Difference of elementary sets is elementary / Difference of intervals are intervals We call a set $E$ in $\mathbb{R}^d$ elementary iff it can be represented as a finite union of boxes. Let $E,F$ be elemnetary sets. I need to demonstrate that $E/F$ is elementary.
My question My proof (below) feels very messy and clumsy. Is there a way to simplify it, or maybe propose a different proof?
My proof
So using trivial set theoretic relations I rewrote the original equation as
\begin{align*}
E/F &= \left(\bigcup_{i=1}^{n} B_i\right) /\left(\bigcup_{j=1}^{m} C_j\right)\\[10pt]
&=\bigcup_{i=1}^{n} \bigcap_{j=1}^{m} B_i / C_j
\end{align*}
From here on it suffices to show that the difference of two boxes is elementary. 
Now to prove this I went brute force.
Result 1. Set difference of two intervals is a union of two intervals.
Let $I_1 :=(a,b), I_2 = (c,d)$. We have permutations (unfortunately together with the cases where $b< a$ or $d<c$)
\begin{align*}
(a, b, c, d) &\implies N = (a,b)\\
(a, b, d, c) &\implies N = (a,b) \\
(a, c, b, d) &\implies N=(a,c) \\
(a, c, d, b) &\implies N=(a,c)\cup(d,b)\\
(a, d, b, c) &\implies N =(a,b)\\
(a, d, c, b) &\implies N=(a,b)\\
(b, a, c, d) &\implies N=\emptyset\\
(b, a, d, c) &\implies N=\emptyset\\
(b, c, a, d) &\implies N=\emptyset\\
(b, c, d, a) &\implies N=\emptyset\\
(b, d, a, c) &\implies N=\emptyset\\
(b, d, c, a) &\implies N=\emptyset\\
(c, a, b, d) &\implies N=\emptyset\\
(c, a, d, b) &\implies N=(d,b)\\
(c, b, a, d) &\implies N=\emptyset\\
(c, b, d, a) &\implies N=\emptyset\\
(c, d, a, b) &\implies N=(a,b)\\
(c, d, b, a) &\implies N=\emptyset\\
(d, a, b, c) &\implies N=(a,b)\\
(d, a, c, b) &\implies N=(a,b)\\
(d, b, a, c) &\implies N=\emptyset\\
(d, b, c, a) &\implies N=\emptyset\\
(d, c, a, b) &\implies N=(a,b)\\
(d, c, b, a) &\implies N=\emptyset\\
\end{align*}
I somehow have a feeling that this step can be justified very simply, but somehow I cannot see how.
Result 2. $I_1\cup J_1 \times I_2 \times \dots \times I_n = I_1\times I_2 \times \dots \times I_n \cup J_1 \times I_2 \times \dots \times I_n$. Can be verified directly using the definition of a box.
Combining both we obtain
\begin{align*}
E/F &=\bigcup_{i=1}^{n} \bigcap_{j=1}^{m} B_i / C_j \\[10pt]
&=\bigcup_{i=1}^{n} \bigcap_{j=1}^{m} \left\{(x_1,\dots,x_d)\in\mathbb{R}^d: x_i \in I_i / J_i\right\} \\
&=\bigcup_{i=1}^{n} \bigcap_{j=1}^{m} \left\{(x_1,\dots,x_d)\in\mathbb{R}^d: x_i \in A_i \cup B_i\right\} \\
&=\bigcup_{i=1}^{n} \bigcap_{j=1}^{m} (A_1 \cup B_1) \times \dots \times (A_n \cup B_n) \\
\end{align*}
And applying Result 2 we see that this is a untion of boxes, and thus elementary.

Note the $C_j^c$ is a union of $2d$ unbounded boxes.
So $E_i \cap C_j^c$ is again  a union of boxes since the intersection of two boxes with edges parallel to the axis  is a box or the empty set.
So finally you will have two finite big unions of  intersections of boxes,which is a finite union of boxes
