In Lemmon's Beginning Logic, page 41 exercise 1g says to prove that from $\neg (P \cap Q)$ then $\neg P \cup \neg Q$ follows.
1 (1) $\neg (P \cap Q)$ (A)
2 (2) $\neg (\neg P \cup \neg Q)$ (A)
3 (3) $\neg P$ (A)
3 (4) $\neg P \cup \neg Q$ (3 vI)
2,3 (5) $(\neg P \cup \neg Q) \cap \neg(\neg P \cup \neg Q)$ (2,4 &I)
3 (6) $\neg \neg (\neg P \cup \neg Q)$ (2,5 RAA)
3 (7) $\neg P \cup \neg Q$ (6 DN)
This is my first attempt, but line 7 needs to rely on line 1 which hasn't been used in the proof. Any hints?