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
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?
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.
Here is a hint.
What is a simpler way to describe $A \cup (A \cap B)$?
What is a simpler way to describe $C \cup (B \cap C)$?
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.
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 $C \cup A$.
