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I have several of these types of problems, and it would be great if I can get some help.

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    $\begingroup$ It would be great if you'd actually write the problem (you can use $\LaTeX$ code, just put it between $ symbols); and what sort of help you're looking for: that is, what is the problem that you're facing when solving these questions? $\endgroup$
    – Asaf Karagila
    Apr 20, 2014 at 14:50
  • $\begingroup$ I have to prove this identity, but i do not know how to do that and do not have any examples similar to this. $\endgroup$
    – user144463
    Apr 20, 2014 at 14:57
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    $\begingroup$ Are you familiar with $X\subseteq Y\iff\forall z\left[z\in X\Rightarrow z\in Y\right]$? $\endgroup$
    – drhab
    Apr 20, 2014 at 15:02
  • $\begingroup$ @user144463 you can try writing an element wise proof i.e assume x belongs to L.H.S and prove it belongs to the R.H.S $\endgroup$
    – happymath
    Apr 20, 2014 at 15:02
  • $\begingroup$ Maybe this hint can help you: You have to prove that if $x\in C$ and $x\notin B$ then $x\in C$ and $x\notin A$. Notice that you already know that $x\in C$, which means you have to proove that $x\notin B$ implies $x\notin A$; can you use $A\subseteq B$ to show this?. $\endgroup$
    – Cure
    Apr 20, 2014 at 15:09

2 Answers 2

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Given that $A \subseteq B$, there exists a $D$ such that $A \cup D = B$. So $C - B \subseteq C - A$ is equivalent to $C - (A \cup D) \subseteq C - A$. It should be clear that this last statement is true because on the left side you're subtracting what you're subtracting on the right side, or possibly more.

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Let $x\in C-B$. This means $x$ is in $C$ but not $B$. But if $x$ is not in $B$, it cannot be in $A$ since $A\subseteq B$. Therefore $x$ is in $C$ but not $A$, or $x\in C-A$.

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