Need Hints Prove "$((\neg \alpha \to \alpha) \to \alpha) $" Using Axiom 1,2,3 and MP and deduction theorem

$((\neg \alpha \to \alpha) \to \alpha) $

Hi, I am trying to prove this.

Can someone gives me some hints to start the question... My friend told me I might need to use deduction theorem here, but I have no clue to approach it..

Axiom 1: A→(B→A).

Axiom 2: (A→(B→C))→((A→B)→(A→C)).

Axiom 3: (¬B→¬A)→((¬B→A)→B).

To clarify: A, B, C, α, and β are propositions (i.e. assigned True or False). → and ¬ have the standard logical meanings.

here's a proof without using the deduction theorem. I will use the Mendelson Axioms, which I think are the basic ones (or at least they are taught on a first course of logic). Hope it helps.

$1.- ¬a \implies ((¬a \implies ¬a) \implies ¬a )$ (Ax1)

$2.- (¬a \implies ((¬a \implies ¬a) \implies ¬a )) \implies ((¬a \implies (¬a \implies ¬a)) \implies (¬a \implies ¬a ))$ (Ax2)

$3.- (¬a \implies (¬a \implies ¬a)) \implies (¬a \implies ¬a ) $ (MP (1,2))

$4.- ¬a \implies (¬a \implies ¬a )$ (Ax1)

$5.- ¬a \implies ¬a $ (MP(3,4))

$6.- ( ¬a \implies ¬a) \implies (( ¬a \implies a) \implies a)$ (Ax3)

$7.- (¬a \implies a) \implies a $ (MP(5,6))

I don't know what Axioms 1, 2, and 3 are. They are not standardized to that extent. But the following may help:

1. Suppose $\neg\alpha \implies \alpha$.

2. Suppose to the contrary that $\neg\alpha$.

3. $\alpha$ (from 1 and 2).

4. $\alpha \land \neg\alpha$ (from 3 and 2)

5. $\neg\neg\alpha$ (from 2 and 4)

6. $\alpha$ (from 5)

7. $[\neg\alpha \implies \alpha]\implies \alpha$ (from 1 and 6)