Is this a valid or invalid inference? 
  
*
  
*$(p \land q ) \iff (r \implies s)$
  
*$(s \vee \neg t)$
So, $\neg s \implies ((r \implies \neg p) \vee \neg t)$

My lecturer has written in our notes that this statement is invalid, but I'm not so sure. I've attached my workings and have found there to be a contradiction whilst using the 'no counterexample' method (i.e. assume the premise to be T whilst the conclusion F, if there is a contradiction then the statement is valid) to find the validity of this statement. Thus, I think this inference is valid. 
My question is: is the above inference valid or not?
Thank you so much in advanced for your help! :)

 A: We will follow the instructor's notes and assume that the inference is invalid.
This means : assume the premises T and the conclusion F.
We may rewrite the conclusion as :

$s \lor \lnot r \lor \lnot p \lor \lnot t \ $;

in order to have it F we need :

$s=$ F and $r=p=t=$ T.

With this truth-assignment, the second premise : $s \lor \lnot t$ would be F.
Thus, the argument is valid.

If instead the sought conclusion is : $¬s \to ((r \to ¬p) \land ¬t)$, with the same approach we have two possibilities in order to "falsify" it; either :

(i) $s=$ F and $t=$ T, or :
(ii) $s=$ F and $r=p=$ T.

In the first case, again, we cannot satisfy the second premise.
In the second case, we have two variables "undefined" : $t$ and $q$.
The two premises are equivalent to :

$($ T $\land q) \leftrightarrow$ F
F $\lor \lnot t$.

Thus, if we set : $q=t=$ F, we can satisfy both premises and we have shown that the argument is invalid.


Conclusion :

Is this a valid or invalid inference?

It depends on the formula (in this case : the conclusion) ...
A: If s is not true, then (not t) has to be true, thus the inference is automatically true.
On a side note, I think that you may have made a typo...
