How can I show that $ \{A \to B, B \to C \} \vdash A \to C $ without using deduction theorem in hilbert-style system??

Question

I've been confused with that for a few days. The axiomatic system is:

1. α→(β→α)
2. α→(β→γ)→((α→β)→(α→γ))
3. (¬β→¬α)→(α→β)

Answer 1

Here is the sequence:

1. A->B [assumption 1]
2. B->C [assumption 2]
3. (B->C)->(A->(B->C)) [instance of axiom 1]
4. A->(B->C) [step 2 and step 3, modus ponens]
5. (A->(B->C))->((A->B)->(A->C)) [instance of axiom 2]
6. (A->B)->(A->C) [step 4 and step 5, modus ponens]
7. A->C [step 1 and step 6, modus ponens]