$A-(B-C) = (A-B) \cup (A \cap B \cap C)$ Prove $A-(B-C) = (A-B) \cup (A \cap B \cap C)$ : 
What I think of doing is showing the LHS is a subset of the RHS and the RHS is a subset of the LHS, then the RHS = LHS. 
$\Rightarrow$
We have $x \in A$, but $x \notin (B-C)$, thus $x \in (A-B)$ and then we see that $x \in (A-B) \cup (A \cap B \cap C)$. And we have the LHS a subset of the RHS. 
$\Leftarrow$
3 cases: 
1.) $x \in (A-B)$ only 
from this we see that the RHS is a subset of the LHS 
2.) $x \in (A \cap B \cap C)$ only 
This is not possible. Exclude this from our discussion. 
3.) $x \in (A-B) \wedge x \in (A \cap B \cap C) $
This is not possible. Exclude this from our discussion. 
Thus, the RHS is a subset of the LHS, shown by case 1.
Since the LHS is a subset of the RHS and the RHS is a subset of the LHS, then the LHS = RHS and 
$A-(B-C) = (A-B) \cup (A \cap B \cap C) $
Is this methodology right. I have more details in the proof I constructed but wanted to know if this method of thinking is good for the proof. Thanks!  
 A: You could use a Venn Diagram:

Unless the goal was to learn the specific axioms of a theory, I wouldn't make the problem harder than it has to be.
A: $$(A-B)\cup(A\cap B\cap C)=(A\cap B')\cup(A\cap B\cap C)=A\cap (B'\cup(B\cap C))=A\cap((B'\cup B)\cap (B'\cup C))=A\cap (B'\cup C)=A\cap (B\cap C')'=A-(B-C)$$
A: Its much easier to work with indicator function. An indicator function $I_A$ of a subset $A$ is defined as:
$$
I_{A}(x)=\begin{cases}
1 & x\in A\\
0 & x\not\in A
\end{cases}
$$
Some basic rules:
$$
A=B\quad \text{iff}\quad I_A=I_B
$$
and
$$
I_{A\cap B}=I_AI_B,\quad I_{A^c}=1-I_A,\quad I_{A-B}=I_{A\cap B^c}=I_A(1-I_B)
$$
and
$$
I_{A\cup B}=1-I_{(A\cup B)^c}=1-I_{A^c\cap B^c}=1-(1-I_A)(1-I_B),\quad I_A^2=I_A
$$
Working from the left side:
\begin{align}
I_{A-(B-C)} &=I_A(1-I_{B-C})=I_A(1-I_B(1-I_C)) \\
&=I_A-I_AI_B+I_AI_BI_C \\
&=I_A - I_{A \cap B} + I_{A \cap B \cap C} \\
&= I_{A - (A \cap B) \cup (A \cap B \cap C)} \\
&= I_{(A-B) \cup (A \cap B \cap C)}
\end{align}
from which the result follows.
