Is this a valid propositional natural deduction proof?

Question

I'm new to logic and I tried to solve an exercise. Since there isn't a given answer, I'd appreciate an indication of whether this is correct

    Prove that p | q , !p |- q 

1   p|q         premise
2   !p          premise
-----------------------
3   !q          assumption
4   p|q         copy 1
    -------------------
5   p           assumption
6   FALSUM      !e 2,5
    -------------------
    -------------------
7   q           assumption
8   FALSUM      !e 3,8 
    -------------------
9   !(p|q)      !e 5-8
10  FALSUM      !e 1,9
-----------------------
11  q           !e 3-10

Answer 1

The proof is not correct.

You have two premises: 1) and 2), and three temporary assumptions:

3) $\lnot q$

5) $p$

7) $q$;

in 9) you have derived $\lnot (p \lor q)$ by $\lnot$-intro, discharging 1).

The final step 11) derives $q$ by RAA (or Double Negation) from 3) and 10) and discharges 3).

In conclusion, the proof is:

$\lnot p, p, q \vdash q$.