How to prove ¬p ∧ ¬q ⊢ ¬(p ∨ q) by natural deduction?

Question

Here's my attempt, but I think it's incorrect because I don't discharge assumption 1:

1. ¬p ∧ ¬q $\qquad$ premise

2. ¬p $\qquad$ by (∧E)

3. ¬q $\qquad$ by (∧E)

4. p $\qquad$ assumption¹

5. p ∨ q $\qquad$ assumption²

6. ⊥ $\qquad$ by (⊥I) on 2 and 4

7. ¬(p ∨ q) $\qquad$ by (→I)²

Answer 1

Your last step as you figured is wrong "$\to$ Intro" twice gets you $p\to \neg(p\lor q)$ but that is not what you want to prove.

I suggest proving $p\to \bot$, proving $q\to \bot$, and then assuming $p\lor q$, you use "or elim" to obtain $\neg(p\lor q)$.