Going from ¬(¬q → s) to ¬q ∧ ¬s, I am confused. Is this using expression for implication, double negation and DeMorgan's? The following is what I thought:
I thought first in terms of the rule that q → s ⇔ ¬q ∨ s
Then with the original negation on the q back in ¬q → s ⇔ q ∨ s
Then with the original negation outside premises back in ⇔ ¬(q ∨ s)
Then DeMorgan's Law ⇔ ¬q ∧ ¬s
I feel that I'm making things more complicated then they need to be.
You are correct:
$$\begin{align} \lnot (\lnot p \rightarrow s)&\iff \lnot(\lnot(\lnot p) \lor s)\tag{1}\\\\ & \iff \lnot (p \lor s)\tag{2}\\\\ & \iff \lnot p \land \lnot s\tag{3}\end{align}$$
and for the reasons you give: $(1)$ by the definition of $\rightarrow$, $(2)$ double negation, and $(3)$ by applying DeMorgan's. Note that each step represents a biconditional, meaning that the implications go both directions.
It's not terribly complicated; you'll likely encounter far more complicated proofs using rules of replacement as well as proofs using rules of inference.