How do I write a proof for: $\neg p\vdash p\to q$

Question

I haven't seen this question asked here, but I stumbled upon this question while practicing. I couldn't/can't figure it out and it's bugging me, even though it's not mandatory for any exercises, I would still like to know the answer.

The question is, using Fitch-Style logic, prove: $\neg p\vdash p\to q$

I've figured out that this has to do with $\neg p \lor q \equiv p \to q$, but using Fitch-Style logic I'm pretty sure I can't just do the following:

1. $\neg p$
2. $\neg p \lor q \space (\lor Intro \space 1) $
3. $p \to q \space (using \neg p \lor q \equiv p \to q)$

So I figured out that I can do:

1. $\neg p$
2. $\neg p \lor q \space (\lor Intro \space 1) $

From there on I guess I can free $q$, but from there on I am clueless and I've been clueless for almost the entire day. Hopefully someone can help.

Answer 1

Your proof would be correct in some systems, but for the openlogic system, you'll need to use the contradiction: