$p\land\neg q\to r, \neg r, p ⊢ q$ -natural deduction I have the following:
$$p\land\neg q\to r, \neg r, p ⊢ q$$
I know that my attempt is incorrect, but I will show it anyways:
Step 1) $p\land\neg q\to r$  ----premise
Step 2) $\neg r$  -----premise
Step 3) $p$ -----premise
Step 4) $\neg q\to r$ ---- e1
Step 5) $\neg \neg q$ ----MT4,2  
Can someone show me the proper steps? I do not think I can use MT in the way shown above, but I cannot find out how to get to q.
OP's remark from a comment: "I was curious, is there a way to bypass DeMorgan's law?"
 A: $$p\land\neg q\to r \iff \neg(p\land\neg q) \vee r \iff (\neg p \vee q \vee r)$$ (ref)
Since $\neg r$ and $p$ are in the premise, $q$ follows.
A: $$¬r \Rightarrow ¬(p \land ¬q) \mbox{ by modus tollens}$$ 
$$¬(p \land ¬q) \iff  ¬p \lor ¬¬q \iff ¬p \lor q$$
$$( ¬p \lor q) \land p \Rightarrow q \mbox{ by definition of the disjunction operator.}$$
$$\therefore p\land\neg q\to r, \neg r, p ⊢ q$$
A: Something like this?
$$\begin{split}
   p\wedge\neg q \to r ,  \neg r &\vdash \neg (p\wedge \neg q)&\quad&\textsf{Premise 1,Premise 2, Modus Tollens}
\\
 \neg (p\wedge \neg q)&\vdash  \neg p\vee q &&1,\textsf{de Morgan's}
 \\ \neg p\vee q, p &\vdash q&&2,\textsf{Premise 3},\textsf{Disjunctive Syllogism} \\\hline p∧¬q→r,¬r,p &⊢q
\end{split}$$

Avoiding de Morgan's
$$\begin{split}
 (p\wedge \neg q)\to r, p, \lnot q&\vdash r &\quad&\textsf{Premise 1,Premise 3, Assumption of $q$, Modus Tolens}
 \\ r, \lnot r &\vdash \bot&&1,\textsf{Premise 2},\textsf{Negation Elimination}\\\hline(p\wedge\neg q)\to r,\lnot r,p,\lnot q&\vdash \bot&&\textsf{Cut}\\\hline  
(p\wedge\neg q)\to r,\lnot r,p&\vdash \lnot\lnot q&&\textsf{Negation Introduction (discharges the assumtion)}\\\hline (p\wedge\neg q)\to r,\lnot r,p& \vdash q &&\textsf{Double Negation Elimination}\end{split}$$
A: The following proof uses neither modus tollens nor De Morgan's law. 
It, however, uses the precedence of logical operators where the conjunction operator (∧) has higher precedence over the conditional operator (→). That is,  $p∧¬q→r$ is the same as $(p∧¬q)→r$.
Given the above, here is a proof:


Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
"Operator Precedence" Introduction to Logic http://intrologic.stanford.edu/glossary/operator_precedence.html
