is this proposition (inference) valid? Is this inference valid or invalid? Why and how to prove this kind of question? 
$$p \rightarrow q, \neg q \rightarrow r  
, r  \vDash p $$
Would a single truth table be enough for all types?
 A: You won't be able to derive $p$ from the hypotheses. There are a couple of ways you can see this, but the easiest might be to realize that, if the hypotheses are true, then you should be able to conclude that $p$ is true. 
Here, $p \to q$, $\lnot q \to r$, and $r$ being true are not enough to coerce $p$ to be true, since these conditions are satisfied if $r$ is true and $q$ is true, while $p$ could be anything.
A: This problem must have been intended as torture, because the answer is: you won't get to p (a horrible thought, isn't it? :). 
The only nonconditional proposition you're given is r, which leads nowhere. The only other option is indirect proof in which you assume ~p and try to derive a contradiction. But that also leads nowhere, because you can't apply ~p to any of the premises.
It would have been different had your setup been, say:
P1. ~p → q
P2. q → ~r
P3. r
By the way, the problem seems not to be a case of ex falso quodlibet ("from a falsehood, anything [follows]"). That would apply if you were able to derive a contradiction, from which (in some circles) you would be entitled to infer anything. The problem is that you're given p → q. The only way you could make an inference in reverse of the arrow would be to somehow get to ~q, but from that you could only infer ~p.
