Prove $(A \backslash B)'\backslash (B \backslash A) = B \backslash A$ Let $A$ and $B$ be subsets of some universal set.
Prove that $(A \backslash B)'\backslash (B \backslash A) = B \backslash A$
Given: 
Definition 3.3.1 states that $A$ and $B$ are sets. The complement of $B$ relative to $A$, written $A \backslash B$ is the set 
$ A \backslash B =[x:x \in A \land x \notin B]$
Definition 3.3.3 states that $U$ is a universal set and $A \subseteq U$. The complement of $A$, written $A'$ is the set $A' = U \backslash A =[x \in U: x\notin A]$
For B relative to A,
Using Definition 3.3.1
$A \backslash B = [x: x \in A \land x \notin B]$
Using Definition 3.3.3
$(A \backslash B)' = [x : x \in A : x \notin B]$
For A relative to B,
Using Definition 3.3.1
$B \backslash A =[x: x \in B \land x \notin A]$ 
Using Definition 3.3.3 
$(B \backslash A)' = [x: x \in B : x \notin A]$

The $\backslash$ can be read as $-$, so the problem may look like this 
$( a-b)' - (b-a) = b-a$
I could try to use one of the Propositionals which is $(B')' = B$
Maybe if B = a - b
$((a-b)')' - (b-a) = b-a$
$(a-b) - (b-a) = b-a$
$(a-b)-b+a = b-a$
$a-b-b+a = b-a$
hmmm that's $2a-2b = b-a$
factoring  out the -2 would result in $-2(-a+b) = b-a$
That clearly doesn't equal to b-a.
Is it possible to prove this problem with venn diagrams instead?
I could draw one for $(A \backslash B)'$, $(B \backslash A)$ and $(A \backslash B)' \backslash ( B \backslash A)$
On one diagram, the A would be shaded
On a second diagram, the B would be shaded.
On the third diagram A and B which would be the middle is shaded. 
Edit: Maybe a counterexample will work since the left side of the problem could  not have the elements of $B \backslash A$, but the right side does. 
Or...
since
$(A \backslash B)'\backslash (B \backslash A) = B \backslash A$
applying definitions. 
$[x : x \in A : x \notin B]' \backslash [x: x \in B \land x \notin A]$ = $[x: x \in B \land x \notin A]$ ???
Edit: I need to provide an improved version of this proof.
Let $A$ and $B$ be subsets of some universal set.
Prove that $(A \backslash B)'\backslash (B \backslash A) = B \backslash A$
Given: 
Definition 3.3.1 states that $A$ and $B$ are sets. The complement of $B$ relative to $A$, written $A \backslash B$ is the set 
$ A \backslash B =[x:x \in A \land x \notin B]$
Here is what I have so far... 
$( B \backslash A ) = [x: x \in B \land x \notin A]$
that is the end result. I need to somehow work from the left and achieve $(B \backslash A ))$
$(B \backslash A)' =[x:x \notin B \lor x \in A]$
$(A \backslash B)' =[x:x \notin A \lor x \in B]$
I may have to use the definition again, but it is messy. 
Set property by default definition is $A \backslash B = A \cap B'$
so from set intersection definition ...
$ [x: x \in A \land x \notin B]$
since $(A \backslash B)'$....$ [x: x \notin A \lor x \in B]$
/////
set property  $B \backslash A = B \cap A'$
set intersection definition... $[x: x\in B \land x \notin A]$
since $(B \backslash A)'$...$ [x: x\notin B \lor x \in A]$
Edit: Going through the definition is a very bad idea. I've used the set property and was able to get $ B \backslash A$
 A: Hint:


*

*$A\setminus B$ should not be understood as $a - b$, one of the reasons being $(a-b)+b = a$, but $(A \setminus B) \cup B = A \cup B$.

*There is a very useful identity for set difference, namely $$A \setminus B = A \cap B\,', \tag{$\spadesuit$}$$ it is often enough to solve most of this kind of homework-related problems.


I hope this helps $\ddot\smile$
A: We think that the statement of the problem is wrong.
As said in the above comments, if we correct it as :

$(A \backslash B)' \backslash (B \backslash A)' = B \backslash A$

we have that the condition for $(A \backslash B)'$ is :
$\lnot (x \in A \land x \notin B)$, that is (using De Morgan) :
$(x \notin A \lor x \in B)$.
With the same approach, the condition for $(B \backslash A)'$ is :
$(x \notin B \lor x \in A)$.
So the "global" condition becomes :
$(x \notin A \lor x \in B) \land \lnot (x \notin B \lor x \in A)$, that is :
$(x \notin A \lor x \in B) \land (x \in B \land x \notin A)$
Now, going back to set properties, this is :
$(A' \cup B) \cap (A' \cap B)$.
But $(A' \cap B)$ is a subset of $(A' \cup B)$, so that their intersection is $(A' \cap B)$.
Now, $x \in (A' \cap B)$ iff $(x \in B \land x \notin A)$ i.e. 

$x \in B \backslash A$. 

