$p\land\neg q\to r, \neg r, p ⊢ q$ -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?"

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
