# $(A\cup B)\cap C = A\cup(B\cap C)$ if and only if $A\subset C$

I tried to prove this statement:

$$[(A\cup B)\cap C = A\cup(B\cap C)]\iff A\subset C$$

I did it in the following way, can anyone tell me if it's correct what I've done?

$\leftarrow$ Assume $A\subset C$, then $\forall x \in A$, $x\in C$

Then, $\forall x \in (A\cup B)\cap C$, $x\in C$ and $\in B$

Similarly, for $\forall x \in A\cup(B\cap C)$, $x\in B$ and $x\in C$

So $(A\cup B)\cap C = A\cup(B\cap C)$

$\rightarrow$ I didn't know how to do the counterpart.

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Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. –  Lord_Farin Jul 3 at 17:34
Okay, understood Lord! –  Sjoerd Smaal Jul 3 at 17:35
@SjoerdSmaal, after the hypothesis, your first statement is not true. "Then, $\forall x\in\ldots$". The correct is "... $x\in C$ and $x\in A$ or $x\in B$". Then, on each case you continue your argument accordingly. –  Sigur Jul 3 at 17:37
For the counterpart, assume the LHS is true and then assume the RHS is false; that is, there exists some $x \in A$ that is not in $C$. Then reach a contradiction. –  A.E Jul 3 at 17:38
@Sigur, yes but that's the same as you wrote down except I didn't write $x\in A$, can you tell me what is wrong? –  Sjoerd Smaal Jul 3 at 17:43
For the counterpart: assume that $(A\cup B)\cap C = A\cup(B\cap C)$ and let $x\in A$ then $x\in A\cup(B\cap C)$ so $x\in (A\cup B)\cap C$ and therefore $x\in C$ hence we find $$x\in A\Rightarrow x\in C$$ and you can conclude.
Can you please explain the second part, where you say $x\in (A\cup B)\cap C$. –  Sjoerd Smaal Jul 3 at 18:03