Bayesian Theorem Update Inference

Question

According to Bayes theorem $P(A|B) = \frac{P(B|A)*P(A)}{P(B)}$

I've found somewhere that: $P(x_t|z_{1:t}) = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})}{P(z_t|z_{1:t-1})}$ but I don't really understand it, is it expressed according to the bayes theorem ?

This is from page 13 of http://www.igi.tugraz.at/pfeiffer/documents/particlefilters.pdf

EDIT after @ConjugatePrior's answer :

According to Bayes theorem $P(x_t|z_{1:t}) = \frac{P(z_{1:t}|x_t)*P(x_t)}{P(z_{1:t})}$ and this is equal to $\frac{P(z_t,z_{1:t-1}|x_t)*P(x_t)}{P(z_t,z_{1:t-1})}$, but since $z_t$ and $z_{t-1}$ are conditionally independent given $x_t$ (according to the above figure), then $P(z_t,z_{1:t-1}|x_t) = P(z_t|x_t)*P(z_{1:t-1}|x_t)$, so we get $\frac{P(z_t|x_t)*P(z_{1:t-1}|x_t)*P(x_t)}{P(z_t,z_{1:t-1})}$ and again according to Bayes theorem we have $P(z_{1:t-1}|x_t) = P(x_t|z_{1:t-1})*P(z_{1:t-1}) / P(x_t)$, so we get $\frac{P(z_t|x_t)*P(x_t|z_{1:t-1})*P(z_{1:t-1})}{P(z_t,z_{1:t-1})} = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})*P(z_{1:t-1})}{P(z_t|z_{1:t-1})*P(z_{1:t-1})} = \frac{P(z_t|x_t)*P(x_t|z_{1:t-1})}{P(z_t|z_{1:t-1})}$

Answer 1

It is, but you're being confused by the extra unmentioned assumption: conditional independence of the past observations and the current observation given the current state. Your slides do not bother to draw the state space model diagram for the state space time series models they're interested in. This is unfortunate because you can read off this relationship directly.

To make it look like your first version of Bayes theorem, rewrite $P(z_t | x_t)$ to condition also on $z_{1\ldots t-1}$ as everything else does. That should look more familiar. Now notice that the conditional independence assertion above means that once you condition on $x_t$, $z_t$ no longer actually depends on its past, the part you just added in. So you can take it back out.