# $(A\cup B)\cap C = A\cup(B\cap C)$ 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.

-
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Jul 3 '13 at 17:34
Okay, understood Lord! – Sjoerd Smaal Jul 3 '13 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 '13 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 '13 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 '13 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 '13 at 18:03
\begin{align} & (A \cup B) \cap C \;=\; A \cup (B \cap C) \\ \equiv & \;\;\;\;\;\text{"set extensionality"} \\ & \langle \forall x :: x \in (A \cup B) \cap C \;\equiv\; x \in A \cup (B \cap C) \rangle \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\cup,\cap\;$, both twice"} \\ & \langle \forall x :: (x \in A \lor x \in B) \land x \in C \;\equiv\; x \in A \lor (x \in B \land x \in C) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: distribute $\;\lor\;$ over $\;\land\;$ in the right hand side of $\;\equiv\;$} \\ & \;\;\;\;\;\phantom"\text{-- alternatively, do this for the left hand side, for the same result"} \\ & \langle \forall x :: (x \in A \lor x \in B) \land x \in C \;\equiv\; (x \in A \lor x \in B) \land (x \in A \lor x \in C) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: factor common conjunct from both sides of $\;\equiv\;$"} \\ & \langle \forall x :: x \in A \lor x \in B \;\Rightarrow\; (x \in C \;\equiv\; x \in A \lor x \in C) \rangle \\ \equiv & \;\;\;\;\;\text{"logic:$\;p \lor q \equiv q\;$ is one of the ways to write $\;p \Rightarrow q\;$"} \\ & \langle \forall x :: x \in A \lor x \in B \;\Rightarrow\; (x \in A \;\Rightarrow\; x \in C) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: combine antecedents"} \\ & \langle \forall x :: (x \in A \lor x \in B) \land x \in A \;\Rightarrow\; x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify antecedent by using $\;x \in A\;$ in the other conjunct"} \\ & \langle \forall x :: x \in A \;\Rightarrow\; x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\ & A \subseteq C \\ \end{align}