Conditional probability interpretation?

Question

I am having a hard time to understand a given solution in problem 1e and 1f) in this problem set http://www.eecs.umich.edu/courses/eecs401/pdfs-w07/401w07hw3_soln.pdf:

The given solution for 1e looks like this:

$$P((F_2\cap D_2)^c| F_1^c,S_1,S_2) = 1 -P(F_2\cap D_2|F_1^c,S_1,S_2)$$

$$=1-P(D_2|F_2,F_1^c,S_1,S_2)P(F_2|F_1^c,S_1,S_2)$$ $$=1-P(D_2|F_1^c)P(F_2|F_1^c,S_1)$$

I get how the first line says that to get the complement, we use 1-P(a) formula. But I am lost for the rest of the explanations.

1. How did the right-hand side in the first part transforms to the
   second line? If they just simply expand the equation, why does the
   $P(D_2|all)$ part has $F_2$ for the given condition, and the $P(F_2|all)$ doesn’t?
2. Where do the $S_1,S_2$ go from line 2 to line 3? While $S_1$ remains in the $P(F_2|all)$ part?

For 1f, how come most of the given conditions disappear in the fifth line?

Edit: I added the images as requested here. This is the context of the question:

These are the parts where I have problems interpreting:

I really want to understand this, so any help is appreciated. Thank you.

Answer 1

1. Please consider $P(D_2 \cap F_2) = P(D_2 \mid F_2)P(F_2)$ and add the conditioning on $(F_1^c,S_1,S_2)$ to the whole expression.
2. Most likely it is due to independence. It looks like $F_2$ is independent to $S_2$ but depends on $S_1$, and $D_2$ is independent to both $(S_1,S_2)$. When events are independent, they drop from the conditioning by definition.