I just wanted to clarify what you were able to deduce and chime in with the others about your question.
The premises you're working with:
1) $l \lor s$
2) $\lnot c$
3) $l \rightarrow b$
4) $s \rightarrow c$
5) $s \rightarrow \lnot w$
You mention that the exercise you were working with "wants us to deduce" some statements; first you start with
$\lnot c \rightarrow \lnot s\quad$ by Modus Tollens (4) (and $\lnot c$ is in (2))
- This should probably be written as two steps (as in 6, 7 below). (See 6, 7, 8, 9 based on your logic, with additional justifications.)
6) $\;\lnot c \;\rightarrow \;\lnot s \quad$ (Modus Tollens, 4)
7) $\;\lnot s\;\qquad\qquad\quad$ (Modus Ponens, by (2) & (6))
8) $\;\therefore \;l\qquad\qquad\quad$ (Disjunctive syllogism of (1) & (7))
9) $\;\therefore \;b\qquad\qquad\quad$ (Modus ponens by (4) & (8))
I realize the question in your post didn't concern your deduction of the above steps, but I just wanted to make sure you weren't taking "shortcuts" and possibly omitting a step and references to previous statements used to justify those statements.
Now, note that, up to this point premise (5) is irrelevant/superfluous to your conclusions above...
You knew that it was appropriate and smart to apply Modus Tollens to conclude statement 6) above; I'm assuming now that others have clarified your question, you realize that given $\lnot s$, one cannot apply Modus Tollens to $s \rightarrow \lnot w$ to conclude therefore $w$, as that is not equivalent to Modus Tollens.
- If you had $w$, you could deduce therefore $\lnot\lnot w$, and by Modus Tollens, get $\lnot s$, (by $\lnot \lnot w$ and premise (5), with Modus Tollens); but why would you want to do that? ...you already have $\lnot s$!
- One other point: you could have actually constructed the 5th premise from the first 4 premises, along with your subsequent deduction of $\;\lnot s$. For example:
7) $\lnot s\qquad\qquad\qquad$(...reasoning above)
7+) $\lnot s \lor \lnot w\quad\;$ (Disjunction Introduction/Addition from (7): that is, if we have established
$\qquad\qquad\qquad\qquad\quad\lnot s$, then we've established $\lnot s \lor $(anything))
5) $s \rightarrow \lnot w\qquad\;$ (Given (7+) and the fact that $\lnot s \lor \lnot w \equiv s \rightarrow \lnot w$).