Answer
$For\,\,all\,\,sets\,\,A\,\,and\,\,B,\,\,if\,\,A\subseteq B\,\,then\,\,A\cap B^{c}=\varnothing \\
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) \\
and\,\,by\,\,def.\,\,of\,\,complement \Rightarrow x\in A \,and\,x\notin B \\
but\,A\subseteq B (\therefore if\,x\in A\,\,\Rightarrow x\in B ) \\
so\,\,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\,\,and\,\,B,\,\,if\,\,A\subseteq B\,\,then\,\,A\cap B^{c}=\varnothing \\
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) \\
and\,\,by\,\,def.\,\,of\,\,complement \Rightarrow x\in A \,and\,x\notin B \\
but\,A\subseteq B (\therefore if\,x\in A\,\,\Rightarrow x\in B ) \\
so\,\,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
$
