Natural Deduction - Choosing the assumptions I have been trying to understand how to use natural deduction rules to solve problems in logic. I understand the different rules. However, I find it the most difficult to determine what can be set as assumptions, in order to solve the problems. For instance, I have understood the simple problems, such as the following: Where it is clear what ought to be the assumption and how to work from there. 

However, when it comes to other problems, I get really confused as to what I am allowed to state as assumptions and what not to state as assumptions. Following is an example: 

Is there a simple rule for proving formulas through natural deduction, with regards to problem solving techniques? For instance, when given problems such as: 

How can I know which assumptions to make, etc.?
Any help is highly appreciated!

Added
I'm still a bit confused as to what I can set as the assumption(s). 
The below example confuses me. In line (I), both the first conjunction [(P->Q)&(R->Q)] of the expression, as well as the second conjunction, [P & R], which is actually the conclusion of the left-most implication, is also set as an assumption. 
As @Henning Makholm explained, I expected to end this proof with two stacked →Is. 
The problem is how to know what expressions to use as assumptions? I keep seeing worked examples, all of them which make sense. However, when I am to solve problems myself, I get confused with the assumptions that can be made. Any advice on this particular matter? 
For instance, I thought one could not assume an expression that is also the conclusion of an implication (P & R), as an assumption? Or is this just the case of the right-most conclusion, in this case Q? 
Evidently, the choice of assumptions confuses me!. 

 A: Usually natural deduction proofs are easiest to construct from the bottom up. Whenever you need to prove something of the form $\varphi\to\psi$, your options are either to produce it using ${\to}E$ on $\sigma\to\varphi\to\psi$ and $\sigma$, or to produce it using ${\to}I$ on $\psi$. The first of these options requires some cleverness in choosing a formula $\sigma$; the second is more or less mechanical.
Whenever you use ${\to}I$ to conclude $\varphi\to\psi$, you get to state $\varphi$ as an assumption in the proof of $\psi$. And these are the only assumtions you ever get to make, at least in most orthodox formulations of Natural Deduction.
So if you're writing a proof from the top down, you have better not make any assumptions that you don't plan to "discharge" using ${\to}I$ further down in the tree.

In your second example is looks like you have the ${\lor}I$ rule wrong. It ought to be just
$$ \frac{P}{P\lor R} {\lor}I $$
with no $R$ premise. And that's good because you cannot assume $R$ in this place -- there is no instance of ${\to}I$ in the tree that concludes something of the form $R\to\cdots$.

In the last example, you;re trying to conclude $((P\to Q)\land (R\to Q)) \to (P \land R) \to Q$. Here the natural approach will be to end the proof with two stacked ${\to}I$s to produce the two arrows. Then your task is to conclude $Q$, while being allowed to assume $(P\to Q)\land (R\to Q)$ and $P\land R$. That's easy enough; would have been slightly more interesting of one of the $\land$s (but not both) had been $\lor$.
