I need to prove that: $ (A_1\cap A_2\cap ...\cap A_n) \Delta (B_1\cap B_2\cap ...\cap B_n) \subset (A_1\Delta B_1) \cup (A_2\Delta B_2)\cup...\cup(A_n\Delta B_n) $ and show that inverse statement is not correct. But, appling the symmetric difference formulas doesn't really helps. For left side for the case of $(A_1\cap A_2) \Delta (B_1\cap B_2)$ I get $(A_1\cap A_2)\setminus(B_1\cap B_2)\cup(B_1\cap B_2)\setminus(A_1\cap A_2)$, and for the right side I get $(A_1\setminus B_1)\cup (B_1\setminus A_1)\cup (A_2\setminus B_2)\cup (B_2\setminus A_2)$, but it doesn't help much. Anyway, I gets worse in case if I add $A_3$ and $B_3$ to the statement, as the formula becomes more complicated.
Addition:
Thanks Martin Sleziak for the formula! I didn't know it before. It seems like using De Morgan's law can help. The left side: $ (A_1\cap A_2 \cap A_3)\Delta(B_1\cap B_2 \cap B_3) = ( A_1\cap A_2 \cap A_3)\cap (B_1\cap B_2 \cap B_3)' \cup ( A_1\cap A_2 \cap A_3)'\cap (B_1\cap B_2 \cap B_3) = ( A_1\cap A_2 \cap A_3)\cap (B_1'\cup B_2' \cup B_3') \cup ( A_1'\cup A_2'\cup A_3')\cap (B_1\cap B_2 \cap B_3) = (B_1'\cap A_1\cap A_2 \cap A_3) \cup (B_2'\cap A_1\cap A_2 \cap A_3) \cup (B_3'\cap A_1\cap A_2 \cap A_3) \cup (A_1'\cap B_1\cap B_2 \cap B_3) \cup (A_2'\cap B_1\cap B_2 \cap B_3)\cup (A_3'\cap B_1\cap B_2 \cap B_3) $ while the right side is: $ (A_1\Delta B_1) \cup (A_2\Delta B_2) \cup (A_3\Delta B_3) = (B_1'\cap A_1) \cup (A_1'\cap B_1) \cup (B_2'\cap A_2) \cup (A_2'\cap B_2) \cup (B_3'\cap A_3) \cup (A_3'\cap B_3) $ As you can see, there are 6 terms on both sides. Moreover, each term from the left side has a pair on the right side: $(A_3'\cap B_1\cap B_2 \cap B_3)$ and $(A_3'\cap B_3)$, and so on. I think, that the following is correct $(A_3'\cap B_3\cap B_2 \cap B_1) \subseteq (A_3'\cap B_3)$, because the additional intersection with $B_2\cap B_3$ can only reduce (or left unchanged) the size of $(A_3'\cap B_3)$. So, the left side of the equation is a subset of the right. Here I show that both left and right sides can be equal in some cases, but the original equation is strict. Why did that happen?