entropy: is H(X+Y) = H(X) + H(X+Y|X) true? Is this equation true for discrete Shannon entropy $H$:
$$
   H(X+Y) = H(X) + H(X+Y|X)
$$
I came across it in a reply to an earlier question, 
strange statement in proof of entropy of a sum,
about a step in a proof of the entropy of a sum of two variables.
They gave this proof that $H(X+Y) = H(X)+H(Y)$ in the case that $X,Y$ are independent:
$$
   H(X+Y) = H(X) + H(X+Y)|X) = H(X) + H(Y|X) 
$$
and $H(Y|X)=H(Y)$ if independent.
Unfortunately the reply was deleted. My fault, I replied in turn that this was not about my question.
But, later I see that the equation they gave is not something I know about, and wish they had not deleted.
I wonder if they deleted because I was rude, or because they realized it is incorrect.
The equation in the title "looks" plausible, but I do not remember it in the textbook. How can we prove it?
$$
\begin{align*}
   H(X+Y) &= \sum_x \sum_y p(X+Y) \log p(X+Y)
\\ 
      &= \cdots ?
\end{align*}
$$
Writing the entropy formula does not seem to help. $p(x+y)$ does not simplify further I think?  It is the convolution of pdfs of X and Y, but that does not help (I think).  
 A: 
Reading more carefully the former question, it appears that this
  identity functions when $(X,Y)$ is identified by $Z$, meaning that a
  value of $Z=z$ is associated with a single possible value of
  $(X,Y)=(x,y)$, with $x+y=z$.

Under this provision, and omitting the minus sign in the definition of the entropy for convenience sake,
\begin{align*}
H(X+Y) &= \sum_{z\in {\cal Z}} \text{pr}(X+Y=z)\log \text{pr}(X+Y=z)\\
&=\sum_{z\in {\cal Z}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(X+Y=z,\,X=x)\log \text{pr}(X+Y=z)\right\}\\
&=\sum_{z\in {\cal Z}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(X+Y=z|X=x)\text{pr}(X=x)\log \text{pr}(X+Y=z)\right\}\\
&=\sum_{z\in {\cal Z}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(Y=z-x|X=x)\text{pr}(X=x)\log \text{pr}(X+Y=z)\right\}\\
&=\sum_{y\in {\cal Y}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(Y=y|X=x)\text{pr}(X=x)\log \text{pr}(X+Y=x+y)\right\}\\
&=\sum_{y\in {\cal Y}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(Y=y|X=x)\text{pr}(X=x)
\underbrace{\log {\text{pr}(X+Y=x+y|X=x)}}_{\log{\text{pr}(Y=y|X=x)}\text{ since }X\text{ known}}
\text{pr}(X=x)\right\}\\
&=\sum_{y\in {\cal Y}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(Y=y|X=x)\text{pr}(X=x)\log \text{pr}(Y=y|X=x)\right\}\\
&+\sum_{y\in {\cal Y}}\,\left\{\sum_{x\in {\cal X}} \text{pr}(Y=y|X=x)\text{pr}(X=x)\log {\text{pr}(X=x)}\right\}\\
&=\sum_{x\in {\cal X}}\,\text{pr}(X=x)\,\left\{\sum_{y\in {\cal Y}} \text{pr}(Y=y|X=x)\log \text{pr}(Y=y|X=x)\right\}\\
&+\sum_{x\in {\cal X}}\text{pr}(X=x)\left\{\sum_{y\in {\cal Y}} \text{pr}(Y=y|X=x)\log \text{pr}(X=x)\right\}\\
&= \sum_{x\in {\cal X}}\,\text{pr}(X=x)H(Y|X=x)]+\sum_{x\in {\cal X}}\text{pr}(X=x)\log \text{pr}(X=x)\\
&= H(Y|X)+H(X)
\end{align*}
A: The claim is false. Let $X$ have a non-constant discrete distribution, and define $Y:=-X$. Then $Z:=X+Y$ is a constant (namely zero), so its entropy $H(X+Y)$ is zero, and the conditional entropies $H(Y\mid X)$ and $H(Z\mid X)$ are also zero, since the values of both $Y$ and $Z$ are completely determined by the value of $X$. If the claim
$$
H(X+Y)=H(X) + H(Y\mid X)
$$
were true, it would follow that $H(X)=0$, which is impossible when $X$ is non-constant.
