puzzle in derivation of entropy of sum I was reading through a derivation of an entropy of a sum example, and cannot understand one step.
The statement to be proven is if $Z=X+Y$ then $H(Z|X) = H(Y|X)$, where H() is discrete Shannon entropy.
The derivation:
$Z = X+Y$, Hence note $p(Z=z|X=x) = p(Y=z-x|X=x)$.
\begin{align}
H(Z|X) &= \sum p(x) H(Z|X=x)
\\
 &= -\sum_x p(x) \sum_z p(Z=z|X=x) \log p(Z=z|X=x)
\\
 &= -\sum_x p(x) \sum_y p(Y=z-x|X=x) \log p(Y=z-x|X=x)
\\
 &= \sum p(x) H(Y|X=x)
\\
 &= H(Y|X)
\end{align}
My issue is the step introducing probability $p(Y=z-x|X=x)$. 
First, I think this probability is always $1$,
because $Y$ is defined to be $z-x$, so how can it be anything else?
Second, an expression $\sum_y p(Y=z-x)$ bothers, because in summing over the index $y$, $y$ is supposed to be an independent thing, but here $Y$ depends on $x$ through $Y=z-x$.  Actually the notation in the derivation does not clarify $y$ verus $Y$ also.
 A: The notation convention here is that the capitalized letter denotes a random variable, and the lower-case letter denotes a possible value for the corresponding random variable. For example if $X$ and $Y$ are the respective results of two rolls of a die, then $x$ would be any one of $1, 2, 3, 4, 5, 6$. Similarly $y$ stands for any one of those same six values.
In your derivation, for a given $x$ the quantity $P(Y=z-x\mid X=x)$ is not always $1$, since $Y$ and $X$ (and $Z$) are random variables and can take a variety of values. Try plugging values like $x=3, z=8$ in the die-rolling example.  If you know that $X=3$, the conditional probability is not $1$ that $Y=8-3$.
For your second question, the expression $\sum_y p(Y=z-x\mid X=x)$ has a typographical error. It should be $\sum_z p(Y=z-x\mid X=x)$. To get the next line you realize that for a given $x$, summing over all possible values of $z$ is the same as summing over all possible values of $y$, so the inner sum is  just $H(Y\mid X=x)$.
