Answer
$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\
(A - B) \cup (A \cap B) = A.\\
Proof:\,\\
(A - B) \cup (A \cap B)=(A \cap B^c) \cup (A \cap B)\\
by\,\,\,(set\,\,dif\! ference\,\,law )\\ =A \cap (B^c \cap B)\\
by\,\,\,(distributive\,\,law)\\
=A \cap (B \cap B^c)\\
by\,\,\,(commutative\,\,law\,\,for\,\,\cap) .\\ =A \cap U\\
by\,\,\,complement\,\,law \\
=A\\
by(identity\,law\,for\,\cap )\\
$
Work Step by Step
$For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\
(A - B) \cup (A \cap B) = A.\\
Proof:\,\\
(A - B) \cup (A \cap B)=(A \cap B^c) \cup (A \cap B)\\
by\,\,\,(set\,\,dif\! ference\,\,law )\\ =A \cap (B^c \cap B)\\
by\,\,\,(distributive\,\,law)\\
=A \cap (B \cap B^c)\\
by\,\,\,(commutative\,\,law\,\,for\,\,\cap) .\\ =A \cap U\\
by\,\,\,complement\,\,law \\
=A\\
by(identity\,law\,for\,\cap )\\
$