Show that $f^{-1}(F-B)=E-f^{-1}B$ where $f:E\to F$ and $B\subset F.$ Show that $f^{-1}(F-B)=E-f^{-1}(B)$ where $f:E\to F$ and $B\subset F.$
Proof: Let $e\in f^{-1}(F-B).$ Then $f(e)\in F-B$ which means that $f(e)\in F$ and $f(e)\not \in B.$ Now since $B\subset F$ we have that $f^{-1}(B)\subset f^{-1}(F)=E.$ And so, since $e\not \in f^{-1}(B)$ we have that $e\in E-f^{-1}B.$ We can therefore say that $e\in E-f^{-1}(B).$ Thus $f^{-1}(F-B)\subset E-f^{-1}(B).$ Now for the other inclusion we consider $e\in E-f^{-1}(B)$ which means that $e\in E$ and $e\not \in f^{-1}(B).$ Thus $f(e)\in F$ and $f(e)\not \in B.$ Since $B\subset F$ we have that $f(e)\in F-B.$ And therefore $e\in f^{-1}(F-B).$ 
I don't know whether this proof is correct or not and would, therefore, appreciate any inputs/suggestions. 

$x\in f^{-1}(F-B) \Leftrightarrow f(x)\in F-B \Leftrightarrow f(x)\not\in B \Leftrightarrow x\not\in f^{-1}(B) \Leftrightarrow x\in E-f^{-1}(B)$
