I'm able to "manufacture" a quite long proof, with ref to :
- Elliott Mendelson, Introduction to mathematical logic (4th ed - 1997); see page 35 for the axioms [those that you are requested to use] and the only rule of inference : Modus Ponens.
With Ax 1 and Ax 2 we can prove [see Mendelson, Lemma 1.8, page 36] : $\vdash \alpha \rightarrow \alpha$.
(1) $\alpha → ((\alpha \rightarrow \alpha) → \alpha)$ --- from Ax 1
(2) $(\alpha → ((\alpha \rightarrow \alpha) → \alpha)) → ((\alpha → (\alpha \rightarrow \alpha)) → (\alpha \rightarrow \alpha))$ --- from Ax 2
(3) $(\alpha → (\alpha \rightarrow \alpha)) → (\alpha \rightarrow \alpha)$ --- from (1) and (2) by MP
(4) $\alpha → (\alpha \rightarrow \alpha)$ --- from Ax 1
(5) $\alpha \rightarrow \alpha$ --- from (3) and (4) by MP
With it, Ax 1 and Ax 2, we can prove the Deduction Theorem [see Mendelson, page 37].
In addition, with Ax 1 and Ax 2, we can prove Hypothetical Syllogism : $\alpha \rightarrow \beta, \beta \rightarrow \gamma \vdash \alpha \rightarrow \gamma$; in this post we can find a proof of it.
We need two additional Lemmas :
Lemma 1 [see Mendelson, Lemma 1.11(a), page 38] : $\vdash \lnot \lnot \beta \rightarrow \beta$
(1) $\vdash (\lnot \beta \rightarrow \lnot \lnot \beta) \rightarrow ((\lnot \beta \rightarrow \lnot \beta) \rightarrow \beta)$ from Ax 3 with $\lnot \beta$ in place of $\alpha$
(2) $\lnot \beta \rightarrow \lnot \lnot \beta$ --- assumed [a]
(3) $\vdash \lnot \beta \rightarrow \lnot \beta$ --- from $\vdash \alpha \rightarrow \alpha$ : see above
(4) $\beta$ --- from (1), (2) and (3) by MP twice
(5) $\vdash (\lnot \beta \rightarrow \lnot \lnot \beta) \rightarrow \beta$ --- from (2) and (4) by Deduction Th, discharging [a]
(6) $\vdash \lnot \lnot \beta \rightarrow (\lnot \beta \rightarrow \lnot \lnot \beta)$ --- Ax 1
(7) $\vdash \lnot \lnot \beta \rightarrow \beta$ --- from (5) and (6) by HS
Lemma 2 [see Mendelson, Lemma 1.11(e), page 38] : $\alpha \rightarrow \beta \vdash \lnot \beta \rightarrow \lnot \alpha$
(1) $\alpha \rightarrow \beta$ --- assumed [a]
(2) $\lnot \beta$ --- assumed [b]
(3) $\vdash \lnot \beta \rightarrow (\lnot \lnot \alpha \rightarrow \lnot \beta)$ --- Ax 1
(4) $\lnot \lnot \alpha \rightarrow \lnot \beta$ --- from (2) and (3) by MP
(5) $\vdash \lnot \lnot \alpha \rightarrow \alpha$ --- Lemma 1
(6) $\lnot \lnot \alpha \rightarrow \beta$ --- from (1) and (5) by HS
(7) $\vdash (\lnot \lnot \alpha \rightarrow \lnot \beta) \rightarrow ((\lnot \lnot \alpha \rightarrow \beta) \rightarrow \lnot \alpha)$ from Ax 3 with $\lnot \alpha$ in place of $\beta$
(8) $\lnot \alpha$ --- from (4), (6) and (7) by MP twice
(9) $\lnot \beta \rightarrow \lnot \alpha$ --- from (2) and (8) by DT, discharging [b]
Now for the proof of : $\vdash (\alpha \rightarrow \lnot \beta) \rightarrow ((\alpha \rightarrow \beta) \rightarrow \lnot \alpha)$
(1) $\alpha \rightarrow \lnot \beta$ --- assumed [a]
(2) $\alpha \rightarrow \beta$ --- assumed [b]
(3) $\lnot \beta \rightarrow \lnot \alpha$ --- from (2) by Lemma 2
(4) $\alpha \rightarrow \lnot \alpha$ --- from (1) and (3) by HS
(5) $\lnot \lnot \alpha \rightarrow \lnot \alpha$ --- from (4) by Lemma 2
(6) $ \vdash (\lnot \lnot \alpha → \lnot \lnot \alpha) → ((\lnot \lnot \alpha \rightarrow \lnot \alpha) → \lnot \alpha)$ --- from Ax 3
(7) $\vdash \lnot \lnot \alpha \rightarrow \lnot \lnot \alpha$ --- from $\vdash \alpha \rightarrow \alpha$
(8) $\lnot \alpha$ --- from (5), (6) and (7) by MP twice
(9) $\vdash (\alpha \rightarrow \lnot \beta) \rightarrow ((\alpha \rightarrow \beta) \rightarrow \lnot \alpha)$ --- from (1), (2) and (8), by two applications of DT, discharging assumptions [a] and [b].
Comment
It can be "istructive" to compare the above proof (I hope it can be simplified) with the Natural Deduction 9-lines proof.
