CHeck my proof of how I arrived at my answer Show proof for: 
(P --> Q), P ^ (~ ~R & ~ ~ Q), (S --> ~ R), ~(P & Q) : (~S ^ ~Q)
is this correct? 


*

*(P --> Q)            Premise

*P ^ ( ~ ~R & ~ ~Q)           Premise

*(S --> ~R)           Premise

*~(P & Q)         Premise

*(P --> (P & Q))      1, Abs (Absorption)

*P ^ (R & Q)              2, DN (Double Negation )

*(~S ^ R)         3, Impl (Implication)

*(~P ^ ~Q)                4, DeM(De Morgan’s Theorem)

*~S              7, DS (Disjunctive Syllogism)

*~Q              8, DS(Disjunctive Syllogism)

*(~S ^ ~Q)           9, 10 Add (Addition)

 A: Steps (7), (9), and (10) are wrong. 


*

*Step 7 problem: $(S \rightarrow \lnot R)$ turns into $\color{blue}{(\lnot S \lor \lnot R)}$, not $\color{red}{(\lnot S \lor R)}$

*Step 9 problem: $(\lnot S \lor R)$ alone will not give you $\lnot S$ by disjunctive syllogism

*Step 10 problem: see the problem for step 9
Here's one way of proving what you want to prove (in a Fitch-style system).
 1 $~~P \rightarrow Q~~~~~~~~~~~~~~~~~~~~~~~$ hyp
 2 $~~P \lor (\lnot\lnot R \land \lnot\lnot Q)~~~$ hyp
 3 $~~S \rightarrow \lnot R~~~~~~~~~~~~~~~~~~~~~$ hyp
 4 $~~\lnot(P \land Q)~~~~~~~~~~~~~~~~~~~$ hyp
 5 $~~~~~|~~ P~~~~~~~~~~~~~~~~~~~~~~~~~~~$ hyp
   $~~~~~|~~ ------$
 6 $~~~~~|~~ Q~~~~~~~~~~~~~~~~~~~~~~~~~~~$ modus ponens 1, 5
 7 $~~~~~|~~ P \land Q~~~~~~~~~~~~~~~~~~~$ $\land$-introduction 6, 5
 8 $~~~~~|~~ \bot~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $\bot$-introduction 7, 4
 9 $~~~~~|~~ \lnot S \lor \lnot Q~~~~~~~~~~~~~~$ $\bot$-elimination 8 
10 $~~~~~|~~ \lnot\lnot R \land \lnot\lnot Q~~~~~~~~$ hyp
   $~~~~~|~~ ------$
11 $~~~~~|~~ \lnot\lnot R~~~~~~~~~~~~~~~~~~~~~~$ $\land$-elimination 10
12 $~~~~~|~~ R~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $\lnot\lnot$-elimination 11
13 $~~~~~|~~ \lnot S~~~~~~~~~~~~~~~~~~~~~~~~$ modus tollens 12, 3
14 $~~~~~|~~ \lnot S \lor \lnot Q~~~~~~~~~~~~~~$ $\lor$-introduction 13
15 $~~\lnot S \lor \lnot Q~~~~~~~~~~~~~~~~~~~~$ $\lor$-elimination 10-14, 5-9, 2
