Set-theoretic proof for $A\setminus(B\setminus C) = (A\setminus B) \cup C$, provided that $C\subseteq A$ I'm trying to prove $A\setminus(B\setminus C) = (A\setminus B)\cup C$ using the fact that $C\subseteq A$. Somehow, I can use the relation $(A\setminus B)^c = A^c \cup B$. However, I can't really see how to start the proof.
Any help on how to start/structure this proof would be appreciated
 A: A funny way to prove set-theoretic equality is to use indicator functions. Given $A, B \subset X$, we have $\mathbb{1}_{A\setminus B} = \mathbb{1}_{A}(1 - \mathbb{1}_B)$. Then we have the following equalities:
$\mathbb{1}_{A\setminus (B\setminus C)} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$
$\mathbb{1}_{(A \setminus B) \cup C} = \mathbb{1}_{(A \setminus B)} + \mathbb{1}_{C} - \mathbb{1}_{(A \setminus B)\cap C} = \mathbb{1}_{A} - \mathbb{1}_{A}\mathbb{1}_{B} + \mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} + \mathbb{1}_{A}\mathbb{1}_{B}\mathbb{1}_{C}$
Because $C \subset A$, $\mathbb{1}_{C} - \mathbb{1}_{A}\mathbb{1}_{C} = 0$, so that both indicator functions are equal.
A: \begin{align}
A-(B-C)&=A\cap (B\cap C^c)^c
\\
&=A\cap (B^c\cup C)
\\
&=(A\cap B^c)\cup (A\cap C)
\\
&=(A-B)\cup C\tag1
\end{align}
$(1)$ is for
$$
C\subset A\implies A\cap C=C
$$
A: Using De Morgan's Laws in addition to the provided set rule:
$ A\setminus (B \setminus C) = A \setminus (( B \setminus C )^{c})^{c} = A \setminus (B^{c} \cup C)^{c} = A \setminus (B \cap C^{c}) =((A \setminus (B \cap C^{c}))^{c})^{c} = (A^{c} \cup (B \cap C^{c}))^{c} =( (A^{c} \cup B) \cap (A^{c} \cup C^{c}))^{c} = (A \setminus B) \cup C$ given $A \supseteq C $  
A: Such a proof is best done using both inclusions: prove that $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$ and that $(A \setminus B) \cup C \subseteq A \setminus (B \setminus C)$.


*

*$A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$
Let $x \in A \setminus (B \setminus C)$. Then $x \in A$ and $x \notin B\setminus C$. If $x \notin B \setminus C$, then $x \notin B$ (case 1) or $x \in C$ (case 2). So in case 1, $x \in A$ and $x \notin B$ hence $x \in A \setminus B$, and therefore $x \in (A \setminus B) \cup C$. In case $2$, $x \in C$ hence $x \in (A \setminus B) \cup C$. In both cases $x \in (A \setminus B) \cup C$, hence $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$. 

*Perhaps try this yourself in the same way as the first inclusion?
A: Does $C\subseteq A$ is an assumption? The identity is false. 
Take $A=B=\varnothing$ and $C=\{1,2,3\}$
