Help proving $\lnot$(G $\rightarrow$ $\lnot$A) $\vdash$ G.

Question

I'm have been working on a natural deduction assignment for a couple of days and I went yesterday to ask my teacher for help but he gave me no helpful information so I'm asking here. I have tried a couple different methods but each one has just ended up in a loop of opening indirect proofs to discharge the assumption of the previous proof.

I'm trying to prove $\lnot$(G $\rightarrow$ $\lnot$A) $\vdash$ G.

The only path I haven't tried yet is using a material conditional equivalency but I don't think ($\lnot$G $\rightarrow$ A) $\equiv$ (G $\lor$ A) is the same as (G $\rightarrow$ $\lnot$A) $\equiv$ ($\lnot$G $\lor$ A).

At this point any advice that can help me find a path through this proof or a hint in the right direction would be amazing.

Answer 1

With Natural Deduction.

1. $\lnot (G \to \lnot A)$ --- premise

2. $\lnot G$ --- assumed [a]

3. $G$ --- assumed [b]

4. $\lnot A$ --- from 2) and 3) using rules for negation

5. $(G \to \lnot A)$ --- from 3) and 4) by $(\to \text I)$ discharging [b]

6. $\lnot \lnot G$ --- from 1) and 5) using rules for negation and discharging [a]

7. $G$ --- from 6) by Double Negation.

Conclusion: we have proved $\lnot (G \to \lnot A) \vdash G$.

Answer 2

This (with P instead of $G$ and Q instead of $\neg A$) is an exercise in my Intro to Formal Logic. I'm pasting below the answer (or rather, one of the answers) from the on-line answers.

You might find the relevant chapters of the book and the on-line answers to exercises useful back-up to your work on natural deduction. Everything is freely downloadable from https://www.logicmatters.net/ifl.