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I'm having trouble proving $(A\setminus B)\cap C ⊂ A\setminus (B\cap C).$

The way I started was

$x \in (A\setminus B)\cap C$ and now $x \in A\setminus B$ and $x \in C$ but I don't know how to go on with this.

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... hence $x \in A$, $x \not\in B$, $x \in C$. Since $x \not\in B$, $x$ is not in any subset of $B$, hence $x \not\in B \cap C \subset B$.

We conclude that $x \in A$ and $x \not\in B \cap C$, hence $x \in A\setminus (B \cap C$).

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You have that $x\in A\setminus B$ and $x\in C$. Then, by definition of difference of sets, $x\in A$ and $x\notin B$ and $x\in C$. Then, $x\in A$ and $x\notin B\cap C$ because an element is in the intersection of two sets if and only if is in both sets. Thus, $x\in A\setminus(B\cap C)$

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It's probably most instructive to do this in two step:

$$ (A\setminus B)\cap C \quad\subseteq\quad A \setminus B \quad\subseteq\quad A \setminus (B\cap C) $$

The truth of the claim does not in fact depend on $C$ being the same set on both sides!

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