Prove that $ (A_1 \cap \dots \cap A_n) \triangle (B_1 \cap \dots \cap B_n) \subset (A_1 \triangle B_1) \cup \dots \cup (A_n \triangle B_n) $ Prove that $ (A_1 \cap \dots \cap A_n) \triangle (B_1 \cap \dots \cap B_n) \subset (A_1 \triangle B_1) \cup \dots \cup (A_n \triangle B_n) $ is true for any sets $A_1, \dots , A_n$ and $B_1, \dots , B_n  $
I tried to solve it using math induction.
n = 1: $A_1 \triangle B_1 \subset A_1 \triangle B_1$ is true
n = m: $ (A_1 \cap \dots \cap A_m) \triangle (B_1 \cap \dots \cap B_m) \subset (A_1 \triangle B_1) \cup \dots \cup (A_m \triangle B_m) $
n = m + 1: $ (A_1 \cap \dots \cap A_m) \triangle (B_1 \cap \dots \cap B_m) \cup A_{m+1} \triangle B_{m+1} \subset (A_1 \triangle B_1) \cup \dots \cup (A_{m+1} \triangle B_{m+1})$
But I have no idea what to do next

Notice that
$$A\triangle B=(A\setminus B) \cup (B\setminus A) = (A\cap B^c)\cup(B\cap A^c)=A^c\triangle B^c $$
So
$$\Big(\bigcap^n_{k=1} A_k\Big)\triangle \Big(\bigcap^n_{j=1}B_j\Big)= \Big(\bigcup^n_{k=1} A^c_k\Big)\triangle \Big(\bigcup^n_{j=1}B^c_j\Big)$$
Now, using distributive properties of union and intersection, i.e.,  $C\cap\Big(\bigcup_\alpha D_\alpha\Big)=\bigcup_\alpha (D_\alpha \cap C)$, we obtain that
$$
\begin{align}
\Big(\bigcup^n_{k=1} A^c_k\Big)\triangle \Big(\bigcup^n_{j=1}B^c_j\Big)&=
\left(\Big(\bigcup^n_{k=1}A^c_k\Big)\setminus\Big(\bigcup^n_{j=1}B^c_j\Big)\right)\cup \left(\Big(\bigcup^n_{j=1}B^c_j\Big)\setminus\Big(\bigcup^n_{k=1}A^c_k\Big)\right)\\
&= \left(\Big(\bigcup^n_{k=1}A^c_k\Big)\cap\bigcap^n_{j=1}B_j\right)\cup
\left(\Big(\bigcup^n_{j=1}B^c_j\Big)\cap\bigcap^n_{k=1}A_k\right)\\
&=\left(\bigcup^n_{k=1}\Big(A^c_k\cap\bigcap^n_{j=1}B_j\Big)\right)  \cup
\left(\bigcup^n_{j=1}\Big(B^c_j\cap\bigcap^n_{k=1}A_k\Big)\right) \\
&\subset 
\left(\bigcup^n_{k=1}\Big(A^c_k\cap B_k\Big)\right)  \cup
\left(\bigcup^n_{j=1}\Big(B^c_j\cap A_j\Big)\right)\\
&=\bigcup^n_{k=1}\big((B_k\setminus A_k)\cup (A_k\setminus B_k)\big)=\bigcup^n_{k=1}A_k\triangle B_k
\end{align}
$$
