How can I solve -P ∧ -Q ⊢ -(P ∨ Q) using deMorgan's Law?

Question

Using propositional logic rules (--E, -I, ^I, ^E, vI, vE, ->I, ->E) how can I solve -P ∧ -Q ⊢ -(P ∨ Q)?

I don't know if I'm going in the correct direction. Would appreciate some help in solving this. Here is what I have gotten:

1(1) -P^-Q A

2(2) --(PvQ) A

1(3) -P 1 ^E

4(4) P A

5(5) PvQ A

4,5(6) P^(PvQ) 4,5 ^I

4,5(7) P 6 ^E

1,4,5(8) -P^P 3,7 ^I

1,4(9) -(PvQ) 5,8 -I

1,2,4(10) -(PvQ)^--(PvQ) 2,9 ^I

Answer 1

You have rules of $\sf\lnot\lnot E,\lnot I, \wedge I, \wedge E, \vee I, \vee E, {\to} I, {\to} E$.

Your premise is $\lnot P\land\lnot Q$, and you are aiming to derive $\lnot(P\vee Q)$.

To introduce that negation, you must assume $P\vee Q$, aiming to derive a contradiction.

Having assumed that disjunction, you should eliminate it. Proof by cases. Well, here's the tricky part. It would be easy to derive $P\land\lnot P$ in one case and $Q\land\lnot Q$ in the other, but you need the same contradiction.

So inside each case assume $P\vee Q$ again and negate that second assumption to conjunct with the initial assumption producing $(P\vee Q)\wedge\neg(P\vee Q)$ in each case.