$(A\cup B)\cap C = A\cup(B\cap C)$ 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.
Could someone please help me out?
 A: 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.
A: Here is how I would do this, by starting with the most complex side, expanding the definitions, and then simplifying using the laws of logic:
\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}
