Prove this intersection subset question. $(A\cap B\subseteq C) \wedge ({A}'\cap B\subseteq C)\Leftrightarrow B\subseteq C$
"A intersection B is a subset of C, and compliment A intersection B is a subset of C if and only if B is a subset of C".
How would I go about proving or disproving this? I've tried drawing Venn diagrams, so I'm quite sure that it is true.
 A: $\Rightarrow)$ Suppose $A\cap B\subseteq C$ and $A'\cap B\subseteq C$
As a result $((A\cap B)\cup (A'\cap B))\subseteq C$
But $((A\cap B)\cup (A'\cap B))$ simplifies to be...implying that...

 $((A\cap B)\cup (A'\cap B)) = (A\cup A')\cap B = \dots$

$\Leftarrow)$ Suppose that $B\subseteq C$
Then since $A\cap B\subseteq B$ that implies that... Also since $A'\cap B\subseteq B$ it also implies that...
A: If $B \subseteq C$, then $A \cap B \subseteq B \subseteq C$ and $A' \cap B \subseteq B \subseteq C$.
Conversely, suppose some element $b$ of $B$ does not lie in $C$. This element $b$ is either in $A$ or $A'$. If it is in $A$, then $A \cap B \not\subseteq C$. If it is in $A'$, then $A' \cap B \not\subseteq C$.
A: Observe that we can write the entire sample space $S$ as the disjoint union of $A$ and its complement $A'$.  Hence, every set $B$ can be written as $B=B \cap S = B \cap (A \cup A') = (B \cap A) \cup (B \cap A')$.  In other words, $B$ consists of elements of $B$ that are in $A$ and of elements of $B$ that are in $A'$. This is intuitively clear.  The truth of the assertion is now easy to see.
