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I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof

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    Should the right hand side be $(A \setminus B) \cap (A\setminus C)$? – 2017-01-17
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    @Matt no, see e.g. De Morgan's rules. – 2017-01-17
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    @UmbertoP. Oops. Yes, in my head I was thinking $x\in B\cap C$ rather than $x\not\in B\cap C$. Never do math before finishing your coffee! – 2017-01-17
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    Prove that A\(B∩C) = (A\B) ∪ (A\C) for arbitrary sets A, B. this is how question given to me sir – 2017-01-17
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    In general, proving equality of two sets $A$ and $B$, requires to show $A \subseteq B$ and $B \subseteq A$ that is to show $\forall x \in A, x \in B$ and similarly, $\forall x \in B, x \in A$ – 2017-01-17
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    Have a look at [de morgan law $A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $](http://math.stackexchange.com/q/597499) and other questions [linked there](http://math.stackexchange.com/questions/linked/597499). – 2017-01-17
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    @MartinSleziak thank you its very useful – 2017-01-17

1 Answers 1

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Select a point $x \in A \setminus (B \cap C)$.

Then $x \in A$, but $x \notin B \cap C$.

The latter condition means that either $x \notin B$ or $x \notin C$ (since it does not belong to both $B$ and $C$). Thus either $x \in A \setminus B$ or $x \in A \setminus C$. That is, $x \in (A \setminus B) \cup (A \setminus C)$.

Look at the implication that was just proved: $$x \in A \setminus (B \cap C) \implies x \in (A \setminus B) \cup (A \setminus C).$$ This is precisely the meaning of $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.

It remains to prove that $ (A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$. You can take it from here.

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    thank you for clear answer sir :) – 2017-01-17