Elementary symmetric difference: $A \Delta (B \cap C)\subseteq(A \Delta (A \cap C)) )\cup ((A\cap C) \Delta (B \cap C))$. I need to prove that $A \Delta (B \cap C)$ = $(A \Delta (A \cap C)) )\cup ((A\cap C) \Delta (B \cap C))$. I have convinced myself that the result is true, but I just can't seem to get the details right to prove that $A \Delta (B \cap C)$ is contained in $(A \Delta (A \cap C)) )\cup ((A\cap C) \Delta (B \cap C))$. Any suggestions are welcome.
 A: Note 1. We can use two definitions for $\Delta$ that are equivalent:
Def 1. $A\Delta B = (A\cup B)- (A\cap B)$
Def 2. $ A\Delta B= (A- B)\cup (B- A)$  

Note 2. there are the equalities below, which can be easily proved:
$$ (1)\quad \quad A\Delta (B \cap C)\underbrace{=}_{\text{Def 1}} [A\cup (B\cap C)]-A\cap B\cap C$$
$$(2)\quad \quad A \Delta (A \cap C)) )\cup ((A\cap C) \Delta (B \cap C))= (A-C)\cup [C\cap (A\Delta B)]$$
(using Def 2 and Intersection distributes over symmetric difference)

Let $x\in A\Delta (B \cap C)$. Then (by $(1)$), $x\notin A\cap B\cap C$ and either $x\in A$ or $x\in (B \cap C)$  


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*Let $x\in A.$ 


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*If $x\notin C$, then $x\in A-C $. Now see $(2)$ $ \checkmark $

*If $x\in C$, then since $x\in A$ and $x\notin A\cap B\cap C,$  we have $x\notin B\cap C$. So $x\in(A\cap C) \Delta (B \cap C)$. Now see Def 1.  $\checkmark$  


*Let $x\in (B\cap C)$. Then since  $x\notin A\cap B\cap C,$  we have $x\notin A.$ So $x\notin A\cap C$. So $x\in (A\cap C) \Delta (B \cap C) $ Now see Def 1. $ \checkmark$

