Question on logical inferences The instruction of this question is:

Encode the following arguments and show whether they are valid or not.
  If not valid give countermodels i.e., truth assignments to the
  propositions which make them false.
If the investigation continues, then new evidence is brought to light.
  If new evidence is brought to light, then several leading citizens are
  implicated. If several leading citizens are implicated, then the
  newspapers stop publicizing the case. If continuation of the
  investigation implies that the newspapers stop publicizing the case,
  then bringing to light of new evidence implies that the investigation
  continues. The investigation does not continue. Therefore, new
  evidence is not brought to light.

My attempt:
Let $p$ denote "the investigation continues"
    $\quad \space \space q$ denote "new evidence is brought to light"
    $\quad \space \space r$ denote "several leading citizens are implicated"
    $\quad \space \space s$ denote "the newspapers stop publicizing the case"  
So, the premises are:
$p \to q $
$q \to r $
$r \to s $
$(p \to s) \to (q \to p)$
$\neg p$
$\therefore \neg q$
I am very much a beginner in logic so I am not sure if this is correct so far or how to prove or disprove this using the standard rules of inference. Any ideas?
 A: looks like you missed a couple.
The fourth statement should be:
(p -> s) -> (q -> p)

So we have:
p→q
q→r
r→s
(p→s)→(q→p)
¬p
∴¬q

So by the first three primitives we have p->s. Hence q -> p. So not p implies not q.
I think this checks out, but I my formal logic training is a little weak.
A: Here's a simplification of Zhen Lin's argument. We know
$$ \tag A (p \to s) \to (q \to p)$$
Contrapose this:
$$ \tag B (q\land\neg p) \to (p\land \neg s)$$
Now I'm going to prove $q\to p$ by contradiction. We assume its negation 
$$ \tag C q\land \neg p $$
and from this and (B) by modus ponens we can then conclude
$ p\land \neg s $ and in particular $p$. However (C) also trivially implies $\neg p$ which is a contradiction.
Thus, given (A) I have proved 
$$\tag D q\to p$$
(And in contrast to Zhen I didn't even need the $\neg p$ premise to do so. On the other hand, my proof is not intuitionistically valid).
Now the story explicitly tells us $\neg p$. Therefore, by modus tollens and (D), we have $\neg q$. Q.E.D.
A: Edit. I assume you meant $(p \to s) \to (q \to p)$ in your fourth premise – but it doesn't actually matter, as we shall see.


*

*Ex falso quodlibet, so
$$\lnot p, p \vdash s$$
and by conditional proof,
$$\lnot p \vdash (p \to s)$$

*By modus ponens,
$$\lnot p, ((p \to s) \to (q \to p)) \vdash (q \to p)$$

*By contraposition,
$$\lnot p, (q \to p) \vdash \lnot q$$

*Putting it all together,
$$\lnot p, ((p \to s) \to (q \to p)) \vdash \lnot q$$
So it turns out the other premises $p \to q, q \to r, r \to s$ are irrelevant for this deduction.
A: You meant, for the fourth premiss, 
$$(p \to s) \to (q \to r)$$
Which might help! Intuitively, the first three premisses in PL give you
$$p \to s$$
Modus ponens with the corrected fourth premiss yields
$$q \to r$$
Modus tollens with the fifth premiss gives the conclusion. So it is just a question of following these steps in your preferred formal system.
A general point about such questions in elementary logic texts, however. You should always ask yourself, when translations involving conditionals or "implies" are concerned, whether the validity of the supposed rendering into PL shows (i) that the original argument is valid, or (ii) that the material conditional is here a bad translation of the intuitive content of the original. (After all, you are being asked whether the original argument is valid, and showing a certain rendering of it into PL is valid only settles the matter if the rendering is a good one: and where vernacular conditionals are involved, things can easily go wrong. So you should always, when answering, indicate whether you think that showing the PL argument is valid establishes what is asked.)
A: The text of the problem does not state that you have to use the rules of interference. To show that $\neg q$ holds when $\neg p$ and $(p \rightarrow s)\rightarrow(q \rightarrow p)$ holds you can also use truth tables. The lines were $\neg p$ and $(p \rightarrow s)\rightarrow(q \rightarrow p)$ holds are the lines 1,2,3,4. For these lines the last column is 1 so  $\neg q$ is true.
$$
\begin{array}{ccccccccccccc}
&n&p&q&r&s&p \rightarrow  q&q \rightarrow  r&r \rightarrow  s&(p \rightarrow s)\rightarrow(q \rightarrow p)& \neg p& \neg q\\\hline
&1&0&0&0&0&1&1&1&1&1&1\\
&2&0&0&0&1&1&1&1&1&1&1\\
&3&0&0&1&0&1&1&0&1&1&1\\
&4&0&0&1&1&1&1&1&1&1&1\\
&5&0&1&0&0&1&0&1&0&1&0\\
&6&0&1&0&1&1&0&1&0&1&0\\
&7&0&1&1&0&1&1&0&0&1&0\\
&8&0&1&1&1&1&1&1&0&1&0\\
&9&1&0&0&0&0&1&1&1&0&1\\
&10&1&0&0&1&0&1&1&1&0&1\\
&11&1&0&1&0&0&1&1&1&0&1\\
&12&1&0&1&1&0&1&1&1&0&1\\
&13&1&1&0&0&1&0&1&1&0&0\\
&14&1&1&0&1&1&0&1&1&0&0\\
&15&1&1&1&0&1&1&1&1&0&0\\
&16&1&1&1&1&1&1&1&1&0&0
\end{array}
$$
