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 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.
Does $C\subseteq A$ is an assumption? The identity is false. 
Take $A=B=\varnothing$ and $C=\{1,2,3\}$