Proving A -> (A -> B) Equivalent to (A -> B)

[EvLer]

Is there a way to formally prove:
A -> (A -> B) equivalent to (A -> B)

do I just assume A and then use modus ponenes (by deduction method)? It just looks a bit odd asserting a premise out of the blue...
I checked the truth table for those, it seems to be true, however, truth table is not going to work as the answer, I need to do this formally.
Thanks in advance.

ps: sorry for trashing forum with my logic threads... from now on, i'll just put my questions (if i have any) here

 

[Hurkyl]

I checked the truth table for those, it seems to be true

More than just "seems"! Any logical equivalence provable via truth table can be deduced with the rules of inference you're using, and vice versa.

 

[arildno]

Proofs by exhaustion (like truth tables) are perfectly rigorous.

 

[EvLer]

well that is the thing... I check my "inference" steps with truth tables, however I need a rule or transformation of some sort and I can't see anything unless I assert the premise (A).
Ok, so here's the full thing:
prove Y -> W
given these premises:
1. Y -> Z'
2. X' -> Y
3. Y -> (X -> W)
4. Y -> Z
-----my proof----
5. X' -> Z' (2,1, hypoth. syllogism)
6. Z -> X (5, contraposition)
7. Y -> X (4,6, hypothet. syllogism)
8. (X' v W)' -> Y' (3, contraposition)
9. (X ^ W') -> Y' (DeMorgan's)
10. X -> (W' ->Y') (9, exportation)
11. X -> (Y ->W) (contraposition)

12. Y -> (Y -> W) (7,11, hypothet. syllogism)
13. Y (?)
14. Y -> W (?)
12-14 is where I need something or can I just assert Y?