Distributing Set Intersections Over an Intersection I was working through some examples, and found this to be true: 
$(A \cap B) \cap (B \cap C) = A \cap B \cap C $
$(A \cap B) \cap(A \cap C) = A \cap B \cap C$
$(A \cap B) \cap(A \cap C) \cap (B \cap C) = A \cap B \cap C$
Is there some proof, or rule for these statements to hold true?
Thank you
 A: Of course you can prove them, starting from the definition of intersection of sets :

$x \in A \cap B$ iff $x \in A$ and $x \in B$

and the "basic fact" that equality between sets amounts to mutual inclusion : $A = B$ iff $A \subseteq B$ and $B \subseteq A$.
In turn, to prove inclusion you have to use its definition :

$A \subseteq B$ iff : if $x \in A$, then $x \in B$.

You can try yourself the above "machinery" proving that :


$(A \cap B) \cap (B \cap C) = A \cap B \cap C$.


A: Note:
$$
A \cap B = B \cap A
$$
so that all three statements are related. Now we can prove the first one (and thus all three) through double inclusion i.e. we prove
$$
(A \cap B) \cap (B \cap C) \subset A \cap B \cap C \text{ and } (A \cap B) \cap (B \cap C) \supset A \cap B \cap C
$$
which shows
$$
(A \cap B) \cap (B \cap C) = A \cap B \cap C
$$
Do you know how to show the above?
Spoiler:

 $x \in (A \cap B) \cap (B \cap C) \implies x \in A, B, C \implies x \in A \cap B \cap C$. $x \in A \cap B \cap C \implies x \in A,B , x \in B,C \implies x \in (A \cap B) \cap (B \cap C)$

A: $(A \cap B) \cap (B \cap C) = A \cap (B \cap B) \cap C  = A \cap B \cap C$, using associativity of $\cap$ (i.e. it doesn't matter where we put brackets), and $B \cap B = B$.
The second on wis similar, but we also use commutativity (the order in $\cap$ does not matter): $(A \cap B) \cap (A \cap C) = A \cap B \cap A \cap C = (A \cap A) \cap B \cap C = A \cap B \cap C$. 
You probably yes how to do the third one this way as well.
A: Intersection is both commutative and associative, so we can change the order of your expressions and remove (and add) parenthesis. Doing this for the first expression, we get:
$(A \cap B) \cap (B \cap C) = A \cap B \cap B \cap C = A \cap (B \cap B) \cap C$.
Now, the intersection of two sets (informally) is the set which has both of the two sets in question. If the two sets are the same, then every element in one is in the other, and so the intersection of a set B with itself is B. Thus, $B \cap B = B$.
The expression in question thus simplifies to $A \cap B \cap C$.
A very similar argument can be used to simplify the second expression and the third one.
