I'm wondering how the equality in Eq.(6) could be derived.

[WGook]

Hi, I have a question about the equation in the paper.
I'm wondering how the equality in Eq.(6) could be derived.
With my knowledge, $p(x)$ becomes:

$$ \begin{align} p(x) &= \int p(x|z_0)p(z_0)dz_0, \\ &\mbox{ and } \\ p(z_0) &= \int p(z_{0:1})dz_{t(1):1},\\ p(z_{0:1}) &= p(z_1) \prod_{i = 1}^Tp(z_{s(i)}|z_{t(i)}), \end{align} $$

if $p(z_{0:1})$ is a Markov chain.

Then, how is it same to $\int_z p(z_1)p(x|z_0) \prod_{i = 1}^T p(z_{s(i)}|z_{t(i)})$?
Also, is $dz_1$ not needed to calculate the marginal distribution with integral in Eq.(6)?