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