prove cardinality rule $|A-B|=|B-A|\rightarrow|A|=|B|$ I need to prove this $|A-B|=|B-A|\rightarrow|A|=|B|$
I managed to come up with this:
let $f:A-B\to B-A$ while $f$ is bijective.
then define $g\colon A\to B$ as follows:
$$g(x)=\begin{cases}
  f(x)& x\in (A-B) \\ 
  x& \text{otherwise} \\ 
\end{cases}$$
but I'm not managing to prove this function is surjective.
Is it not? or am I on the right path? if so how do I prove it?
Thanks

Note that
$$\begin{align}
|A| = |A \cap B| + |A \cap B^c| = |B \cap A| + |B \cap A^c| = |B|.
\end{align}$$
Here $E^c$ denotes the compliment of the event $E$ in the universal space $X$.
