A proof in the Hilbert-style axiom system

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PWiz
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Homework Statement


Provide a complete formal proof that ## \vdash ((A \rightarrow B) \rightarrow C)
\rightarrow (B \rightarrow C)##.

Homework Equations


I am only allowed to use modus ponens and these four 'sentential logic' axioms:
A1 ## \neg \alpha \rightarrow (\alpha \rightarrow \beta)##
A2 ##\beta \rightarrow (\alpha \rightarrow \beta)##
A3 ##(\alpha \rightarrow \beta) \rightarrow ((\neg \alpha \rightarrow \beta) \rightarrow \beta)##
A4 ##(\alpha \rightarrow (\beta \rightarrow \gamma )) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \gamma ))##

The Attempt at a Solution


I have no idea where to begin. I'm thinking about using axiom 2, but I don't know how I would proceed from there. The problem would become easy if I was allowed to 'add an antecedent', but I am not allowed to directly do that. Any help is appreciated. Please note that I am a new student to logic, and I have only studied zero-order logic until now, so kindly provide hints that I will be able to understand. (The homework is due tomorrow!)
 
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These kind of proofs are much easier if you have the deduction theorem. What the deduction theorem says is that:

If you can prove [itex]Y[/itex] using [itex]X[/itex] as an axiom, then you can prove [itex]X \rightarrow Y[/itex]​

You don't have the deduction theorem, but you can use it as a strategy for finding the proof, as follows:

A proof of [itex]S[/itex] is a sequence of statements [itex]S_1, S_2, ..., S_n[/itex] such that each statement is either an axiom, or follows from two previous statements by modus ponens, and such that the last statement is [itex]S[/itex]. So:
  1. See if you can construct a proof of [itex]C[/itex] where the first statement is [itex](A \rightarrow B) \rightarrow C[/itex] and the second statement is [itex]B[/itex]. So this original proof will have [itex]C[/itex] as its last sentence.
  2. Now, modify your proof as follows: Starting with the third statement, replace each statement [itex]S[/itex] by the modified statement [itex]B \rightarrow S[/itex]. Now, see if you can add additional steps so that you prove each modified statement without using [itex]B[/itex] as an axiom. After these modifications, you will have a proof where [itex]B \rightarrow C[/itex] is the last statement.
  3. Now, modify your proof again: Starting with the new third statement, replace each statement [itex]S[/itex] by the modified statement [itex]((A \rightarrow B) \rightarrow C) \rightarrow S[/itex]. Now, see if you can add additional steps so that you can prove each modified statement without using [itex](A \rightarrow B) \rightarrow C[/itex] as an axiom. After these modifications, you will have a proof where [itex]((A \rightarrow B) \rightarrow C) \rightarrow (B \rightarrow C)[/itex] is the last statement.
  4. Now, get rid of the first two statements, since you are no longer using them. Now you have a proof of [itex]((A \rightarrow B) \rightarrow C) \rightarrow (B \rightarrow C)[/itex]
 
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