Proving a Conclusion Comprised of Conditionals (Logic) I'm a new logic student, so I apologize if this question is improperly formatted, or if I am using incorrect terminology. I'll try my best to explain the question I have as clearly as possible. If I make a mistake, please let me know and I'll be happy to make a change to my question.
I'm attempting to prove the logical statement $((P\rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow (P \rightarrow ( Q \rightarrow R ))$ to be valid. I find myself hitting a dead-end when I assume conditionals and attempt to show the consequents.
Here is what I have tried. The conclusion states
$$\therefore ((P\rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow (P \rightarrow ( Q \rightarrow R )).$$
I begin by attempting to show
$$(1) \;\;\; ((P\rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow (P \rightarrow ( Q \rightarrow R ))$$
by first assuming the antecedent
$$(2) \;\;\; (P \rightarrow Q) \rightarrow (P \rightarrow R) $$
and attempting to show the consequent
$$(3) \;\;\; P \rightarrow (Q \rightarrow R) .$$
To do that, I assume the antecent of $(3)$ once more:
$$(4) \;\;\; P $$
and attempt to show
$$(5) \;\;\; Q \rightarrow R .$$
However, at this point, I get stuck. In most of my exercises, the most common way to go about these types of problems is by breaking the sentences down into their most basic parts, then handling the innermost conditional using a direct/indirect derivation. However, I don't immediately see a way to do that in this scenario. Perhaps I'm missing something bigger here.
This is an exercise for a class, so any hints would be greatly appreciated.
Note: According to the exercise, the rules of repetition, double negation, modus ponens, and modus tollens are only permitted. I'm hoping to get a hint that can help me get to the solution using only these rules; however, if there are any alternate methods of proving this statement outside the bounds of these rules, I'd be interested in seeing them as well.
 A: I managed to determine the answer on my own. Here is the proof.
The conclusion states
$$\therefore ((P\rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow (P \rightarrow ( Q \rightarrow R )).$$
I begin by attempting to show
$$(1) \;\;\; ((P\rightarrow Q) \rightarrow (P \rightarrow R)) \rightarrow (P \rightarrow ( Q \rightarrow R ))$$
by first assuming the antecedent
$$(2) \;\;\; (P \rightarrow Q) \rightarrow (P \rightarrow R) $$
and attempting to show the consequent
$$(3) \;\;\; P \rightarrow (Q \rightarrow R) .$$
To do that, I assume the antecent of $(3)$ once more:
$$(4) \;\;\; P $$
and attempt to show
$$(5) \;\;\; Q \rightarrow R .$$
To do this, I assume the antecedent of $(5)$
$$(6) \;\;\; Q $$
and attempt to show
$$(7) \;\;\; R. $$
To do this, I assume the negation of $(7)$ for an indirect derivation
$$(8) \;\;\; \neg R $$
and attempt to show
$$(9) \;\;\; P \rightarrow Q. $$
To do this, I assume the antecedent of $(9)$
$$(10) \;\;\; P $$
(which happens to be the assumption on line $(4)$) and use the repitition rule to bring down line $(6)$:
$$(11) \;\;\; Q .$$
Line $(9)$ is cancelled by a conditional derivation by $(11)$.
By modus ponens $(9)$ $(2)$, we show
$$(12) \;\;\; P \rightarrow R.$$
By modus ponens $(12)$ $(4)$, we show
$$(13) \;\;\; R.$$
This allows me to cancel the show lines $(9)$, $(7)$, $(5)$, $(3)$ and use the conditional derivation on $(3)$ to successfully show $(1)$.
A: 
I'm attempting to prove the logical statement ((P→Q)→(P→R))→(P→(Q→R)) to be valid. I find myself hitting a dead-end when I assume conditionals and attempt to show the consequents.

Here are the assumptions you need to make for the conditional introductions:$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{}{\fitch{(p\to q)\to(p\to r)}{\fitch{p}{\fitch{q}{~\vdots\\r}\\q\to r}\\p\to (q\to r)}\\((p\to q)\to(p\to r))\to(p\to (q\to r))}$$ So you just need to derive $r$ under those assumptions.  Just raise one more conditional introduction subproof, and the rest is a few conditional eliminations.
