Prove using natural deduction for minimal logic that $\neg\neg\neg A \rightarrow \neg A$.
I'm trying to prove this argument using only the rules of minimal logic.
So far I have this.
$$1. ⊢ \neg\neg\neg A \rightarrow \neg A$$ $$2. \neg\neg\neg A ⊢ \neg A$$ $$3. \neg\neg\neg A, A ⊢ \bot$$ $$4.1 \neg\neg\neg A, A ⊢ A (DONE)$$ $$4.2 \neg\neg\neg A, A ⊢ \neg A $$
And from the branch 4.2 I don't know how to proceed.
Can someone help?
Until step $3$, your attempt goes in the right direction. Your step $4$ is problematic, because$-$as you noticed$-$in the branch $4.2$ it is not clear what to do. Indeed, your branch $4.2$ is very similar to your step $2$, which roughly means that from step $2$ to step $4$ your attempt did not achieve any progress.
The solution is to remark that your step $3$ (which expresses a contradiction between $\lnot\lnot\lnot A$ and $A$) can be obtained in several ways, not only via your step $4$. Your step $4$ derives step $3$ by putting together $A$ and $\lnot A$, but my suggestion is to derive step $3$ by putting together $\lnot \lnot \lnot A$ and $\lnot \lnot A$. The derivation you obtain is below (I draw it as a tree, built bottom up).
\begin{align} \dfrac{ \dfrac{ \dfrac {\dfrac{}{\lnot \lnot \lnot A, A \vdash \lnot \lnot \lnot A}\text{ax} \qquad \dfrac{\vdots}{\lnot \lnot \lnot A, A \vdash \lnot \lnot A}} {\lnot \lnot \lnot A, A \vdash \bot} \lnot_e } {\lnot \lnot \lnot A \vdash \lnot A} \lnot_i } {\vdash \lnot \lnot \lnot A \to \lnot A} \to_i \end{align}
It remains to find a derivation for $\lnot \lnot \lnot A, A \vdash \lnot \lnot A$ (the vertical dots above), which is quite easy:
\begin{align} \dfrac{ \dfrac{\dfrac{}{\lnot \lnot \lnot A, A, \lnot A \vdash \lnot A}\text{ax} \qquad \dfrac{}{\lnot \lnot \lnot A, A, \lnot A \vdash A}\text{ax}}{\lnot \lnot \lnot A, A, \lnot A \vdash \bot}\lnot_e } {\lnot \lnot \lnot A, A \vdash \lnot \lnot A} \lnot_i \end{align}
We have only used rules that are admissible in minimal logic.