Proving that that ${(R \setminus S)\setminus T} \subseteq R \setminus (S \setminus T)$ How might I prove that ${(R \setminus S)\setminus T} \subseteq R \setminus (S \setminus T)$? I am not sure the best place to start other than assuming $x\in(R \setminus S)\setminus T$ and trying to work toward the other end, but I am stuck on how to progress. Helpful thoughts or comments would be much appreciated. 
 A: Hint: Use two steps and show
$$(R\setminus S)\setminus T\subseteq R\setminus S\subseteq R\setminus(S\setminus T)$$
A: You are only dealing with three sets, so you can easily draw a Venn diagram, similar to this one.
Instead of $A, B,$ and $C$ you have $R,S,$ and $T$.
All that remains for you is to label the various intersections appropriately.
This will help you understand what it going on.
In order to actually write the proof, all you must do is show that any element of $(R \backslash S) \backslash T$ is also an element of $R \backslash (S \backslash T)$. This can be done directly (if $x \in (R \backslash S) \backslash T$ then $x \in R$, $x \notin S$, and $x \notin T$ .. etc .. therefore $x \in R \backslash (S \backslash T)$).
A: Here is a brass tacks approach. Let $x \in (R - S) - T$. So $x \in R - S$ and $x \not\in T$. Hence, $x \in R$ and $x \not\in S$ and $x \not\in T$.
We must show that $x \in R - (S - T)$, so we must show $x \in R$ and $x \not\in S -T$. We know $x \in R$, so we must show $x \not\in S - T$. 
At this point, you can show it by contradiction, or by a certain logical equivalence.
A: An element in $(R\smallsetminus S)\smallsetminus T$ is an element of  $R$, not in $S$ nor in $T$. 
An element in $R\smallsetminus (S\smallsetminus T)$ is an element in $R$, not in $S\smallsetminus T$. This means it is not in $S$, except if it is also in $T$.
Thus, isn't the inclusion clear?
A: I think the easiest way to prove your claim is by making two observations at the outset:
$$
(R\setminus S)\setminus T=(R\cap S^C)\cap T^C\tag{1}
$$
and
$$
R\setminus(S\setminus T)=R\cap(S\cap T^C)^C=R\cap(S^C\cup T)=(R\cap S^C)\cup(R\cap T)\tag{2}.
$$
Now your element-chasing proof is extremely easy:
\begin{align}
x\in (R\setminus S)\setminus T&\implies x\in(R\cap S^C)\cap T^C\tag{by $(1)$}\\[0.5em]
&\implies x\in(R\cap S^C)\land x\in T^C\tag{by defn. of $\cap$}\\[0.5em]
&\implies x\in(R\cap S^C)\tag{$(p\land q)\to p$}\\[0.5em]
&\implies x\in(R\cap S^C)\lor x\in(R\cap T)\tag{$p\to(p\lor q)$}\\[0.5em]
&\implies x\in(R\cap S^C)\cup(R\cap T)\tag{by defn. of $\cup$}\\[0.5em]
&\implies x\in R\setminus(S\setminus T).\tag{by $(2)$}
\end{align}
Since $x\in (R\setminus S)\setminus T\implies x\in R\setminus(S\setminus T)$, we have $(R\setminus S)\setminus T\subseteq R\setminus(S\setminus T)$, as desired. $\blacksquare$
