How to prove ⊢ (B → ¬B) → ¬B using an informal deduction

Question

So I need to prove ⊢ (B → ¬B) → ¬B

I can use the Modus ponus (MP) rule, and deduction theorem (DT). And I have these 3 axioms:

1. α → (β → α) --- (A1)

2. (α → (β → γ)) → ((α → β) → (α → γ)) --- (A2)

3. (¬β → ¬α) → ((¬β → α) → β) --- (A3)

The first step I did was to use DT to simplify what I need to proof to (B → ¬B) ⊢ ¬B. And then I'm stuck.

Does anyone know how to go on to solve the formal proof? Or any hints on how to start would be greatly appreciated! Also, as a general rule, how do you know which axioms to use to solve the proof? Is there any particular pattern that I should look out for? If not, it seems very trial and error to me and I always get stuck at the start of the proof.

Thank you so much!

Answer 1

Note that the third axiom says that if the negation of a proposition implies some other proposition and it's negation, then the original proposition holds. Note also that the negation of $\lnot$B is $\lnot$$\lnot$B.

So, assume $\lnot$$\lnot$B. Deduce B (hint assume $\lnot$B next). And I hope you can take it from there.

Answer 2

First:

$1. \ B \ Assume$

$2. \ B \to (B \to B) \ Ax. 1$

$3. \ B \to B \ MP \ 1,2$

$4. B \ MP \ 1,3$

So: $B \vdash B$

By DT: $\vdash B \to B$ (ID)

Next:

$1. \neg \neg B \ Assume$

$2. \neg \neg B \to (\neg B \to \neg \neg B) \ Ax. 1$

$3. \neg B \to \neg \neg B \ MP \ 1,2$

$4. (\neg B \to \neg \neg B) \to ((\neg B \to \neg B) \to B \ Ax.3$

$5. (\neg B \to \neg B) \to B \ MP \ 3,4$

$6. \neg B \to \neg B \ ID$

$7. B \ MP \ 5,6$

So: $\neg \neg B \vdash B$

So, by DT: $\vdash \neg \neg B \to B$ (DN)

Then:

$1. A \to B \ Assume$

$2. B \to C \ Assume$

$3. A \ Assume$

$4. B \ MP \ 1,3$

$5. C \ MP \ 2,4$

So: $A \to B, B \to C, A \vdash C$

So, by DT: $A \to B, B \to C \vdash A \to C$ (HS)

Finally:

$1. B \to \neg B \ Assume$

$2. (\neg \neg B \to \neg \neg B) \to ((\neg \neg B \to \neg B) \to \neg B) \ Ax.3$

$3. \neg \neg B \to \neg \neg B \ ID$

$4. (\neg \neg B \to \neg B) \to \neg B \ MP \ 2,3$

$5. \neg \neg B \to B \ DN$

$6. \neg \neg B \to \neg B \ HS \ 1,5$

$7. \neg B \ MP \ 4,6$

So: $B \to \neg B \vdash \neg B$

So, by DT: $ \vdash (B \to \neg B) \to \neg B$