You only criticized my first proof though. I shipped two proofs gladly! The second one just uses inverses of the collatz function and graph theory.
In general, for any odd n, both n and 6n+2 in a single step lead to 3n+1.
I can inverse f(x)=f(x+1) easily as g(x)=f(x)-1 since g(f(x)) = x do you understand my functional inversion approach now better?