What is the proof of reductio ad absurdum (RAA) in a Łukasiewicz axiom system for propositional logic with only modus ponens? I am working with the following Lukasiewicz axiom system:

Axiom Schema 1: $\alpha  \rightarrow (\beta \rightarrow \alpha  )$
Axiom Schema 2: $(\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \gamma)) $
Axiom Schema 3: $(\lnot \alpha \rightarrow \lnot \beta) \rightarrow (\beta \rightarrow \alpha)$
Rule of Inference (Modus ponens): From $\alpha$ and $\alpha \rightarrow \beta$, infer $\beta$.

I am looking for a proof of the following:

Reductio ad absurdum (RAA): If $\Gamma \cup \{ \alpha \}$ is inconsistent, then $\Gamma ⊢ \lnot \alpha$.

I have seen a proof of the first-order case, but unless I am mistaken it relies on the deduction theorem for first-order logic and the completeness theorem for propositional logic, so I am trying to find a proof of the propositional case which does not depend on the completeness theorem for propositional logic. The deduction theorem can (and I am assuming must?) be assumed.
Proofs with similar axiom systems (using only $\lnot$ and $\rightarrow$) are fine, but slightly less ideal.
References, suggestions, and edits are welcomed.
 A: 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.
