What's the proof for $a-(b-c)=(a-b)\cup(a\cap c)$? I have to prove that: 
$$a-(b-c)=(a-b)\cup(a\cap c)$$
I do know that in $a-(b-c)$ that $x\in a$ and $x\notin b-c$.
And also I know in $(a-b)\cup(a\cap c)$ that either $x\in a$ and $x\notin b$ or $x\in a$ and $x\in c$ but something is missing and I'll be happy to get some help :)
 A: 
$x \in a \setminus (b \setminus c)$ is equivalent to "$x \in a$ and $x \notin b \setminus c$."

That's right. You then need to show that $x$ is either in $a \setminus b$ or $a \cap c$. Try handling the two cases $x \in c$ and $x \notin c$ separately.

 Suppose $x \in c$ as well. Since we already know $x \in a$, we have $x \in a \cap c$.

${}$

 Otherwise, $x \notin c$. Then we know $x \notin b$ as well (otherwise, we would have $x \in b \setminus c$, which would contradict our assumption $x \notin b\setminus c$). So, $x \in a$ and $x\notin b$, that is, $x \in a \setminus b$.



$x \in (a \setminus b) \cup (a \cap c)$ is equivalent to "either $x \in a$ and $x \notin b$, or $x \in a$ and $x \in c$."

That's right. Again, try casework. First assume $x \in a$ and $x \notin b$, and show that this implies $x \in a \setminus (b \setminus c)$. Then handle the other case.

 If $x \in a$ and $x \notin b$, then clearly $x \notin b \setminus c$ holds as well. So $x \in a$ and $x \notin b \setminus c$, which is equivalent to $x \in a \setminus (b \setminus c)$.

${}$

 If $x \in a$ and $x \in c$, then $x \notin b \setminus c$ holds, and the same reasoning leads to $x \in a \setminus ( b \setminus c)$.

A: From this expression −(−)=(−)∪(∩), you know that :
∈, the question here is wether ∈c or ∉c.
If ∈c :
Then you got that ∈ by hypothesis, and ∈, because you are getting rid off all elements of c in b you know that ∈ and ∈, because you re supposing that ∈. 
If ∉c :
Then you got that ∈ and ∉b, this is a simple deduction from the assertion that was given to you.
Thus −(−)=(−)∪(∩) is True.
A: When in doubt, element chase... and when stuck do cases.
Suppose $x \in a - (b-c)$.
Then $x \in a$ and $x \not \in b -c$.
Now either $x \in b$ or $x \not \in b$.
Case 1:  $x \in b$.  Now either $x \in c$ or $x \not \in c$.
Case 1a: if $x \in c$ then $x \in a$ and $x\in c$ so $x \in a\cap c$
So it is true that either $a \in a\cap c$ or $x \in a-b$ so $x \in (a-b)\cup (a \cap c)$.
Case 1b: if $x \not \in c$ then as $x \in b$ then $x \in b-c$ which is a contradiction.
Case 2: $x \not \in b$.  Then as $x \in a$ then $x\in a-b$ so it is true that either $a \in a\cap c$ or $x \in a-b$ so $x \in (a-b)\cup (a \cap c)$.
So $a-(b-c) \subset (a-b)\cup (a\cap c)$.
.....
Suppose $y \in (a-b) \cup (a\cap c)$.
Case 1:  $y \in a-b$.  Then $y \not \in b$ so $y \not \in b-c$. But $y \in a$ so $y \in a-(b-c)$.
Case 2:  $y \in a\cap c$.  So $y \in a$ but as $y \in c$ it can't be the case that $y\not \in c$ and $y \in b$.  So $y \not \in b-c$.  So $y\in a-(b-c)$.
So $(a-c)\cup (a\cap c) \subset a-(b-c)$.
