Prove $P \lor Q, P \rightarrow R \lor S, R \rightarrow W \land Z, Z \land \lnot S \rightarrow \lnot W \vdash \lnot S \rightarrow Q$ using ND The problem is to prove
$$P \lor Q, P \rightarrow R \lor S, R \rightarrow W \land Z, Z \land \lnot S \rightarrow \lnot W \vdash \lnot S \rightarrow Q$$
using natural deduction.
I'm studying natural deduction for my upcoming exam in Logic and I have come across this problem. Unfortunately, I cannot seem to find a path to its solution. What I have done is the following and I don't know how to proceed any further:

  
*
  
*$P \lor Q$
  
*$P \rightarrow R \lor S$
  
*$R \rightarrow W \land Z$
  
*$Z \land \lnot S \rightarrow \lnot W$
  
*$P \qquad H$
  
  

    
*$R \lor S \qquad E \rightarrow 2,5$ 
    
*$R \qquad H$
    
    

      
*$W \land Z \qquad E\rightarrow 3,7$
      
*$W \qquad E\rightarrow 8$
      
*$Z \qquad E\rightarrow 8$
      
    


My professor only uses the Fitch notation, so I would appreciate you using it in your answer.
EDIT: Using the suggestions in the answers provided and studying the rules some more, I have come up with this solution. Can you tell me whether it is acceptable?
$1 \quad P \lor Q
\\ 2 \quad P \rightarrow R \lor S
\\ 3 \quad R \rightarrow W \land Z
\\ 4 \quad Z \land \lnot S \rightarrow \lnot W
\\ 5 \qquad \lnot S \qquad assumption
\\ 6 \qquad \qquad P \qquad assumption
\\ 7 \qquad \qquad R \lor S \qquad implication \ elim \ 2,6
\\ 8 \qquad \qquad R \qquad disjunctive \ syll \ 7,5
\\ 9 \qquad \qquad W \land Z \qquad implication \ elim \ 3,8
\\ 10 \qquad \qquad Z \qquad conjunction \ elim \ 9
\\ 11 \qquad \qquad Z \land \lnot S \qquad conjunction \ intro \ 10,5
\\ 12 \qquad \qquad \lnot W \qquad implication \ elim \ 4,11
\\ 13 \qquad \qquad W \qquad conjunction \ elim \ 9
\\ 14 \qquad \qquad false \qquad negation \ elim \ 13,12
\\ 15 \qquad \lnot P \qquad negation \ intro \ 14
\\ 16 \qquad Q \qquad disjunctive \ syll \ 1,15
\\ 17 \quad \lnot S \rightarrow Q \qquad implication \ intro \ 5,16$
 A: We have to start from $P \lor Q$ and use $∨$elim [see here for the ND rules used].
The branch starting with $Q$ will be :
$1_Q)$ $Q$ --- assumed for $\lor$-elim


A) $\lnot S \to Q$ --- from 1) by $\to$-intro.


For the "branch" starting with $P$ we have :
$1_P)$ $P$ --- assumed for $\lor$-elim
2) $R \lor S$ --- from 1) and 2nd premise by $\to$-elim
3) $S$ --- assumed [a] from 2) for $\lor$-elim
4) $\lnot S$ --- assumed [b]
5) $\bot$ --- contradiction : from 3) and 4) by $\lnot$-elim
6) $Q$ --- from 5) by $\bot$-elim

7) $\lnot S \to Q$ --- from 4) and 6) by $\to$-intro, discharging [b].

Now for :
8) $R$ --- assumed [c] for $\lor$-elim
9) $W \land Z$ --- from 8) and 3rd premises by $\to$-elim
10) $W$ --- from 9) by $\land$-elim
11) $Z$ --- from 9) by $\land$-elim
12) $\lnot S$ --- assumed [d]
13) $Z \land \lnot S$ --- from 11) and 12) by $\land$-intro
14) $\lnot W$ --- from 13) and 4th premise by $\to$-elim
15) $\bot$ --- contradiction : from 10) and 14) by $\lnot$-elim
16) $Q$ --- from 15) by $\bot$-elim
17) $\lnot S \to Q$ --- from 12 and 16) by $\to$-intro, discharging [d]


B) $\lnot S \to Q$ --- from 2), 3)-7) and 8)-9) by $\lor$-elim, discharging [a] and [c].


Now from A), B) and the 1st premise: $P \lor Q$, we conclude by $\lor$-elim with:

$\lnot S \to Q$,

discharging assumptions $1_Q)$ and $1_P)$.
A: The question asks us to prove $\lnot S \to Q$ from the following hypothesis:


*

*$P \lor Q$

*$P \to R \lor S$

*$R \to W \land Z$

*$Z \land \lnot S \to \lnot W$


Let's think about how this could go wrong. It is claimed that if $S$ is false and all four hypotheses hold, then $Q$ is true. So, for a contradiction, assume $S$ is false and $Q$ is false. 
If $Q$ is false, then $P$ is true by item 1. Then $R \lor S$ is true by item $2$. We have assumed $S$ is false, so $R$ must be true.
Now, by item 3, $W$ and $Z$ must be true. Finally, by item 4, because $Z$ is true and $S$ is false, $W$ must be false. This is a contradiction, which is what we wanted.
That argument can be turned into a proof by natural deduction. However, instead of assuming $P$ at line 5 of your deduction above, you want to assume $\lnot S$. This is because you are trying to prove an implication whose hypothesis is $\lnot S$, and so you want to put yourself in a position to use implication introduction to do that. Assuming $P$ will not help as directly.
Then, at the next step, you can assume $\lnot Q$. Eventually, you will be able to derive a contradiction, which will let you deduce $Q$. 
So the outline looks like this:


*

*Hypothesis 1

*Hypothesis 2

*Hypothesis 3

*Hypothesis 4

*$\lnot S$ (assumption)
5.1. $\lnot Q$ (assumption)
5.2. ...
5.3. ...
5.4. contradiction
5.5. $Q$ (contradiction rule, discharge 5.1)

*$\lnot S \to Q$ (implication introduction, discharge 5)

To handle the implication $P\lor Q, \not Q \vdash P$, you can use something like this.


*

*$P \lor Q$  - assumption

*$Q$  - assumption

*$\lnot Q$ - assume for $\lor$ elimination from 1
3.1 Contradiction with 2
3.2 Conclude $P$, discharge 3

*$P$ - assume for $\lor$ elimination from 1
4.1 Conclude $Q$ from 2, discharge 4

*$Q$ - $\lor$ elimination from 1, 3, 4
A: $$\begin{array} {l|l}
\lnot S & \text{New Assumption} \\
\quad P & \text{New Assumption}  \\
\quad \quad \vdots \\
\quad \quad \bot \\
\quad \lnot P & \lnot \text{Intro} \\
\quad P \lor Q & \text{Given} \\
\quad \vdots \\
\quad Q & \lor\text{Elim} \\
\lnot S \implies Q & \implies \text{Intro} \\
\end{array}$$
Can you fill in the $\vdots$ ?
A: After an edit the OP asks the following:

Using the suggestions in the answers provided and studying the rules some more, I have come up with this solution. Can you tell me whether it is acceptable?

One way to tell if a proof is likely acceptable is to put it in a proof checker. Since the OP is using a similar Fitch-style natural deduction system as presented by forallx, the proof checker associated with that text could be used to see if the proof is acceptable.  Here is the result:


Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
