Chapter 6 - Set Theory - Exercise Set 6.2 - Page 366: 34

$for\,\,all\,\,sets\,\,A,B\,and\,\,C,if\,B\cap C\subseteq A\,\\ then\,(C-A)\cap (B-A)=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,(C-A)\cap (B-A)\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(C-A)\cap (B-A)\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,(C-A)\cap (B-A) \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ so\,x\in (C-A)\cap (B-A)\\ (by\,def.\,\,of\,\,inter\! section\,and\,set\,dif\! ference) \\ \Rightarrow x\in C\,and\,x\notin A\,and\,x\in B\,and\,x\notin A \\ so\,x\in C \,and\,x\in B \,and\,x\notin A \Rightarrow \\ x\in B\cap C\,and\,x\notin A \\ but\,B\cap C\subseteq A(this\,mean\,if\,x\in B\cap C\Rightarrow x\in A)\\ so\,(x\in A\,and\,x\notin A)\\ but\,\,this\,\,is\,\,a\,\,contradiction\\ \therefore (C-A)\cap (B-A)=\varnothing $

$for\,\,all\,\,sets\,\,A,B\,and\,\,C,if\,B\cap C\subseteq A\,\\ then\,(C-A)\cap (B-A)=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,(C-A)\cap (B-A)\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(C-A)\cap (B-A)\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,(C-A)\cap (B-A) \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ so\,x\in (C-A)\cap (B-A)\\ (by\,def.\,\,of\,\,inter\! section\,and\,set\,dif\! ference) \\ \Rightarrow x\in C\,and\,x\notin A\,and\,x\in B\,and\,x\notin A \\ so\,x\in C \,and\,x\in B \,and\,x\notin A \Rightarrow \\ x\in B\cap C\,and\,x\notin A \\ but\,B\cap C\subseteq A(this\,mean\,if\,x\in B\cap C\Rightarrow x\in A)\\ so\,(x\in A\,and\,x\notin A)\\ but\,\,this\,\,is\,\,a\,\,contradiction\\ \therefore (C-A)\cap (B-A)=\varnothing $