Proofs using theorems instead of axioms I'm not sure how to prove these basic theorems in propositional calculus. Instead of using the standard axioms, we're supposed to use:


*

*Deduction Theorem (if $\Phi, \alpha \vdash \beta$ then $\Phi \vdash \alpha \to \beta$), 

*Reductio (if $\Phi, \alpha \vdash \, $, then $\Phi \vdash \lnot \alpha$),

*Cut Rule (if $\Phi \vdash \alpha$ and $\Psi, \alpha \vdash \beta$ then $\Phi \cup \Psi \vdash \beta$), 

*Inconsistency Effect (if $\Phi \vdash \, $, then $\Phi \vdash \beta$ for every formula $\beta$), and 

*the Principle of Indirect Proof (if $\Phi, \lnot \alpha \vdash \, $, then $\Phi \vdash \alpha$), 


as all the axioms can be deduced using these theorems.
I don't really know how to start the proofs without using the axioms:
i) prove that $\lnot(\alpha \to \beta) ⊢ \alpha$
ii) prove that $\lnot\alpha \vdash \alpha \to \beta$
Any suggestions on how to start these proofs or any insight at all would be greatly appreciated!
Thanks!
 A: Now that we have the source of your problem, we can help you...
See: Moshe Machover, Set Theory, Logic and Their Limitations Cambridge UP (1996), page 116-on for the definitions and some results about propositional calculus.

Definition 8.1.
A set of two formulas $\{ \alpha,  \lnot \alpha \}$, one of which is the negation of the other, is called a contradictory pair.
A set $\Phi$ of formulas is said to be [propositionally] inconsistent - 
  in symbols: "$\Phi \vdash_0$" - if both members of some contradictory pair 
  are propositionally deducible from $\Phi$.

We have to note some results: Problem 8.12 [page 126]: $\alpha \vdash_0 \lnot \lnot \alpha$, for all $\alpha$, and Lemma 8.14: $\lnot \lnot \alpha \vdash_0 \alpha$, for all $\alpha$.
At this point of the book, the proof system regarding $\vdash_0$, based on $\lnot$ and $\to$ and the five axioms of page 117 plus modus ponens, has been enlarged with addiotnal (derived) rules:

Theorem 7.2: Deduction Theorem. 
Theorem 6.13: Cut Rule: If $\Phi \vdash_0 \delta_i$ for each $i = 1, 2,\ldots, k$ and $\Psi \cup \{ \delta_0, \ldots, \delta_k \} \vdash_0 \alpha$, 
  then $\Phi \cup \Psi \vdash_0 \alpha$. 
Inconsistency Effect: If $\Phi \vdash_0$, then $\Phi \vdash_0 \beta$, for every formula $\beta$.
Reductio: If $\Phi, \alpha\vdash_0$, then $\Phi \vdash_0 \lnot \alpha$.
Indirect proof: If $\Phi \lnot \alpha \vdash_0 $, then $\Phi \vdash_0 \alpha$. 


Now we have: Problem 8.19 [page 128]: 

Prove: (i) $\lnot \alpha \vdash_0 \alpha \to \beta$; (iv) $\lnot (\alpha \to \beta) \vdash_0 \alpha$.

We assume to use, in addition to modus ponens, also the derived rules above, as well as the already available results.
For (i):
1) $\vdash_0 \lnot \alpha \to (\alpha \to \beta)$ --- Axiom scheme iv
2) $\lnot \alpha$ --- premise
3) $\alpha \to \beta$ --- from 1) and 2) by mp.
According to Definition 6.8 [page 117], the above is a propositional deduction of $\alpha \to \beta$ from the set of formulas $\Phi= \{ \lnot \alpha \}$ and we can write (according to Definition 6.9): $\lnot \alpha \vdash_0 \alpha \to \beta$.

For (iv):
1) $\lnot (\alpha \to \beta)$ --- premise
2) $\lnot \alpha$ --- premise
3) $\alpha \to \beta$ --- from 2) and previous result (Problem 8.19 (i)).
Up to now we have:

$\lnot (\alpha \to \beta), \lnot \alpha \vdash_0 (\alpha \to \beta)$

and obviously:

$\lnot (\alpha \to \beta), \lnot \alpha \vdash_0 \lnot (\alpha \to \beta)$.

This means: $\lnot (\alpha \to \beta), \lnot \alpha \vdash_0$.
Finally, we apply Reductio to get:
4) $\lnot (\alpha \to \beta) \vdash_0 \lnot \lnot \alpha$,
followed by Lemma 8.14: $\lnot \lnot \alpha \vdash_0 \alpha$, to conclude:

$\lnot (\alpha \to \beta) \vdash_0 \alpha$.



Note. How to prove (i) with MP, DT and Inconsistency (without axioms)?
1) $\alpha$ --- premise
2) $\lnot \alpha$ --- premise
Form 1) and 2) we have: $\Phi= \{ \alpha, \lnot \alpha \} \vdash_0$.
Thus, we can apply Inconsistency to get:
3) $\lnot \alpha, \alpha \vdash^* \beta$,
concluding, by DT, with:

$\lnot \alpha \vdash^* \alpha \to \beta$.


Having proved $\lnot \alpha \vdash^* \alpha \to \beta$, we can use it in the proof of (iv) above (line 3)) to get:

$\lnot (\alpha \to \beta) \vdash^* \alpha$.

A: As correctly suggested by Derek Elkins in his comment, I strongly conjecture that your system should have an axiom rule of the form $\alpha \vdash \alpha$ (or $\Phi, \alpha \vdash \alpha$) for every formula $\alpha$. I do not know the exact formulation of the cut rule in your system, anyway it should be equivalent to the formulation of the cut rule I used in my derivations.
First, I answer your question (ii). The following is a derivation of $\lnot A \vdash A \to B$ in your system. 


*

*$\lnot A \vdash \lnot A$ -- axiom

*$A \vdash A$  -- axiom

*$\lnot A, A \vdash \ $  -- cut rule (modus ponens) of 1. and 2.

*$\lnot A, A \vdash B$  -- inconsistency effect (ex falso quodlibet) from 3.

*$\lnot A \vdash A \to B$  -- deduction theorem from 4.


Concerning your question (i), the following is a derivation of $\lnot(A \to B) \vdash A$ in your system. It uses my answer to your question (ii).


*

*$\lnot (A \to B) \vdash \lnot (A \to B)$ -- axiom. 

*$\lnot A \vdash A \to B$  -- see derivation above, question (ii)

*$\lnot (A \to B), \lnot A \vdash \ $ -- cut rule (modus ponens) of 1. and 2. 

*$\lnot (A \to B) \vdash A$ -- (principle of indirect proof (reductio ad absurdum) from 3.

