Cantor-Schroder-Bernstein Contradiction I need help figuring out where to start a proof that says I should use a proof by contradiction.
$f\colon A\to B$ and $g\colon B\to A$ be functions and each is 1-1. Let $D$ be the range of $f$ (i.e., $D=f(A)$).
Let $x$ and $y$ be natural numbers with $x>y$.  Prove: $g(fg)^x (B-D)\cap g(fg)^y (B-D)=\emptyset$.
 A: Maybe start with the simplest case: $x=1$ and $y=0$.  (I'm going to consider $0$ to be a "natural number"; not everybody does so.)  Then the statement to prove is
$$ gfg(B-D)\cap g(B-D) = \emptyset $$
To prove that some set is empty by contradiction, I'd start by assuming the opposite, that is, suppose the set is nonempty.  That gives me something concrete to work with: an element in the set.  Then I look for things that I can say about that element.
So, to start:

Suppose, on the contrary, that $gfg(B-D)\cap g(B-D)\ne\emptyset$.  Let $x\in gfg(B-D)\cap g(B-D)$.

Then... what?  What can you say about $x$?  What have we already said about $x$?
A: If $g(fg)^x(B-D)∩g(fg)^y(B-D)≠∅$ let a belong to the intersection of the two sets. Then there are u, v in B-D such that $g(fg)^x(u)=g(fg)^y(v)=a$. Then $(fg)^x(u)=(fg)^y(v)$ because we can multiply each side of the first equality by $g^{-1}$, which is 1-1, and get the second equality. But then $(fg)^{-y}(fg)^x(u)=(fg)^{-y}(fg)^y(v)$ and thus 
$(fg)^{x-y}(u)=v$. But fg(u) is not in B-D and so also every $(fg)^n(u)$ is not in D-B for every natural number n (because fg is 1-1 and thus it preserves the disjointedness of sets), but v is in B-D so we get a contradiction.
Incidentally, (fg)^x is a (commonly used) misnomer because fg need not be applied again to get to the xth image of a member form B-D.
