# 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.

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. $\blacksquare$

Hint: Use two steps and show $$(R\setminus S)\setminus T\subseteq R\setminus S\subseteq R\setminus(S\setminus T)$$

• Isn't that more of a proof than a hint? :-) – Brian Tung Jul 29 '15 at 20:00
• @BrianTung I assum ethe OP would need to show these as well by considering elements – Hagen von Eitzen Jul 29 '15 at 20:09
• Fair enough. :-) – Brian Tung Jul 29 '15 at 20:12

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.

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 $x \in R \backslash (S \backslash T)$).

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.

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?