With exams just around the corner I am really struggling with formal proofs.
Using natural deduction rules give a formal proof of Q from the premises
(¬P → Q)∧(R → ¬Q)
¬Q → ¬S
¬S → (R ∧ ¬P)
So the strategy I have been trying is to get to a point where I can prove ¬P → Q, but for that to work I first need to prove ¬P but I'm afraid that is where it all falls apart, I have no idea where to start to prove ¬P
This is the final answer I came up with thanks to help from @shane
1. (¬P → Q)∧(R → ¬Q)
2. ¬Q → ¬S
3. ¬S → (R ∧ ¬P)
4. (¬P → Q) ∧elim: 1
| 5. ¬Q
| 6. ¬S →elim: 2,5
| 7. R ∧ ¬P →elim: 6,3
| 8. ¬P ∧elim: 7
| 9. Q →elim: 8,4
| 10. ⊥ ⊥intro: 5,9
11. ¬¬Q ¬intro: 5-10
12. Q ¬elim 11