \cup (A \cup (B \cap C))  
&= (A \cap B) \cup C \cup A \cup (B \cap C) \\
&= ((A \cap B) \cup A)\; \cup \; ((B \cap C) \cup C) \\ 
(A \cap B) \cup A &= A \\
(B \cap C) \cup C &= C \\
\therefore ((A \cap B) \cup C) \cup (A \cup (B \cap C)) &= A \cup C
\end{align}
Trying to push along your track of applying the distributive law gets messy... I had a look and basically you eventually have to "undistribute" and apply something like the above anyway.
A: The union of two sets, each of which is the union of some two sets, is the union of four sets:  $(W \cup X) \cup (Y \cup Z) = W \cup X \cup Y \cup Z$, where the unions can be taken in any order.  
In your case, we are given the expression $((A \cap B) \cup C) \cup (A \cup (B \cap C))$. By the argument in the previous paragraph, this is equal to $(A \cap B) \cup C \cup A \cup (B \cap C)$.  Observe that the first subset $A \cap B$ is a subset of the third subset $A$; the fourth subset $ B \cap C$ is a subset of the second subset $C$.  Hence, the first and fourth subsets can be ignored, and the expression evaluates to $C \cup A$. 
When the number of different variables in the expression is 3 or fewer, the answer can be obtained by drawing a Venn diagram.  In your expression there are only three different variables $A, B$ and $C$, so you can draw the Venn diagram and verify that you get $C \cup A$.  
