Prove contrapositive rule from the propositional calculus logic

Question

It's been a while since I studied math, but I am looking back at a mathematical logic book, and found this problem I am currently unable to solve. I assume I am missing something obvious.

It's problem 1.47 (d) from An Introduction to Mathematical Logic by Mendelson.

Prove $\vdash_L (\neg C \Rightarrow \neg B) \Rightarrow (B \Rightarrow C)$

He gives the following axioms for the theory

(A1) $B \Rightarrow (C \Rightarrow B)$
(A2) $((B \Rightarrow (C \Rightarrow D)) \Rightarrow ((B \Rightarrow C) \Rightarrow (B \Rightarrow D)))$
(A3) $(((\neg C) \Rightarrow (\neg B)) \Rightarrow (((\neg C) \Rightarrow B) \Rightarrow C))$

It seems likely, that I want to start off with A3 since it has the premise, and applying (A2) gives me something looking pretty promising.

1. $(\neg C \Rightarrow \neg B) \Rightarrow ((\neg C \Rightarrow B) \Rightarrow C)$ (A3)
2. $[(\neg C \Rightarrow \neg B) \Rightarrow ((\neg C \Rightarrow B) \Rightarrow C)] \Rightarrow [((\neg C \Rightarrow \neg B) \Rightarrow (\neg C \Rightarrow B)) \Rightarrow ((\neg C \Rightarrow \neg B) \Rightarrow C))]$ (A2)
3. $((\neg C \Rightarrow \neg B) \Rightarrow (\neg C \Rightarrow B)) \Rightarrow ((\neg C \Rightarrow \neg B) \Rightarrow C))$ (Modus Ponens, 1, 2)
4. $C \Rightarrow (B \Rightarrow C)$ (A1)

The two things I see from here:

One is that if I could get something like $((\neg C \Rightarrow B) \Rightarrow C) \Rightarrow (B \Rightarrow C)$, then I could use transitivity of $\Rightarrow$ (proven in 1.47 (b)) along with statement (1). This statement is a tautology, so in theory should be provable, but I haven't been able to figure it out.

Two is that with (4) and the consequent of (3), I could also use transitivity to get my result. Although I am not sure what hack I need to employ to get there since that consequent is obviously not a tautology.

In general I know that $\neg C \Rightarrow \neg B \Leftrightarrow C \vee \neg B \Leftrightarrow \neg B \vee C \Leftrightarrow B \Rightarrow C$, which should imply the above. However, it seems the exercise here is to only use the axioms, statements preceding the $\vdash_L$ of which there are none, and Modus Ponus.

Anyone have any hints on this?

=== SOLUTION ===

I figured out the solution after someone showed me the problem was solved with the deduction theorem, and playing around with the proof of that theorem a bit.

Since I don't have a the deduction theorem, I use something basically just as good, which is the ability to "Export" a formula. That is, if we were to build a statement, $S_i$ with rule $\text{(Rule)}$, we could immediately apply $S_i \Rightarrow (R \Rightarrow S_i) \ \text{(A1)}$ for any formula $R$, and $\text{(MP)}$ to get $R \Rightarrow S_i$. We do this in one line, and mark it as $\text{(Rule,Export)}$.

Now for the proof.

Let $X := \neg C \Rightarrow \neg B$, and $Y := B \Rightarrow C$. We want to show $X \Rightarrow Y$.

Also let $Z := \neg C \Rightarrow B$, $A_1 := B \Rightarrow (Z \Rightarrow C)$, and $A_2 := ((B \Rightarrow Z) \Rightarrow Y)$.

\begin{align*} S_{1} &:= X \Rightarrow ((Z \Rightarrow C) \Rightarrow A_1) &\text{(A1, Export)} \\ S_{2} &:= S_{1} \Rightarrow ((X \Rightarrow (Z \Rightarrow C)) \Rightarrow (X \Rightarrow A_1)) &\text{(A2)} \\ S_{3} &:= (X \Rightarrow (Z \Rightarrow C)) \Rightarrow (X \Rightarrow A_1) &\text{(MP, $S_{1}$, $S_{2}$)} \\ S_{4} &:= X \Rightarrow (Z \Rightarrow C) &\quad\text{(A3)} \\ S_{5} &:= (X \Rightarrow (A_1 \Rightarrow A_2))\Rightarrow ((X \Rightarrow A_1) \Rightarrow (X \Rightarrow A_2)) &\text{(A2)} \\ S_{6} &:= X \Rightarrow (A_1 \Rightarrow A_2) &\quad\text{(A2, Export)} \\ S_{7} &:= (X \Rightarrow A_1) \Rightarrow (X \Rightarrow A_2) &\text{(MP, $S_{6}$, $S_{5}$)} \\ S_{8} &:= X \Rightarrow A_1 &\text{(MP, $S_{4}$, $S_{3}$)} \\ S_{9} &:= (X \Rightarrow A_2) \Rightarrow ((X \Rightarrow (B \Rightarrow Z)) \Rightarrow (X \Rightarrow Y)) &\text{(A2)} \\ S_{10} &:= X \Rightarrow A_2 &\text{(MP, $S_{8}$, $S_{7}$)} \\ S_{11} &:= (X \Rightarrow (B \Rightarrow Z)) \Rightarrow (X \Rightarrow Y) &\text{(MP, $S_{10}$, $S_{9}$)} \\ S_{12} &:= X \Rightarrow (B \Rightarrow Z) &\text{(A1,Export)} \\ S_{13} &:= X \Rightarrow Y &\text{(MP, $S_{12}$, $S_{11}$)} \\ \end{align*}