$f(A - f^{-1}(B)) = f(A)-B $ proof Let $f:E\to F$ be a function and $A\in P(E) $ and $B \in P(F)$.
I'm asked to prove : 
$$f(A-f^{-1}(B)) = f(A)-B \tag1$$
(Where for any set $X$ and $Y$ we have $X-Y := \{x \ | \ x\in X \ \land x\notin Y \}$)
so $(1)$ is equivalent to $$f(A-f^{-1}(B)) \subset (f(A)-B) \land (f(A)-B)\subset f(A-f^{-1}(B))  $$
The problem is in proving $f(A-f^{-1}(B)) \subset (f(A)-B)$.
I tried to prove it in this way:
let $y\in f(A-f^{-1}(B))$
$$\Rightarrow \exists x\in A-f^{-1}(B) : \ y=f(x) \\
\Rightarrow \exists x\in A: x\notin f^{-1}(B) \ \land \ y=f(x) \\
\Rightarrow \exists x \in A : y=f(x) \ \land \ y\notin f(f^{-1}(B))$$
I'm stuck here.
Is there a better way to prove $(1)$
 A: Using your notation:
\begin{align}
&\Rightarrow \exists x\in A-f^{-1}(B) : \ y=f(x) \\
&\Rightarrow \exists x\in A: x\notin f^{-1}(B) \ \land \ y=f(x) \\
&\Rightarrow \exists x\in A: f(x)\notin B \ \land\ y=f(x) \\
&\Rightarrow  y\notin B\ \land\ y\in f(A)
\end{align}

Suppose $y\in f(A-f^{-1}(B))$; then, for some $x\in A-f^{-1}(B)$, we have $y=f(x)$. Since $x\in A$, we have $y\in f(A)$; since $x\notin f^{-1}(B)$, we have $f(x)\notin B$; therefore $y=f(x)\in f(A)-B$.
A: The only thing about your proof you have to change is the last step. You go from $x \notin f^{-1}(B)$ to $y \notin f(f^{-1}(B))$, but what you need is $y \notin B$.
Suppose that $y \in B$. Then $f(x) = y \in B$, which implies $x \in f^{-1}(B)$, since $f^{-1}(B) = \{ z \in E \mid f(z) \in B\}$. This is a contradiction since we know that $x \notin f^{-1}(B)$.
Combining $f(x) = y$ and $x \in A$ gives you that $y \in f(A)$; and combining that with $y \notin B$ gives you that $y \in (f(A) - B)$.
Note: $x \notin f^{-1}(B)$ also implies that $f(x) = y \notin f(f^{-1}(B))$, but $f(f^{-1}(B)) \subseteq B$ and this inclusion may be strict, which is why you can't use it in the proof.
