help verifying my answer for this" premise-conclusion" question For each of the premise-conclusion pairs below, give a valid step-by-step argument (proof) along with the name of the inference rule used in each step. 
(a) Premise: {¬p ∨ q → r, s ∨ ¬q, ¬t, p → t, ¬p ∧ r → ¬s}, conclusion: ¬q

I did this:
1)  $ p → t  $      Premise
2)  $¬t $        Premise
3) $¬p $        Modus Tol.
4)$¬p ∨ q → r$  premise
5) $q → r $     disj. syll. from 3 ,  4
6) $q ∨ ¬ r $      implication law ( don't know if I'm allowed to use that here)
7) $¬ ( q ∨ ¬ r ) $  Don't know the name of this rule. I'm thinking negation?
8) $¬ q ∧ r  $    
9)$ ¬ q   $     by simplification. 
 A: $\neg p \vee q \to r\; $ is $\; \big((\neg p) \vee q\big) \to r \;$ and not $(\neg p) \vee (q\to r)\;$. 
Operator precedence goes to the disjunction before the implication.
$\begin{array}{lll}
1) & p \to t  & \text{ Premise }
\\
2) &  ¬t & \text{ Premise }
\\
3) &  ¬p  & \text{ Modus Tolens}
\\
4) & ¬p∨q→r  & \text{Premise}
\\
5') & \neg p \vee q & 3, \text{disjunction introduction}
\\
6') & r & 4, 5', \text{modus ponens}
\\
\vdots & \vdots & \vdots
\end{array}$
Can you finish from this?

PS: You also have the implication equivalence incorrect in your original 6; it's the negation of the antecedent disjunct with the consequent, not the antecedent disjunct with the negation of the consequent.
$$\; a\to c \iff \neg a \vee c \;$$
PPS: You don't need implication equivalence to solve this question.  But learn what it is for when you do.
A: You misused the ~p to derive "q->r" given that you have ~p and can only derive ~p from that specific disjunction.  I'm not sure what you did in step six either.
Perhaps try:
Premise: p -> t
Premise: ~t
[~p v t], DeMorgan's law, therefore ~p given premise ~t and disjunctive syllogism.
Premise: [~p v (q -> r)], 
use negation [~(~p v (q -> r))] and distribute negation for:
[p & ~(q -> r)] and distribute after converting conditional to disjunction to derive:
[~(~q v r)] then [q & ~r].
Thus we have [p, ~p, q, and ~r].
with p proven, we can revisit your fourth premise, derive the conditional and use the ~r derived from my proof with modus tollens to prove ~q.
