Prove $\beta \vdash \alpha \rightarrow \beta$ using Cut, DT, IE, RAA and PIP

Question

How am I proving this from Machover's text using Cut, DT, IE, RAA, and PIP. Does it involve using (i) $\lnot\alpha \vdash \alpha \rightarrow \beta$? For (i) I assumed $\lnot\alpha and \alpha$ then using IE to introduce $\beta$ and DT to finish off but here I don't know how to reach cleaner end form I've tried assuming $\lnot\alpha$ and referring to (i) to get the conditional but how do I get rid of $\lnot\alpha$ to end in a form I can use DT on $\beta$

CUT: $\alpha \vdash \beta$ and $\beta \vdash \phi$ then $\alpha \vdash \phi$

DT: $\phi, \alpha \vdash \beta$ then $\phi \vdash \alpha \rightarrow \beta$

IE: $\Gamma \vdash$ then $\Gamma \vdash \beta$

RAA: $\Gamma, \alpha \vdash$ then $\Gamma \vdash \lnot\alpha$

PIP: $\Gamma, \lnot\alpha\vdash$ then $\Gamma\vdash\alpha$

Answer 1

The proof presupposes a deducibility relation [see page 117] for which the basic properties hold.

In addition, we assume to have the Deduction Theorem available (as well as some additional rules).

See Problem 7.4 [page 124] :

Let $\vdash^*$ be the deducibility relation in a calculus that has modus ponens as a - not necessarily sole - rule of inference.

Show that if $\text {Cut}$ and $\text {DT}$ hold for $\vdash^*$, then $\vdash^* \alpha \to (\beta \to \alpha)$ [...].

Thus :

$\alpha, \beta \vdash^* \alpha$;

$\beta \vdash^* \alpha \to \beta$ --- from 1) by $\text {DT}$.

With a further application of $\text {DT}$ we will have $\vdash^* \beta \to (\alpha \to \beta)$, i.e. Axiom scheme i of $\vdash_0$ system [ref. page 117].

Finally consider Problem 8.18 [page 128], where system $\vdash^*$ has modus ponens as well as $\text {Cut}, \text {DT}, \text {IE}$ and $\text {PIP}$. The problem asks to derive Peirce's law, that amounts to conclude the proof that the system $\vdash^*$ is complete for classical logic.