Are $A, B, C \vdash D$ and $A, B\vdash C \rightarrow D$ interchangeable? For an assignment we have to make a proof in the Hilbert system.
And my proof hinges on the following operation being allowed:
$A, B, C \vdash D\tag 1$
Becomming:
$A, B\vdash C \rightarrow D\tag 2$
Intuitively this makes sense, because $A$ is an assumption, so moving it to the right of the turnstile still requires it to hold.
Note: $A$ is not the only assumption made.
 A: Yes, they are interchangeable in Hilbert's system (and indeed for any proof system that is sound and complete).
From 
$A, B\vdash C \rightarrow D\tag 2$
to
$A, B, C \vdash D\tag 1$
is trivial: Start the proof with $A,B$, and $C$ as premises. Given (1), you can derive $C \rightarrow D$ from $A$ and $B$. And then do Modus Ponens using $C$ to get D.
From 
$A, B, C \vdash D\tag 1$
to
$A, B\vdash C \rightarrow D\tag 2$
is of course much more interesting and challenging, but it always works. 
The Deduction Theorem says that for any set of statements $\Gamma$, and statements $\varphi$ and $\psi$: if $\Gamma \cup \{ \varphi \} \vdash \psi$, then $\Gamma \vdash \varphi \rightarrow \psi$.  
I would recommend looking up the Deduction Theorem, and a proof thereof.  But the basic idea is to transform the derivation $\Gamma \cup \{ \varphi \} \vdash \psi$ into one that starts with just $\Gamma$ as its premises, and then for every statement $\chi$ that occurs in the derivation $\Gamma \cup \{ \varphi \} \vdash \psi$ will derive $\varphi \rightarrow \chi$ ... using induction you can show that you can always do this. And since the last line of the first derivation is $\psi$, the last line of the transformed derivation becomes $\varphi \rightarrow \psi$, and hence you're done.
Simple example:
Original derivation of $\{ P\rightarrow Q, Q \rightarrow R, P \} \vdash R$:


*

*$P\rightarrow Q$ Premise

*$Q\rightarrow R$ Premise

*$P$ Premise

*$Q$ MP 1,3

*$R$ MP 2,4
Transform to $\{ P\rightarrow Q, Q \rightarrow R \} \vdash P \rightarrow R$:


*

*$P\rightarrow Q$ Premise

*$(P\rightarrow Q) \rightarrow (P \rightarrow (P\rightarrow Q))$ Axiom 1

*$P \rightarrow (P\rightarrow Q)$ MP 1,2 ('Conditionalized' 1 from original derivation)

*$Q\rightarrow R$ Premise

*$(Q\rightarrow R) \rightarrow (P \rightarrow (Q\rightarrow R))$ Axiom 1

*$P \rightarrow (Q\rightarrow R)$ ('Conditionalized' 2 from original derivation)
...


*$P \rightarrow P$  (this takes a few steps to derive, but you've probably seen that proof.  This is the conditionalized form of line 3 from original derivation)

*$(P \rightarrow (P\rightarrow Q)) \rightarrow ((P \rightarrow P) \rightarrow (P \rightarrow Q))$ Axiom 2

*$(P \rightarrow P) \rightarrow (P \rightarrow Q)$ MP 3,12

*$P \rightarrow Q$ MP 11, 13 (all that work and we're actually back to premise 1 ... hey, no one said this process was the most efficient one!  I just wanted to show you how this works in general ... and please note that now we have the conditionalized form of line 4 of the original derivation)

*$(P \rightarrow (Q\rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ Axiom 2

*$(P \rightarrow Q) \rightarrow (P \rightarrow R)$ MP 6,15

*$P \rightarrow R$ MP 14, 16 (conditionalized form of line 5 of the original derivation)
So yeah, keep using Axioms 1 to conditionalize all premises and axioms, and axiom 2 to emulate all the inferences in the original derivation but where you infer the conditionalized form.
