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For all sets A and B, (A ∪ B) – B = A – B

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$\begin{aligned} LHS &= (A\cup B) - B = (A\cup B )\cap B' &\quad& [A-B = A\cap B'] \\ & = (A\cap B')\cup (B\cap B') \\ & = (A-B)\cup P && [ P = (A\cap A')\text{ and }A - B = A\cap B'] \\ & = (A-B) = RHS\end{aligned}$

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