Symmetric difference of sum of sets included in sum of their symmetric differences I've got the following problem. I am supposed to prove that:
$$(A1\cup A2)\oplus(B1\cup B2)\subseteq (A1\oplus B1) \cup (A2\oplus B2)$$
where $\oplus$ is the symmetric difference.
I know that from definition:
$$X\oplus Y = (X\setminus Y)\cup (Y\setminus X)$$ what can also be written as: $$X\oplus Y = (X\cap Y')\cup (X'\cap Y)$$
I tried messing with the left side and get to the right side and vice versa using various set transformations, but I had no luck. I've also tried other approach, using:
$$(A1\cup A2)\to x\in A1 \lor x\in A2...$$
but still I got stuck and got nothing similar to the other side.
This is kinda upsetting especially that I got graphical solution quickly (refer to link). I would be grateful for any help, clues or tips on how to approach this problem.
There is also a second question asking if:
$$(A1\cap A2)\oplus(B1\cap B2)\subseteq (A1\oplus B1) \cup (A2\oplus B2)$$
is true for any $A1, A2, B1, B2$. For what i tried to do, again, graphically, I came to conclusion it is indeed true.
My tries to do things graphically - imgur
 A: Take any $x\in (A_1\cup A_2)\oplus (B_1\cup B_2)=\left((A_1\cup A_2)\setminus(B_1\cup B_2)\right)\bigcup\left((B_1\cup B_2)\setminus (A_1\cup A_2)\right)$. Then either $x\in (A_1\cup A_2)\setminus(B_1\cup B_2)$ or $x\in(B_1\cup B_2)\setminus (A_1\cup A_2)$.
Case $1$: $x\in (A_1\cup A_2)\setminus(B_1\cup B_2)$
Then $x\in A_1\cup A_2$ and $x\not\in B_1\cup B_2$. It follows that $x\in A_1\setminus (B_1\cup B_2)$ or $x\in A_2\setminus(B_1\cup B_2)$. Since $A_1\setminus(B_1\cup B_2)\subseteq A_1\setminus B_1$, we have that if $x\in A_1\setminus (B_1\cup B_2)$, then $x\in A_1\setminus B_1$. Similarly, $A_2\setminus (B_1\cup B_2) \subseteq A_2\setminus B_2$, so if $x\in A_2\setminus(B_1\cup B_2)$, then $x\in A_2\setminus B_2$. Therefore, $x\in (A_1\setminus B_1)\cup (A_2\setminus B_2)$.
Case $2$: $x\in(B_1\cup B_2)\setminus (A_1\cup A_2)$.
By an almost identical argument, you can show $x\in(B_1\setminus A_1)\cup(B_2\setminus A_2)$.
Therefore, $x\in(A_1\setminus B_1)\cup (A_2\setminus B_2)\cup (B_1\setminus A_1)\cup(B_2\setminus A_2)$.
But $(A_1\setminus B_1)\cup (A_2\setminus B_2)\cup (B_1\setminus A_1)\cup(B_2\setminus A_2)=(A_1\oplus B_1)\cup (A_2\oplus B_2)$.
