I just have some questions in regards to how equations of sets work. I've been looking at these systems of set equations and the way they work has been progressively more confusing especially with how I keep on recieving information that seems to be all over the place.
Example: I'm currently solving a system of 2 set equations:
$$X \cup (A\cap B) = A$$
$$B\backslash X = B\backslash A$$
(Note: I'm supposed to find all X that will solve this equation)
Anyway, I've done some steps, and my equation looks like spaghetti so if I make a mistake whilst typing this out, or if there's something wrong in general I apologise.
We were given some pointers how to solve this, and this is the exact reason why I'm lacking some knowledge since we didn't actually solve any of these problems in class for some reason, but yeah:\

*

*Apparently first I'm supposed to change the equation L = R to an equation of the form $L\oplus R = \emptyset $\

*And then I'm supposed to add all equations together so that all equations of form $S_i=\emptyset$ end up in the equation: $\cup_{i}S_i = \emptyset $

*I'm supposed to put the equation into a form of unions of intersections and/or their complements.
(I won't list more, as this is about where I run into problems)

$$X\cup (A\cap B) = A \rightarrow (X\cup(A\cap B))\oplus A = \emptyset$$ and
$$B\backslash X = B\backslash A \rightarrow (B\cap X^c)\oplus (B\cap A^c) = \emptyset$$
I don't really understand why we do this next thing, but essentially, when you have that $\oplus$ symbol, you can do the following:
$$(X\cup(A\cap B))\oplus A = \emptyset \rightarrow ((X\cup (A\cap B))^c\cap A)\cup ((X\cup(A\cap B))\cap A^c)$$
and the same for the other one
$$(B\cap X^c)\oplus (B\cap A^c) = \emptyset \rightarrow ((B\cap X^c)^c\cap(B\cap A^c))\cup ((B\cap X^c)\cap(B\cap A^c)^c)$$
Then from there we have to make it into a single equation
$$((X\cup (A\cap B))^C\cap A)\cup((X\cup(A\cap B))\cap A^c)\cup ((B\cap X^c)^c\cap(B\cap A^c))\cup ((B\cap X^c)\cap(B\cap A^c)^c) =\emptyset $$
And then comes the part where I tend to get stuck, which is simplification:
$$((X^c\cap (A^c \cup B^c))\cap A)\cup ((X\cup(A\cap B))\cap A^c)\cup ((B^c\cup X)\cap(B\cap A^c))\cup ((B\cap X^c)\cap (B^c\cup A))$$
From this point out I tend to run into a lot of problems as this wasn't properly explained to us because as I've said we haven't really solved any problems of this type, for example, just looking at the first bracket I get:
$$((X^c \cap A^c)\cup (X^c \cap B^c) \cap A)$$
(at least this is what I think I get, I'm not really sure about it)
Here are my questions (finally):
First: Can I move the $\cap A$ into the 2nd bracket of this expression I obtained without messing anything up?
Second; is my distribution of $X^c\cap (A^c\cup B^c)$ even correct?
Third: How do I change the 2nd bracket (from the super long equation) in a way that I end up with intersections inside brackets instead of unions?
(Bracket I'm referring to:)
$$(X\cup (A\cap B))\cap A^c$$
Honestly if someone provides the entire solution that'd be the best possible outcome, but after typing this out I really don't wish that on anyone.\
Anyway, those are all the actual questions I have in regards to this specific problem, I had a couple more before this, but I figured them out whilst writing this.
If by any chance you made it to the bottom thanks I guess. I have an exam tomorrow, I've been studying for like 2 weeks with like 6 hours a day and I just can't seem to figure out some things I probably should've figured out by now, so I'll be posting multiple questions, so if you have time please check out the rest as well when they are up.
 A: The operation $X\oplus Y$ is called the symmetric difference of sets $X$ and $Y$ (often denoted instead by $X\triangle Y$), and the reason you can always do the given replacement is because it is defined as
$$X\oplus Y\ :=\ (X\setminus Y)\cup (Y\setminus X)$$
which is the same as $(X\cap Y^c)\cup(Y\cap X^c)$ by the definition of set difference $\setminus$.
An element $x$ is in $X\oplus Y$ iff either $x\in X$ or $x\in Y$ but not both.
It seems that you're missing the weapon of the distributivity property, which states that both $\cap$ and $\cup$ behave to each other like multiplication to addition: $x\cdot(y+z)=x\cdot y+x\cdot z$:
$$X\cap(Y\cup Z)=(X\cap Y)\,\cup\,(X\cap Z)\\
X\cup(Y\cap Z)=(X\cup Y)\,\cap\,(X\cup Z) $$
Draw the Venn diagram for these equalities and also prove them by picking an element from each side and showing that it's also in the other side.
Actually, already for the symmetric difference we have
$$X\oplus Y\ =\ (X\cap Y^c)\,\cup\,(Y\cap X^c)\ =\ \big((X\cap Y^c)\cup Y\big)\,\cap\,\big((X\cap Y^c)\cup X^c\big)\ =\\
=\ \big((X\cup Y)\,\cap\, (Y^c\cup Y)\big)\,\cap\,\big((X\cup X^c)\cap (Y^c\cup X^c)\big)\ = \\
=\ (X\cup Y)\,\cap\,(Y^c\cup X^c)\ =\ (X\cup Y)\,\cap\,(X\cap Y)^c\ =\ \boxed{ (X\cup Y)\setminus(X\cap Y)}
$$
which result is straightforward to see from the Venn diagram or from the characterization by elements.
Important properties of the symmetric difference:

*

*$X\oplus X=\emptyset$

*$X\oplus\emptyset=X$

*$X\oplus(Y\oplus Z)=\{$elements that are present in an odd number of the three sets $X,Y,Z\}=(X\oplus Y)\oplus Z$

*intersection distributes over it: $X\cap(Y\oplus Z)=(X\cap Y)\oplus (X\cap Z)$.

*$X^c\oplus Y^c = X\oplus Y$ (Try to prove these.)

As a consequence, if we have an equation of sets $L=R$, then
it is indeed equivalent to $L\oplus R=\emptyset$, because
$$L=R\,\implies\, L\oplus R=R\oplus R=\emptyset\, \implies\, \\
L=L\oplus\emptyset=L\oplus(R\oplus R)=(L\oplus R)\oplus R=\emptyset\oplus R=R\,.$$
Now, to the equations. I think, by using distributivity and properties like $X\cap X^c=\emptyset$ you can finish up your work.
However, we can apply simplifications to make life easier, especially the above mentioned distributivity can be handy, e.g. here we 'extract $B$' just like how $b$ is extracted from the sum of numbers $bx'+ba'=b(x'+a')$:
$$(B\cap X^c)\oplus(B\cap A^c)=B\cap(X^c\oplus A^c)=B\cap(X\oplus A)=\\
=B\cap\big((X\setminus A)\cup (A\setminus X)\big)=\ (B\cap X\cap A^c)\,\cup\, (B\cap A\cap X^c)$$
written as a union of intersections.
So the second equation holds iff
$$X\,\cap\,B\cap A^c=\emptyset\text{ and }X^c\,\cap\,A\cap B=\emptyset$$
that is, iff
$$X\subseteq (B\cap A^c)^c=B^c\cup A\text{ and }X^c\subseteq(A\cap B)^c$$
iff
$$A\cap B\ \subseteq \ X\ \subseteq\ A\cup B^c\,.$$
