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