Asume you have a boolean function $f$ which takes $n$ parameters and gives $m$ results. In addition, you have a boolean function $g$ takes $p$ parameters and gives out $n$ results.
You could calculate as follows: $f(g) = h$ resulting in a function which takes $p$ params and returns $m$ results.
Is there a general way to recover $g$ from $h$? In other words, is there a general way to calculate $f^{-1}$ such that $f^{-1}(h)$ gives $g$?
Is it as simple as inverting $f$ to get $\bar f$ and applying that to $h$? i.e. $\bar f(h) = g$.
[I realize I am probably making things imprecise by saying $f$, $g$, and $h$ are functions that return multiple values. I guess they could all be functions of 1 input and 1 output.]
It occurs to me if $f^{-1}(f) = 1$ then we could say $f^{-1} = \bar f \oplus$ since $f^{-1} \oplus f = 1$.
I still don't know if $\bar f \oplus h = g$. I guess thats the question. Or is there some other way to extract $g$ from $h$?