TITLE: Prove that $A\cap (B\setminus C)=(A \cap B)\setminus(A \cap C)$.
QUESTION [5 upvotes]: Problem: Prove that $A\cap (B\setminus C)=(A \cap B)\setminus(A \cap C)$.

I've tried it on my own:
\begin{align}
x&\in A\cap (B\setminus C) \\
&\Leftrightarrow (x\in A) \wedge (x\in B\setminus C) \\
&\Leftrightarrow (x\in A) \wedge (x\in B \wedge x\notin C) \\
&\Leftrightarrow (x\in A \wedge x\in B) \wedge (x\notin C) \\
&\Leftrightarrow x\in (A\cap B)\setminus C\\
&\Leftrightarrow \dots
\end{align}
What would the next step be? I've no idea how to get from $x\in (A\cap B)\setminus C$ to $x\in(A \cap B)\setminus(A \cap C)$.
If I'm trying to do the right side, we get
\begin{align}
x&\in(A \cap B)\setminus(A \cap C)\\
&\Leftrightarrow (x\in A \wedge x\in B)\wedge x\notin (A \cap C)\\
&\Leftrightarrow (x\in A \wedge x\in B)\wedge  (x\notin A \wedge x\notin C)\\
&\Leftrightarrow \dots
\end{align}
How do I make it simpler?

REPLY [2 votes]: This is why I like characteristic functions, see this answer.
With Characteristic functions we have
$$\chi_{(A \cap B)\setminus(A \cap C)}= \chi_{A \cap B}-\chi_{A \cap B}\chi_{A \cap C}
=\chi_A \chi_B -\chi_A \chi_B\chi_A \chi_C \\
=\chi_A \chi_B -\chi_A \chi_B\chi_C= \chi_A( \chi_B -\chi_{B} \chi_C)=\chi_A \chi_{B \setminus C}=\chi_{(A \cap (B \setminus C)} $$