Looking for a proper proof that two sets with sub sets are equivalent Basically I would just like to know how to prove the following equation
$$ A \subseteq B \cup C \iff A \cap \overline B \subseteq C $$
I understand that I have to prove that the left-hand side should equal the right-hand side, and I can perform all the basic logical equivalencies, like Association ad Distribution, but I am unsure how to express the subset ($ \subseteq $) in the proof. I'm tempted to treat the subset sign as intersection ($ \cap $), but since my proof doesn't work... I'm pretty sure this is the wrong approach.
 A: Just "sweat the definitions" [almost always the best strategy faced with little problems like this.]

Assume $A \subseteq B \cup C$.
By definition, $A \subseteq B \cup C$ means that, take any $x$ in the relevant universe, if $x \in A$ then $x \in B \cup C$.
That is to say, if $x \in A$ then either $x \in B$ or $x \in C$ or both.
Propositional logic tells you then that if $x \in A$ and $x \notin B$, then we must have $x \in C$.
But that tells you that if $x \in A \land x \in \overline{B}$, then $x \in C$. [Assuming here that, for any $x$ in the relevant universe, $x \in \overline{B}$ iff $x \notin B$.]
So, by definition again (since $x$ was arbitrary), $A \cap \overline{B} \subseteq C$.

That gives you one direction of the biconditional you need to prove. You can prove the other direction in an exactly similar way.
A: I completely agree with Peter Smith's advice to "sweat the definitions".  Here is the way in which I would do this: start at the most complex side, expand the definitions, and see whether you can get to the other side.
Note that below, $x$ ranges over the 'universe' that is implicit in this question.
\begin{align}
& A \cap \overline B \subseteq C \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall x :: x \in A \cap \overline B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cap$"} \\
& \langle \forall x :: x \in A \land x \in \overline B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\overline{\phantom\square}$"} \\
& \langle \forall x :: x \in A \land x \not\in B \;\Rightarrow\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"expand $\Rightarrow$, using De Morgan on its left hand side"} \\
(*) \; \phantom\equiv & \langle \forall x :: x \not\in A \;\lor\; x \in B \;\lor\; x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"reintroduce $\Rightarrow$ at the first $\lor$ -- suggested by the shape of our goal"} \\
& \langle \forall x :: x \in A \;\Rightarrow\; x \in B \lor x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cup$"} \\
& \langle \forall x :: x \in A \;\Rightarrow\; x \in B \cup C \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& A \subseteq B \cup C \\
\end{align}
This completes the proof.
So by expanding the definitions, and also writing $\Rightarrow$ which often is difficult to calculate with, we arrived at $(*)$.  Then it was not difficult to see how we could get to the goal.  Alternatively, one could also start at both ends and see that both end up at $(*)$.
