Is it true that $E[Y|W]=E[Y|E(X|W)]$, given that $W$ is $X$ measured with error In Carroll's 2006 book "Measurement Error in Nonlinear Models", and on p38 it is stated without proof that:

[One can] estimate the regression of $Y$ on $(Z, X)$ and then to substitute into this model $\{Z,E(X|W)\}$ [ ... ] [which is equivalent to] to substituting (Z, W) into the fitted model for the regression of $Y$ on $(Z,W)$.

$W$ is a measurement of the predictor $X$, but measured with error. $Z$ is a covariate measured without error.
Which is the following equivalence.
$$ E[Y|W] = E[Y|E(X|W)]$$
(I have remove $Z$ from the subsequent calculations, as they are not needed in my problem).
Is there a proof for this? I have found the following paper that mentions this, but it is vague.

Babanezhad, Yaghmaei (2010) "Bias Terms of Measurement Error in Treatment by expected Estimating Equations"

 A: The relation $E[Y | W] = E[Y | E(X | W)]$
does not hold in a general context, e.g. when if $Y$ is assumed
to be independent of $W$ conditional on $X$. 
Assume that $X$ is standard normal and that $W := X^2$, so that $E(X |
W) = 0$. If we take $Y = X^2 + \varepsilon$ where $\varepsilon$ is a
noise independent of $X$, then $E[Y | W] = W = X^2$, while $E[Y | E(X
| W)]$ is simply $E(Y) = 1$.
Now, I confess that in this example $W$ can not be regarded as $X$
with a measurement error, so further assumptions should be imposed.
As suggested by @X'ian, the relation would hold if we could be sure
that $E[X|W]$ is a linear function of $W$, say $\alpha W$ for some
$\alpha$. This linearity is granted for example by assuming that the
vector $[X,\,W]^\top$ has an elliptically symmetric distribution. In
the present context, only $\alpha =1$ would make sense.
A: The equation
$$E[Y\mid W] = E[Y\mid E(X\mid W)]\;\;\;  \text {does not hold in general}$$
but let's see a very "well-behaved" example where it does hold. Assume that $\{Y, W\}$ follow a bivariate normal distribution, and that $\{X,W\}$ also follow a bivariate normal. For simplicity assume all three variables have zero unconditional expected values. I will use for compactness $E(X \mid W) \equiv Z$ (which is not the "Z" in the question). Then
$$E(Y \mid W) = \frac {{\rm Cov}(Y,W)}{\sigma^2_W} W =  \frac {E(YW)}{\sigma^2_W}W  \tag{1}$$
$$E(X \mid W) \equiv Z  = \frac {{\rm Cov}(X,W)}{\sigma^2_W} W = \frac {E(XW)}{\sigma^2_W} W\tag{2}$$
where "$\sigma$" denotes here the standard deviation. $Z$ follows a normal distribution, and being just a scaled version of $W$, it too has a joint bivariate normal distribution with $Y$. So 
$$E(Y \mid Z) = \frac {{\rm Cov}(Y,Z)}{\sigma^2_Z} Z = \frac {E(YZ)}{\sigma^2_Z} Z \tag{3}$$
We are wondering "Is $(1) = (3)\;?$". For it to hold we must have
$$\text{If}\;\;\; (1) = (3) \implies \frac {E(YW)}{\sigma^2_W} W = \frac {E(YZ)}{\sigma^2_Z} Z \tag{4}$$
From $(2)$ we have
$$\sigma^2_Z = \left(\frac {E(XW)}{\sigma^2_W}\right)^2 \sigma^2_W = \frac {[E(XW)]^2}{\sigma^2_W}  \tag {5}$$
Inserting $(5)$ into $(4)$, and inserting the expression for $Z$ from $(2)$ we obtain the requirement
$$\text{If}\;\;\; (1) = (3) \implies \frac {E(YW)}{\sigma^2_W} W = \frac {E(YZ)\cdot \sigma^2_W}{[E(XW)]^2} \frac {E(XW)}{\sigma^2_W} W$$
and simplifying we arrive at
$$\text{If}\;\;\; (1) = (3) \implies \frac {E(YW)}{\sigma^2_W}  = \frac {E(YZ)}{E(XW)}   \tag{6}$$
Now, assume that $W = X+U$ where $U$ is zero-mean  normal independent of $X$ and $Y$ (this is consistent with the assumptions of pairwise bivariate normality). Then
$$ E(XW) = E(X^2) = \sigma^2_X$$
Inserting  into $(6)$, while also substituting for $Z$ (from $(2)$) we obtain
$$\text{If}\;\;\; (1) = (3) \implies \frac {E(YW)}{\sigma^2_W}  = \frac {E\left(Y\frac {E(XW)}{\sigma^2_W} W\right)}{\sigma^2_X}  $$
$$\implies  \frac {E(YW)}{\sigma^2_W}  = \frac {E(XW)}{\sigma^2_W}\frac {E\left(Y W\right)}{\sigma^2_X}$$
$$\implies 1 = \frac {E(XW)}{\sigma^2_X}$$
which holds so here indeed
$$E[Y\mid W] = E[Y\mid E(X\mid W)]$$
does hold.
