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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?

Patrick Clot
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  • Little wonder that line 7 doesn;t depend on line 1, since it immediately follows from the assumption made on line 3, which is never being discharged. So effectively, all you have shown is that $\neg P$ implies $\neg P \lor \neg Q$ – Bram28 Mar 25 '24 at 21:45
  • After the contradiction in step (5) you have to derive $P$ by DN (discharging Assupmtion 3); then you have to assume $\lnot Q$ and repeat, deriving $Q$ by DN again. Now you have $P$ and $Q$ and thus $P \land Q$, that contradicts (1) and now you can use DN again to discharge (2). – Mauro ALLEGRANZA Mar 26 '24 at 07:52

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