Simple algebra of sets "proof" I just wanted to ask about this equality
$A\cap B\cap C = (A\cap B)\cap (A\cap C)$
Can this easily be proven by using the associative property of sets which states?
$(A\cap B)\cap (A\cap C) = (A\cap A)\cap (B\cap C) = A\cap (B\cap C) = A\cap B\cap C$
Or is this not a valid mathematical proof
 A: Henno Brandsma tells you how to write down a formal proof in his comment:
$(A\cap B)\cap (A\cap C) =^{\mathcal A}\; A \cap (B \cap (A \cap C)) =^{\mathcal A}\; A \cap ((B \cap A) \cap C) =^{\mathcal C}\; A \cap ((A \cap B) \cap C) =^{\mathcal A}\; A \cap (A \cap (B \cap C) =^{\mathcal A}\; (A \cap A) \cap (B \cap C) =^{\mathcal I}\; A  \cap (B \cap C)$
where
$=^{\mathcal A}\;$ Apply the Associative Law
$=^{\mathcal C}\;$ Apply the Commutative Law
$=^{\mathcal I}\;$ Apply the Idempotent Law

You can provide more detail at each step:
$\tag 1 X \cap (Y \cap Z)= (X \cap Y) \cap Z$
Applying (1) (from right to left) with $X = A$, $Y = B$, and $Z = A \cap C$, we have
$(A\cap B)\cap (A\cap C) = A \cap (B \cap (A \cap C))$
etc.

You can also provide a shorter proof if you accept the claim that due to associativity and commutativity of the intersection of sets, you can just 'ignore /place parentheses and  rearrange the sets' to your hearts content:
$(A\cap B)\cap (A\cap C) = A \cap B \cap A \cap C =$
$\tag 1 A \cap A \cap B \cap C = (A \cap A)  \cap B \cap C$
By the Idempotent Law, (1) is equal to  $A  \cap B \cap C$
A: Elements $a$ from $A \cap B$ are elements that occur in the set $A$ as well as in the set $B$, formally $a\in A \wedge a \in B$.
Similary $a \in A \cap C => a \in A \wedge a \in C$.
Now you can do the same logical statement for $(A \cap B) \cap (A \cap C)$. Set $P = A \cap B$ and $Q = A \cap C$ cand you have
$a \in P \cap Q => (a \in P) \wedge (a \in Q) => (a \in A \wedge a \in B) \wedge (a \in A \wedge A \in C)$.
In this logical statement, the statement $a \in A$ occurs double (one is overfluent), therefore
$a \in P \cap Q => a \in A \wedge a \in B \wedge a \in C => a \in A \cap B \cap C$.
In the last conclusion I used again the definition of $\cap$.
q.e.d. 
A: $A\cap B\cap C = (A\cap B)\cap (A\cap C)$
First step
$A\cap B\cap C \subset (A\cap B)\cap (A\cap C)$
$\forall x\in A\cap B\cap C \implies \left \lbrace \begin{array}l &x\in A \quad (1)\\\text{and}&\\&x\in B\quad (2)\\\text{and}&\\&x\in C \quad (3)\end{array}\right.$
$\left \lbrace\begin{array}l &(1) \;\text{and} \; (2) \implies x\in A\cap B \quad(4)\\\text{and}&\\&(1) \;\text{and} \; (3) \implies x\in A\cap C \quad(5)\end{array}\right.\quad (4) \;\text{and} \; (5)\implies x\in (A\cap B)\cap(A\cap C)$
So we have proved $\boxed{A\cap B\cap C \subset (A\cap B)\cap (A\cap C)}$
Second step
$A\cap B\cap C \supset (A\cap B)\cap (A\cap C)$
$\forall x\in (A\cap B)\cap (A\cap C) \implies ....$
Conclusion
$ X\subset Y \quad \text{and} \quad  X\supset Y \iff X=Y$

Otherwise :
$\begin{array}l(A\cap B)\cap (A\cap C)&=\;\left[A\cap (B\cap A)\right]\cap C\quad &\text{associative}\\&=\;\left[A\cap (A\cap B)\right]\cap C \quad &\text{commutative}\\&=\;(A\cap A)\cap (B\cap C) \quad &\text{associative}\\&=\;A\cap (B\cap C)\quad &A\cap A=A\\&=\;A\cap B\cap C\quad &\text{associative}\end{array}$
