Prove (p → ¬q) is equivalent to ¬(p ∧ q) I need to prove the above sequent using natural deduction. I did the first half already i.e. I proved $(p\rightarrow\neg q)\rightarrow \neg (p \wedge q)$, but I'm stuck on where to start for the reverse i.e. proving $\neg (p \wedge q) \rightarrow (p\rightarrow\neg q)$. I figured I would start by assuming $\neg (p \rightarrow \neg q)$ and then working towards a contradiction, but I'm still at a dead end. Can someone point me in the right direction? Thanks.
Can we use that $a \to b$ is same as $\neg a \lor b$? If yes then:
$$\neg (p \wedge q) \rightarrow (p\rightarrow\neg q) $$
is same as
$$(\neg p \lor \neg p) \rightarrow (\neg p \lor \neg q)
$$