How do you prove $B \setminus \cap_{i \in I} A_{i} = \cup_{i\in I} B \setminus A_{i}$? I'm stuck on this proof.  Especially in the $\implies$ direction.  I've written the following but feel it is more just a restatement of the original equation than a proof.  It goes from very specific (intersection) to general (union) and this is turning my brain into mush.  Any help is MUCH appreciated.  Here is what I have:
Consider $ x \in B \setminus \cap_{i \in I} A_{i}$. Then $x \in B$ and$ \forall A \in F$, $x \not \in A$.  $\exists A\in F$ $x\in B$ and $x\not \in $A and therefore $x \not \in \cup_{i \in I} B \setminus A_{i}$. 
Thanks again!
 A: I will help you do one direction and the other direction is very similar, so I'll leave it to you.  If you get stuck, just let me know.
Suppose $x \in B \setminus \cap_{i \in I} A_{i}$.  In words, this means $x \in B$ and $x \not \in \cap_{i \in I} A_{i}$.  Restated, this means $x$ is in $B$ and $x$ is not in $A_{i}$ for some $i$ (since if $x$ is in $A_{i}$ for every $i$, $x$ would be in the intersection all the $A_{i}$'s).
Since $x$ is in $B$ and $x$ is not in $A_{i}$ for some $i$, that means $x$ is in $B \setminus A_{i}$ for some $i$, right?  Then $x$ is in the union of $B \setminus A_{i}$, i.e., $x \in \cup_{i \in I} B \setminus A_{i}$, since being in one of the sets $B \setminus A_{i}$ implies you are in the union of all of them.  So the $\implies$ direction is done.  The backward direction is similar in spirit, so try it for yourself.
A: I would simply calculate which elements $\;x\;$ are in both sides of the equality, by expanding the definitions.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$  In this case both sides look equally complex, so we arbitrarily choose to start with the left hand side, and work towards the right hand side: for all $\;x\;$,
$$\calc
    x \in B \setminus \cap_{i \in I} A_i
\op\equiv\hint{definition of $\;\setminus\;$}
    x \in B \;\land\; \lnot (x \in \cap_{i \in I} A_i)
\op\equiv\hint{definition of $\;\cap_{\cdot \in \cdot}\;$}
    x \in B \;\land\; \lnot \langle \forall i : i \in I : x \in A_i \rangle
\op\equiv\hint{logic: DeMorgan -- to simplify}
    x \in B \;\land\; \langle \exists i : i \in I : \lnot (x \in A_i) \rangle
\op\equiv\hints{logic: move part not using $\;i\;$ inside of $\;\forall i\;$}\hint{-- to bring $\;B\;$ and $\;A_i\;$ closer together as in our goal}
    \langle \exists i : i \in I : x \in B \;\land\; \lnot (x \in A_i) \rangle
\op\equiv\hint{definition of $\;\setminus\;$ -- working toward the right hand side}
    \langle \exists i : i \in I : x \in B \setminus A_i \rangle
\op\equiv\hint{definition of $\;\cup_{\cdot \in \cdot}\;$}
    x \in \cup_{i\in I} B \setminus A_i
\endcalc$$
Therefore, by set extensionality, $\;B \setminus \cap_{i \in I} A_i \;=\; \cup_{i\in I} B \setminus A_i\;$.

Note how this proof proves both directions at the same time.  See EWD1300 for details about this proof format and the notations which I used.
