Answer
$a-\,\,\\
an\,element\,argument\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\
means\,\,(A-B)\cap B=\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,(A-B)\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(A-B)\cap B\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,(A-B)\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
proof:\\
x\in (A-B)\cap B \Rightarrow x\in (A-B)\,\,and\,\,B \\(def.\,\,of\,intersection)\\
\Rightarrow x\in A \,and\,x\notin B \,and\, x\in B \\ (def.\,\,of\,\,set\,\,dif\! ference)\\
but\,\,x\in B \,and\, x\notin B \\
(this\,is\,a\,\,contradiction)\\
so\,(A-B)\cap B=\varnothing
$
$b-\,\,\\
an\,algebraic\,\,\,argument\,\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\
means\,\,(A-B)\cap B=\varnothing \\
=(A\cap B^c)\cap B\\ (set\,\,dif\! ference\,\,law)\\
=A\cap \left ( B^c\cap B \right ) \\ (associative\,\,law)\\
=A\cap \left ( B\cap B^c \right ) \\ (commutative\,\,law\,\,for\,\,\cap)\\
=A\cap \varnothing \\(complement\,law)\\
=A \\(def.of\varnothing )$
$c-\,\,\\
algebraic\,\,\,argument\,is\,easier\,than\,element\,\,\,argument$
Work Step by Step
$a-\,\,\\
an\,element\,argument\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\
means\,\,(A-B)\cap B=\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,(A-B)\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(A-B)\cap B\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,(A-B)\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
proof:\\
x\in (A-B)\cap B \Rightarrow x\in (A-B)\,\,and\,\,B \\(def.\,\,of\,intersection)\\
\Rightarrow x\in A \,and\,x\notin B \,and\, x\in B \\ (def.\,\,of\,\,set\,\,dif\! ference)\\
but\,\,x\in B \,and\, x\notin B \\
(this\,is\,a\,\,contradiction)\\
so\,(A-B)\cap B=\varnothing
$
$b-\,\,\\
an\,algebraic\,\,\,argument\,\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\
means\,\,(A-B)\cap B=\varnothing \\
=(A\cap B^c)\cap B\\ (set\,\,dif\! ference\,\,law)\\
=A\cap \left ( B^c\cap B \right ) \\ (associative\,\,law)\\
=A\cap \left ( B\cap B^c \right ) \\ (commutative\,\,law\,\,for\,\,\cap)\\
=A\cap \varnothing \\(complement\,law)\\
=A \\(def.of\varnothing )$
$c-\,\,\\
algebraic\,\,\,argument\,is\,easier\,than\,element\,\,\,argument$