How to deduce using rules of inference I'm very confused on how to use the rules of inference to tackle this problem. Could I get someone guidance to help me solve this? Thanks in advance. 

Given:
  
  
*
  
*$p \implies q$
  
*$\neg p \implies r$
  
*$r \implies s$
  
  
  Prove that $\neg q \implies s$.

 A: You can not get help unless you say what rules of inference you are permitting .there are many deductive systems .
A: In mathematical logic, we have the concept of a 'contrapositive': the contrapositive of the statement $a \implies b$ is $\neg b \implies \neg a$, and the two are equivalent. To intuitively understand why, replace $a$ with 'this is a tree' and $b$ with 'this contains wood'. To rigorously prove this can be a good exercise in basic logic.
Then, from the givens, we have $\neg q \implies \neg p$. We also have $\neg p \implies r$ and $r \implies s$, so using this chain of transitive inferences, we get $\neg q \implies s$ as wanted.
A: You can obtain the argument by just looking at the given statements. You need to prove $\neg q \implies s$. So suppose $\neg q$ is true. We need to prove $s$ is true. Identify where in the given three propositions can we find $\neg q$ is true. We can take the contrapositive of the first proposition. 
Observe from the first proposition $p \implies q$ that we have $\neg q \implies \neg p$. By hypothesis $\neg q$ is true, and so by the first proposition $\neg p$ is true. Using the second proposition, we get that $r$ is true, and from the third proposition, this implies $s$ is also true. Hence, if $\neg q$ is true, then $s$ is true, as was to be shown.
The rules of inference we used are contraposition (which says $a \implies b$ is equivalent to $\neg b \implies \neg a$) and hypothetical syllogism (which says: if $a \implies b$ and $b \implies c$, then $a \implies c$). 
