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

$$¬r \Rightarrow ¬(p \land ¬q) \mbox{ by modus tollens}$$ 
$$¬(p \land ¬q) \iff  ¬p \lor ¬¬q \iff ¬p \lor q$$
$$( ¬p \lor q) \land p \Rightarrow q \mbox{ by definition of the disjunction operator.}$$
$$\therefore p\land\neg q\to r, \neg r, p ⊢ q$$
