Proof of double negation elimination using natural deduction

I am trying to prove that $$\neg A \to \bot \vdash A$$, I can only use double negation elimination to get the conclusion, but I don't know how to prove double negation elimination.

Below is my proof for $$\neg A \to \bot \vdash A$$:

1 $$\neg A\to\bot$$

2 $$\neg A$$ (assumption)

3 $$\bot$$ ($$\to$$e 1, 2)

4 $$\neg\neg A$$ ($$\neg$$i 2-3)

5 $$A$$ (DNE 4)

The rules that can be used would be the basic rules of natural deduction:

• What do you mean precisely when you say that you want to prove double negation elimination? What rules are you allowed to use? Jan 4, 2022 at 17:49
• $\wedge$e, i. $\vee$e, i. $\to$e, i. $\neg$e, i. $\bot$e, i. Jan 4, 2022 at 18:35
• What precisely are your rules $\lnot E$, $\lnot I$, $\bot E$, and $\bot I$? Not everyone uses the same names for the same rules. Please edit these into your question. Jan 4, 2022 at 22:06

The rules you list comprise a natural deduction system for intuitionistic logic. The sequent $$\lnot A\to \bot\vdash A$$ is not valid in intuitionistic logic, so it is not provable in your system.
To prove $$\lnot A\to \bot\vdash A$$, you need to be working with a natural deduction system for classical logic, which means adding an additional rule to your system. One common choice is to add double negation elimination itself; this makes your proof a valid one. Other common choices are adding reductio ad absurdum (proof by contradiction) or the law of excluded middle.