So you want to prove the following theorem:
Theorem: If $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then $\Gamma \vdash \neg \phi$
Proof:
First, I'll assume that you can use the Deduction Theorem, which states that for any $\Gamma$, $\varphi$, and $\psi$:
If $\Gamma \cup \{ \varphi \} \vdash \psi$, then $\Gamma \vdash \varphi \rightarrow \psi$
So if $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then by the Deduction Theorem we have $\Gamma \vdash \phi \to \psi$ and $\Gamma \vdash \phi \to \neg \psi$
This means that if can show that $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$, then we're there.
This is not easy, but here goes:
First, let's prove: $\phi \to \psi, \psi \to \chi, \phi \vdash \chi$:
\begin{array}{lll}
1&\phi \to \psi & Premise\\
2& \psi \to \chi & Premise\\
3&\phi& Premise\\
4&\psi& MP \ 1,3\\
5&\chi& MP \ 2,4\\
\end{array}
By the Deduction Theorem, this gives us Hypothetical Syllogism (HS): $\phi \to \psi, \psi \to \chi \vdash \phi \to \chi$
Now let's prove the general principle that $\neg \phi \vdash (\phi \to \psi)$:
\begin{array}{lll}
1. &\neg \phi& Premise\\
2. &\neg \phi \to (\neg \psi \to \neg \phi)& Axiom \ 1\\
3. &\neg \psi \to \neg \phi& MP \ 1,2\\
4. &(\neg \psi \to \neg \phi) \to (\phi \to \psi)& Axiom \ 3\\
5. &\phi \to \psi& MP \ 3,4\\
\end{array}
With the Deduction Theorem, this means $\vdash \neg \phi \to (\phi \to \psi)$ (Duns Scotus Law)
Let's use Duns Scotus to show that $\neg \phi \to \phi \vdash \phi$
\begin{array}{lll}
1. &\neg \phi \to \phi& Premise\\
2. &\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))& Duns \ Scotus\\
3. &(\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))) \to ((\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi)))& Axiom \ 2\\
4. &(\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi))& MP \ 2,3\\
5. &\neg \phi \to \neg (\neg \phi \to \phi)& MP \ 1,4\\
6. &(\neg \phi \to \neg (\neg \phi \to \phi)) \to ((\neg \phi \to \phi) \to \phi)& Axiom \ 3\\
7. &(\neg \phi \to \phi) \to \phi& MP \ 5,6\\
8. &\phi& MP \ 1,7\\
\end{array}
By the Deduction Theorem, this means $\vdash (\neg \phi \to \phi) \to \phi$ (Law of Clavius)
Using Duns Scotus and the Law of Clavius, we can now show that $ \neg \neg \phi \vdash \phi$:
\begin{array}{lll}
1. &\neg \neg \phi& Premise\\
2. &\neg \neg \phi \to (\neg \phi \to \phi)& Duns \ Scotus\\
3. &\neg \phi \to \phi& MP \ 1,2\\
4. &(\neg \phi \to \phi) \to \phi& Clavius\\
5. &\phi& MP \ 3,4\\
\end{array}
By the Deduction Theorem, this also means that $\vdash \neg \neg \phi \to \phi$ (DN Elim or DNE)
Finally, we can show the desired $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$:
\begin{array}{lll}
1. &\phi \to \psi& Premise\\
2. &\phi \to \neg \psi& Premise\\
3. &\neg \neg \phi \to \phi& DNE\\
4. &\neg \neg \phi \to \psi& HS \ 1,3\\
5. &\neg \neg \phi \to \neg \psi& HS \ 2,3\\
6. &(\neg \neg \phi \to \neg \psi) \to (\psi \to \neg \phi)& Axiom \ 3\\
7. &\psi \to \neg \phi& MP \ 5,6\\
8. &\neg \neg \phi \to \neg \phi& HS \ 4,7\\
9. &(\neg \neg \phi \to \neg \phi) \to \neg \phi& Clavius\\
10. &\neg \phi& MP \ 8,9\\
\end{array}
Now, you can actually get a little more quickly to $\neg \neg \phi \vdash \phi$ as follows:
\begin{array}{lll}
1&\neg \neg \phi&Premise\\
2&\neg \neg \phi \to (\neg \neg \neg \neg \phi \to \neg \neg \phi)&Axiom \ 1\\
3&\neg \neg \neg \neg \phi \to \neg \neg \phi&MP \ 1,2\\
4&(\neg \neg \neg \neg \phi \to \neg \neg \phi) \to (\neg \phi \to \neg \neg \neg \phi) & Axiom \ 3\\
5& \neg \phi \to \neg \neg \neg \phi & MP \ 3,4\\
6&(\neg \phi \to \neg \neg \neg \phi) \to (\neg \neg \phi \to \phi) & Axiom \ 3\\
7& \neg \neg \phi \to \phi & MP \ 5,6\\
8&\phi&MP \ 1,7\\
\end{array}
However, since the proof of $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$ relies on Clavius, I took the road that I did.
Best Answer
Assume $(\alpha \to (\beta \to \gamma))$ ... (1)
Thus, by Deduction Theorem on (2), (5): $(\alpha \to \beta) \to (\alpha \to \gamma)$ ... (6)
And finally by one more Deduction Theorem on (1), (6): $(\alpha \to (\beta \to \gamma)) \to ((\alpha \to \beta) \to (\alpha \to \gamma))$