Symmetric difference inclusion proof In a script for measure theory the proof of this inclusion is an exercise for the reader:
$\left(A_{1} * A_{2}\right) \Delta\left(B_{1} * B_{2}\right) \subset\left(A_{1} \triangle B_{1}\right) \cup\left(A_{2} \Delta B_{2}\right)$
whereby $*$ can be $\cup, \cap, \setminus.$
I proved this inclusion for $\cup, \cap$ but I have no clue how to prove it for $\setminus$.
I know that the symmetric difference $A \triangle B$ of two sets $A, B \subset M$ is defined as follows:
$$
A \triangle B:= (A \backslash B) \cup (B \backslash A) = (A \cup B) \backslash (A \cap B) 
$$
but I cannot apply my first steps of the other two proofs. Nothing seems to work. Has anyone a hint or an idea how to start?
 A: Let $x \in (A_1 \setminus A_2) \Delta (B_1 \setminus B_2) = ((A_1 \setminus A_2) \setminus (B_1 \setminus B_2)) \cup ((B_1 \setminus B_2) \setminus (A_1 \setminus A_2))$. So either $x \in (A_1 \setminus A_2) \setminus (B_1 \setminus B_2)$ or $x \in (B_1 \setminus B_2) \setminus (A_1 \setminus A_2)$. We will only treat the first case, because the second case is similar. So let's unfold what $x \in (A_1 \setminus A_2) \setminus (B_1 \setminus B_2)$ means: we have $x \in A_1 \setminus A_2$ and also $x \not \in B_1 \setminus B_2$. The latter means that either $x \not \in B_1$ or $x \in B_2$. We make a case distinction based on this:

*

*If $x \not \in B_1$ then we use $x \in A_1 \setminus A_2$ to see that also $x \in A_1$, so we conclude $x \in A_1 \setminus B_1$.

*If $x \in B_2$ then we use $x \in A_1 \setminus A_2$ to see that also $x \not \in A_2$, so we conclude $x \in B_2 \setminus A_2$.

Now we write
$$
(A_1 \Delta B_1) \cup (A_2 \Delta B_2) = 
(A_1 \setminus B_1) \cup (B_1 \setminus A_1) \cup (A_2 \setminus B_2) \cup (B_2 \setminus A_2).
$$
So $(A_1 \Delta B_1) \cup (A_2 \Delta B_2)$ can be written as the union of four sets. Our first case shows that $x$ is contained in the first of those sets, while our second case shows that $x$ is contained in the third of those sets. Either way we get $x \in (A_1 \Delta B_1) \cup (A_2 \Delta B_2)$ as required.
As said before: we skipped the case $x \in (B_1 \setminus B_2) \setminus (A_1 \setminus A_2)$, because this is similar to what we did. It might be a nice exercise to try and work this out for yourself, as you have now seen how to do it. In this case $x$ will be either in the second or the fourth set in our union of four sets.
A: A uniform proof for all set operations at once:
Assume $x$ is NOT in the RHS, that is $x \notin A_1 \triangle B_1$ and $x \notin A_2 \triangle B_2$. This means precisely that $x \in A_1 \iff x \in B_1$ (i.e. the statements "$x \in A_1$" and "$x \in B_1$" are simultaneously true or false) and $x \in A_2 \iff x \in B_2$. We have to prove that $x$ is not in the LHS, which is the same as proving the equivalence
$$\tag{$\dagger$} x \in A_1 * A_2 \iff x \in B_1 * B_2.$$
Now let $\diamond$ be the logical connective such that $*$ is defined as
$$z \in A * B \iff (z \in A) \diamond (z \in B)$$
(so: $\vee$ for the union, $\wedge$ for the intersection, $p \diamond q \equiv p \wedge \neg q$ for the difference etc.). Then
$$\begin{align*}
x \in A_1 * A_2 & \iff (x \in A_1) \diamond (x \in A_2) \\[1ex]
& \iff (x \in B_1) \diamond (x \in B_2) \\[1ex]
& \iff x \in B_1 * B_2.
\end{align*}$$
So $(\dagger)$ holds, which ends the proof.
A: Set
$$
C=\left(A_{1} \Delta B_{1}\right) \cup\left(A_{2} \Delta B_{2}\right)
$$
1.)
We verify the statement in the case when * is $\cup$, that is,
$$
\left(A_{1} \cup A_{2}\right) \Delta\left(B_{1} \cup B_{2}\right) \subset C
$$
By definition, $x \in\left(A_{1} \cup A_{2}\right) \Delta\left(B_{1} \cup B_{2}\right)$ if and only if $x \in A_{1} \cup A_{2}$ and $x \notin B_{1} \cup B_{2}$ or conversely $x \in B_{1} \cup B_{2}$ and $x \notin A_{1} \cup A_{2}$. Since these two cases are symmetric, it suffices to consider the first case. Then either $x \in A_{1}$ or $x \in A_{2}$ while $x \notin B_{1}$ and $x \notin B_{2}$. If $x \in A_{1}$ then
$$
x \in A_{1} \backslash B_{1} \subset A_{1} \triangle B_{1} \subset C
$$
that is, $x \in C .$ In the same way one treats the case $x \in A_{2}$, which proves the statement.
2.)
For the case when * is $\cap$, we need to prove
$$
\left(A_{1} \cap A_{2}\right) \Delta\left(B_{1} \cap B_{2}\right) \subset C
$$
Observe that $x \in\left(A_{1} \cap A_{2}\right) \Delta\left(B_{1} \cap B_{2}\right)$ means that $x \in A_{1} \cap A_{2}$ and $x \notin B_{1} \cap B_{2}$ or conversely. Again, it suffices to consider the first case. Then $x \in A_{1}$ and $x \in A_{2}$ while either $x \notin B_{1}$ or $x \notin B_{2}$. If $x \notin B_{1}$ then $x \in A_{1} \backslash B_{1} \subset C$, and if $x \notin B_{2}$ then $x \in A_{2} \backslash B_{2} \subset C$
3.)
For the case when * is $\backslash$, we need to prove
$$
\left(A_{1} \backslash A_{2}\right) \triangle\left(B_{1} \backslash B_{2}\right) \subset C
$$
Observe that $x \in\left(A_{1} \backslash A_{2}\right) \triangle\left(B_{1} \backslash B_{2}\right)$ means that $x \in A_{1} \backslash A_{2}$ and $x \notin B_{1} \backslash B_{2}$ or conversely. Consider the first case, when $x \in A_{1}, x \notin A_{2}$ and either $x \notin B_{1}$ or $x \in B_{2}$. If $x \notin B_{1}$ then combining with $x \in A_{1}$ we obtain $x \in A_{1} \backslash B_{1} \subset C$. If $x \in B_{2}$ then combining with $x \notin A_{2}$, we obtain
$$
x \in B_{2} \backslash A_{2} \subset A_{2} \triangle B_{2} \subset C
$$
which finishes the proof.
Thanks to Mark Kamsma
