Verifying an equation involving entropy and mutual information I would like to show the following:

Consider a joint distribution of $X,Y$ with density $p(x,y)$. For arbitrary functions $f,g:\mathbb{R}\to\mathbb{R}$ we have:
$$ H(X) + H(Y-f(X)) = H(Y) + H(X-g(Y)) - I(X-g(Y),Y) + I(Y-f(X),X)$$
where $H$ is the entropy and $I$ is the mutual information.

"Expanding" the right hand side, I got
$$H(Y) + H(X-g(Y)) - H(X-g(Y)) - H(Y) + H(X-g(Y),Y) + H(Y-f(X)) + H(X) - H(Y-f(X),X)$$
which simplies to
$$H(X) + H(Y-f(X)) + \color{red}{H(X-g(Y))+H(Y|X-g(Y)) - H(Y-f(X))-H(X|Y-f(X))}$$
What properties should I use to get rid of the red part?

$$H(Y) + H(X-g(Y)) - H(X-g(Y)) - H(Y) + H(X-g(Y),Y) + H(Y-f(X)) + H(X) - H(Y-f(X),X)$$
The first 4 terms cancel each other out and it becomes:
$$H(Y-f(X)) + H(X) + H(X-g(Y),Y) - H(Y-f(X),X)$$
However, 
\begin{align}
H(X-g(Y),Y)&=H(Y)+H(X-g(Y)|Y)\\
&=H(Y)+H(X)
\end{align}
and
\begin{align}
H(Y-f(X),X)&=H(X)+H(Y-f(X)|X)\\
&=H(X)+H(Y)
\end{align}
Hence,
$$H(X-g(Y),Y) - H(Y-f(X),X)=H(Y)+H(X)-\left(H(X)+H(Y)\right)=0$$
and it becomes 
$$H(Y-f(X)) + H(X)$$
which is the LHS.
