Natural deduction proof for $(a\to(b\to c))\to((a\to b)\to(a\to c))$ I am trying to prove that 
$(a\to(b\to c))\to((a\to b)\to(a\to c))$
holds in natural deduction, in particular when I work backwards from a Fitch style proof I can only get so far:

How can I prove it?
 A: You have only to remove the unnecessary LEM application : lines 3,9,10,11, and you will get the correct derivation :
1) $a \to (b \to c)$ --- assumed [a]
2) $a \to b$ --- assumed [b]
3) $a$ --- assumed [c]
4) $b$ --- from 2) and 3), by $\to$-elim
5) $b \to c$ --- from 1) and 3), by $\to$-elim
6) $c$ --- from 5) and 4), by $\to$-elim
7) $a \to c$ --- from 3) and 6), by $\to$-intro, discharging [c]
8) $(a \to b) \to (a \to c)$ --- from 2) and 7) by $\to$-intro, discharging [b]

9) $\vdash (a \to (b \to c)) \to ((a \to b) \to (a \to c))$ --- from 1) and 8) by $\to$-intro, discharging [a].

A: $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$
How exactly does one go about proving a conditional statement?   A Conditional Proof is usually used.   Step one: assume the antecedant.   Step two: derive the consequent.   Step three: introduce the conditional.
Here we have a nesting of conditionals, so we need to nest Conditional Proofs.
$$\fitch{}{\fitch{a\to(b\to c)}{\fitch{a\to b}{\fitch{a}{\vdots\\c}\\a\to c}\\(a\to b)\to(a\to c)}\\(a\to(b\to c))\to((a\to b)\to(a\to c))}$$
Well, you understand that, I see, but your trouble is how do we get from $a$ to $c$?   Now, the assumptions we have made are $a$ and two condtionals with $a$ as an antecedant.   This indicates that both those conditionals may be eliminated, and doing so gives $b$ and $b\to c$. 
One more elimination and we are done.
$$\fitch{}{\fitch{~~1.~a\to(b\to c)~;\textsf{assumption}}{\fitch{~~2.~a\to b~;\textsf{assumption}}{\fitch{~~3.~a~;\textsf{assumption}}{~~4.~b~: \textsf{conditional elimination (2,3)}\\~~5.~b\to c~: \textsf{conditional elimination (1,3)}\\~~6.~c~: \textsf{conditional elimination (4,5)}}\\~~7.~a\to c~: \textsf{conditional introduction (3-6)}}\\~~8.~(a\to b)\to(a\to c)~: \textsf{conditional introduction (2-7)}}\\~~9.~(a\to(b\to c))\to((a\to b)\to(a\to c))~: \textsf{cond. intro. (1-8)}}$$
A: In line 7. you Hypothesize B. 
Don't Hypothesize when you can prove.
You have 4: $A\to B$. and 5: $A$ so you know $B$ via "elim 4, 5".
Replace "7. B Hyp" with "7. B elim 4,5"
Then 9, and 10 are unnecessary.  (Not to mention there is no rule and you can't conclude $C$ from $\lnot B$.  It's just that $\lnot B$ is impossible if $A$ and $A\to B$. And if $\lnot A$ or $\lnot (A \to B)$ then it doesn't matter if $C$ is true or not.)
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Oh, yeah remove 11 as 8 is suplants it.  And remove 3 as it isn't necessary (and it can't prove the result anyway).
Remove 3.  Replace 7. with "B elim 4,5" remove 9,10,11.  Then your proof is good.  And shorter.
