Answer
$For\,\,all\,\,subsets\,\,A\,\,of\,\,a\,\,universal\,\,set\,U,\,\,A\cap A^{c}=\varnothing \\
proof:\,\,\\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap A^{c}\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap A^{c}\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\cap A^{c} \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
so\,\,suppose\,\,A\cap A^{c}\neq \varnothing \\
x\in \,\,A\cap A^{c} \\
x\in A\,\,and\,\,x\in A^{c}\,\,(by\,\,def.\,\,of\,\,inter\! section )\\
by\,\,def.\,\,of\,\,complement \\
x\in A\,\,and\,\,x\notin A (this\,\,is\,\,a\,\,contradiction) \\
so\,\,A\cap A^{c}=\varnothing
$
Work Step by Step
$For\,\,all\,\,subsets\,\,A\,\,of\,\,a\,\,universal\,\,set\,U,\,\,A\cap A^{c}=\varnothing \\
proof:\,\,\\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap A^{c}\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap A^{c}\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\cap A^{c} \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
so\,\,suppose\,\,A\cap A^{c}\neq \varnothing \\
x\in \,\,A\cap A^{c} \\
x\in A\,\,and\,\,x\in A^{c}\,\,(by\,\,def.\,\,of\,\,inter\! section )\\
by\,\,def.\,\,of\,\,complement \\
x\in A\,\,and\,\,x\notin A (this\,\,is\,\,a\,\,contradiction) \\
so\,\,A\cap A^{c}=\varnothing
$
