Answer
$For\,\,all\,\,sets\,\,A,\,B,\,\,and\,C\,\,\\
if\,A \subseteq B\,then\,\,A \cap B^c = \varnothing \\
this\,\,is\,\,true\,\,\\
To\,\,prove\,\,that\,\,a\,\,set\,\,A \cap B^c\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A \cap B^c\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A \cap B^c \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction\\
\,x\in A \cap B^c\Rightarrow x\in A \,and\, x\in B^c (by\,def.\,of\,inter\! section) \\
\Rightarrow x\in A \,and\, x\notin B \,(by\,def.\,of\,complement)\\
but\,A\subseteq B (means\,if\,x\in A \Rightarrow x\in B)\\
\Rightarrow x\in A \,and\, x\notin B \Rightarrow x\in B\,and\, x\notin B \\
this\,\,is\,\,a\,contradiction\\
so\,A \cap B^c = \varnothing
$
Work Step by Step
$For\,\,all\,\,sets\,\,A,\,B,\,\,and\,C\,\,\\
if\,A \subseteq B\,then\,\,A \cap B^c = \varnothing \\
this\,\,is\,\,true\,\,\\
To\,\,prove\,\,that\,\,a\,\,set\,\,A \cap B^c\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A \cap B^c\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A \cap B^c \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction\\
\,x\in A \cap B^c\Rightarrow x\in A \,and\, x\in B^c (by\,def.\,of\,inter\! section) \\
\Rightarrow x\in A \,and\, x\notin B \,(by\,def.\,of\,complement)\\
but\,A\subseteq B (means\,if\,x\in A \Rightarrow x\in B)\\
\Rightarrow x\in A \,and\, x\notin B \Rightarrow x\in B\,and\, x\notin B \\
this\,\,is\,\,a\,contradiction\\
so\,A \cap B^c = \varnothing
$
