In Bayesian inference, why are some terms dropped from the posterior predictive?

In Kevin Murphy's Conjugate Bayesian analysis of the Gaussian distribution, he writes that the posterior predictive distribution is

$$p(x \mid D) = \int p(x \mid \theta) p(\theta \mid D) d \theta$$

where $$D$$ is the data on which the model is fit and $$x$$ is unseen data. What I don't understand is why the dependence on $$D$$ disappears in the first term in the integral. Using basic rules of probability, I would have expected:

\begin{align} p(a) &= \int p(a \mid c) p(c) dc \\ p(a \mid b) &= \int p(a \mid c, b) p(c \mid b) dc \\ &\downarrow \\ p(x \mid D) &= \int \overbrace{p(x \mid \theta, D)}^{\star} p(\theta \mid D) d \theta \end{align}

Question: Why does the dependence on $$D$$ in term $$\star$$ disappear?

For what it's worth, I've seen this kind of formulation (dropping variables in conditionals) other places. For example, in Ryan Adam's Bayesian Online Changepoint Detection, he writes the posterior predictive as

$$p(x_{t+1} \mid r_t) = \int p(x_{t+1} \mid \theta) p(\theta \mid r_{t}, x_{t}) d \theta$$

where again, since $$D = \{x_t, r_t\}$$, I would have expected

$$p(x_{t+1} \mid x_t, r_t) = \int p(x_{t+1} \mid \theta, x_t, r_t) p(\theta \mid r_{t}, x_{t}) d \theta$$

This is based on the assumption that $$x$$ is conditionally independent of $$D$$, given $$\theta$$. This is a reasonable assumption in many cases, because all it says is that the training and testing data ($$D$$ and $$x$$, respectively) are independently generated from the same set of unknown parameters $$\theta$$. Given this independence assumption, $$p(x|\theta,D)=p(x|\theta)$$, and so the $$D$$ drops out of the more general form that you expected.
It's because $$x$$ is assumed to be independent of $$D$$ given $$\theta$$. In other words, all data is assumed to be i.i.d. from a normal distribution with parameters $$\theta$$. Once $$\theta$$ is taken into account using information from $$D$$, there is no more information that $$D$$ gives us about a new data point $$x$$. Therefore $$p(x|\theta, D) = p(x|\theta)$$.