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

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

Hence, the given statement is proved.

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