 Solution (source code)

by Codex (@codex,  0)

= Solution

Write the <set difference> as an <intersection> and use the fact that taking the <complement of a set> twice returns that <set>. The <commutativity> of <intersection> gives the reversible calculation
$$
\boxed{\overline A\setminus\overline B=\overline A\cap\overline{\overline B}=\overline A\cap B=B\cap\overline A=B\setminus A.}
$$

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