How to apply Bayes's theorem to the following derivation? Denote $\mathbf{Y}^{(t)}$ and $\mathbf{X}^{(t)}$ for the t dimensional column vectors $\mathbf{Y}^{(t)}=(Y_1,Y_2,...,Y_t)^T$ and $\mathbf{X}^{(t)} = (X_1, X_2,...,X_t)^T$.
Given assumption (1) that $Y_t$ given $(X_t, \mathbf{X}^{(t-1)},\mathbf{Y}^{(t-1)})$ is independent of $(\mathbf{X}^{(t-1)},\mathbf{Y}^{(t-1)})$ with conditional probability density
$$p(y_t | x_t):= p(y_t|x_t, \mathbf{x}^{(t-1)},\mathbf{y}^{(t-1)})$$
and assumption (2) that $X_{t+1}$ given $(X_t, \mathbf{X}^{(t-1)},\mathbf{Y}^{(t)})$ is independent of $(\mathbf{X}^{(t-1)},\mathbf{Y}^{(t)})$ with conditional density function
$$p(x_{t+1}|x_t):= p)(x_{t+1} | x_t, \mathbf{x}^{(t-1)},\mathbf{y}^{(t)})$$
Question: How can I make use of the assumption that the distribution of $Y_t$ given $(X_t,\mathbf{X}^{(t-1)},\mathbf{Y}^{(t-1)})$ does not depend on $(\mathbf{X}^{(t-1)},\mathbf{Y}^{(t-1)})$ and obtain that 
$$p(x_t|\mathbf{y}^{(t)}) = \frac{p(y_t|x_t)p(x_t|\mathbf{y}^{(t-1)})} { p(y_t|\mathbf{y}^{(t-1)})}$$
 A: This equation is proved just by assuming that $y_t$ is independent of $\mathbf{y}^{(t-1)}$ given $x_t$, that is:
$$p(y_t|x_t,\mathbf{y}^{(t-1)})=p(y_t|x_t)\space (*)$$
This assumption is a looser assumption than your first assumption, that is, you can conclude the above assumption from your first assumption. 
We know that:
$$p(x_t|\mathbf{y}^{(t)}) = \frac{p(x_t,\mathbf{y}^{(t)})} { p(\mathbf{y}^{(t)})}$$
Since $\mathbf{y}^{(t)}=(y_1,...,y_t)$ you can write the above equatin as:
$$p(x_t|\mathbf{y}^{(t)}) = \frac{p(x_t,\mathbf{y}^{(t-1)},y_t)} { p(\mathbf{y}^{(t-1)},y_t)}$$
According to the chain rule for probabilities we have:
$$p(x_t|\mathbf{y}^{(t)}) = \frac{p(\mathbf{y}^{(t-1)})p(x_t|\mathbf{y}^{(t-1)})p(y_t|x_t,\mathbf{y}^{(t-1)})} { p(\mathbf{y}^{(t-1)})p(y_t|\mathbf{y}^{(t-1)})}=\frac{p(x_t|\mathbf{y}^{(t-1)})p(y_t|x_t,\mathbf{y}^{(t-1)})} { p(y_t|\mathbf{y}^{(t-1)})}$$
By the assumption $(*)$, the proof is completed:
$$p(x_t|\mathbf{y}^{(t)}) = \frac{p(y_t|x_t)p(x_t|\mathbf{y}^{(t-1)})} { p(y_t|\mathbf{y}^{(t-1)})}$$
