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