How to prove this argument valid? I was just wondering if some helpful person wouldnt mind helping me with this discrete maths question that has had be stuck for about a day now.
The argument is:
p or q
q implies ~p
q implies r
conclusion: r

I cant for the life of me figure out how to prove this.
Thanks heaps for any help you can give.
Thanks
Corey 
 A: The argument looks invalid.
Counterexample in natural language:
Look at this being, call it Purrty: http://thumbs.dreamstime.com/z/cat-1092150.jpg
Then consider three statements, $P$: Purrty is a cat. $Q$: Purrty is a dog. $R$: Purrty is a canine. 
Clearly $Q$ implies not $P$ and $Q$ implies $R$.
Also clear is that $P$ or $Q$, since Purrty is a cat, so Purrty is a cat or a dog. This does not imply Purrty is a canine ($R$) (since it isn't) even though all three statements are assumed true.
Counterexample in mathematics:
Suppose for example we have three sets $A$, $B$ and $C$.
Then let $P$ be the statement $x\in A$ and $Q$ the statement $x\in B$ and $R$ be the statement $x \in C$.
Then $P$ or $Q$ is the statement $x \in A\cup B$.
$Q$ implies not $P$ is the statement $A$ and $B$ are disjoint.
$Q$ implies $R$ is the statement $B\subseteq C$.  
But we can easily construct sets that all these are true, and yet $R$ isn't.
e.g. $A = (1,2)$, $B = (4,5)$ and $C=(3,5)$. We see that the second and third statements are true.
Now take $x=1.5$. Then $x \in (1,2) \cup (4,5)$. So $P$ or $Q$ is true. We also know $A$ and $B$ are disjoint (so $Q$ implies not $P$ is true) and $B \subseteq C$, so $Q$ implies $R$ is true). This  however, does not imply $x \in (3,5)$ ($R$).
Since this argument does not result in true conclusions everytime we give it true premises, it is invalid.
Edit: A good way to see it logically is via the truth table as mentioned in the comments. 
A: The argument is invalid. All you can show is that: $p$ and $q$ cannot both be true and $r$ is true whenever $q$ is.
A valid argument would be
$$\left|\begin{align*}
p\vee q\\ \color{red}{\neg} q\to \neg p\\ q\to r \\ \hline  r
\end{align*}\right.$$
Are you sure that you did not miscopy the question?
A: We can apply Resoulution to the three clauses (the premises) :

1) $p \lor q$
2) $\lnot q \lor \lnot p$ (equivalent to : $q \to \lnot p$)
3) $\lnot q \lor r$ (equivalent to : $q \to r$)

and the negation of the purported conclusion :

4) $\lnot r$.

Applying the procedure we get :
i) $p \lor r$ --- from 1) and 3)
ii) $\lnot q \lor r$ --- from i) and 2)
iii) $\lnot q$ --- from ii) and 4).
We have concluded the procedure without having derived the empty clause; thus, the argument is invalid.
