Equation 13.22 from Probabilistic Robotics below:
Here's how I get from first line to second line:
$$ p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} \\ p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) = p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) \\ p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) = p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t}) $$ Just setting up conditional probabilities above, then I'm subbing back to the first equation:
$$ \\ p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) = \eta \ p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) $$
how do I get to the third line from here?