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