Answer
$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\
(A - B) \cap (A \cap B) = \varnothing .\\
Proof:\,\\
(A - B) \cap (A \cap B)=(A \cap B^c) \cap (A \cap B)\\
by\,\,\,(set\,\,dif\! ference\,\,law )\\
=A\cap (B^c\cap (A\cap B))\\ by(associative\,\,law)\\
=A\cap ((B^c\cap A)\cap B))\\by(associative\,\,law)\\
=A\cap ((A\cap B^c)\cap B))\\by(commutative\,\,law)\\
=A\cap (A\cap (B^c\cap B))\\by(associative\,\,law)\\
=(A\cap A)\cap (B^c\cap B)\\by(associative\,\,law)\\
=A \cap (B^c\cap B) \\by(idempotent\,\,law)\\
=A \cap (B\cap B^c) \\by(commutative\,\,law)\\
=A \cap \varnothing \\by(complement\,\,law)\\
=A\\by(def.\,\,of.\varnothing )$
Work Step by Step
$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\
(A - B) \cap (A \cap B) = \varnothing .\\
Proof:\,\\
(A - B) \cap (A \cap B)=(A \cap B^c) \cap (A \cap B)\\
by\,\,\,(set\,\,dif\! ference\,\,law )\\
=A\cap (B^c\cap (A\cap B))\\ by(associative\,\,law)\\
=A\cap ((B^c\cap A)\cap B))\\by(associative\,\,law)\\
=A\cap ((A\cap B^c)\cap B))\\by(commutative\,\,law)\\
=A\cap (A\cap (B^c\cap B))\\by(associative\,\,law)\\
=(A\cap A)\cap (B^c\cap B)\\by(associative\,\,law)\\
=A \cap (B^c\cap B) \\by(idempotent\,\,law)\\
=A \cap (B\cap B^c) \\by(commutative\,\,law)\\
=A \cap \varnothing \\by(complement\,\,law)\\
=A \\by(def.\,\,of.\varnothing )$
