prove $(A \rightarrow B) \rightarrow (\neg B \rightarrow \neg A)$ in Hilbert System I'm looking for a way to prove :
$$(A \rightarrow B) \rightarrow (\neg B \rightarrow  \neg A)$$
From the axioms :
A1) $(A) \rightarrow ( B \rightarrow A  )$
A2) $(A \rightarrow ( B \rightarrow C  )) \rightarrow((A\rightarrow B)\rightarrow(A\rightarrow C ))$
A3) $A \rightarrow  (B \rightarrow  (A \wedge B ))$
A4) $(A \wedge B )\rightarrow A$
A5) $(A \wedge B )\rightarrow B$
A6) $(A \rightarrow B )\rightarrow ((C \rightarrow B )\rightarrow ((A\vee C)\rightarrow B))$
A7) $A \rightarrow (A \vee B)$
A8) $A \rightarrow (B \vee A)$
A9) $ \neg \neg A \rightarrow A  $
and MP
I'm studying in computer science and I don't know any think about logic course.
Sorry for easy question and bad english. 
 A: You can't prove this because there's a model for this theory in which it's not true. Let $\to$,$\lor$ and $\land$ have their usual meanings, and let $\neg$ be the identity operation. Then all the axioms hold, but the theorem you want to prove doesn't.
A: You mention two different axiom systems.
The one is your last comment [(A1), (A2) and (A3) : correcting a typo in (A3); it must be : $(((¬A)→(¬B))→(((¬A)→B)→A))$] is the one used in the "classic" Elliott Mendelson, Introduction to Mathematical Logic (4th ed - 1997): see page 35.
It is sound and complete respect to tautologies; thus, $(A→B)→(¬B→¬A)$ is provable [see Lemma 1.11.e, page 38].
The axiom system in your original question is derived from the 10-axiom system $\mathsf L_4$ of Mendelson [see Example page 42], that is from the axiom system of S.C.Kleene, Introduction to Metamathematics (1952).
But it is lacking of axiom (9) :

$(A→B)→((A→\lnot B)→\lnot A)$.

Thus, the axiom system is incomplete and you cannot prove $(A→B)→(¬B→¬A)$, as stated by Joriki.
A: Dem:
$A1: A\rightarrow B$  Hipothesis
$A2: x\rightarrow ( x'')$
$A3: B\rightarrow B''$ Substitute x with B in A2
$A4: A \rightarrow B'' $   S-Rule(A1,A3)
$A5: A' \lor B'' $  Abbreviation $A\rightarrow B''$ 
$A6: B'' \lor A' $  Deduction rule on A5: $A\lor B \vdash B\lor A $
$A7: B' \rightarrow A' $ Abbreviation A6 
