I'm trying to prove the following statement:
$$A\cap B \subseteq C \iff A \subseteq \overline{B} \cup C$$
I need to do it using a formal proof.. I've tried to do it for some time now and couldn't find anything close..
Thanks!
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$\Rightarrow$:
$$ A = \underbrace{(A\cap B)}_{\subseteq C} \cup \underbrace{(A\cap\overline{B})}_{\subseteq \overline{B}} \subseteq C\cup \overline{B}$$
$\Leftarrow$:
$$A\cap B \subseteq (\overline{B}\cup C)\cap B = (\overline{B}\cap B)\cup (C\cap B) = C\cap B\subseteq C$$
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I suppose that $\bar B$ is the complementary of $B$.
Suppose $A\cap B\subset C$ and let $x\in A$. If $x\in A\cap B$, by hypothesis, $x\in C$ and thus $x\in \bar B\cup C$. If $x\notin A\cap B$, then $x\in A\cap \bar B$ and thus $x\in \bar B$. Therefore $x\in \bar B\cup C$.
Reciprocally, suppose that $A\subset \bar B\cup C$ and let $x\in A\cap B$. In particular, $x\in A$ and $x\in B$. By hypothesis, $x\in \bar B\cup C$. If $x\in \bar B$, we have a contradiction with $x\in B$, therefore $x\in C$, what prove the claim.
