$p\land\neg q\to r, \neg r, p ⊢ q$ -natural deduction

Question

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?"

Answer 1

Something like this?

$$\begin{split} p\wedge\neg q \to r , \neg r &\vdash \neg (p\wedge \neg q)&\quad&\textsf{Premise 1,Premise 2, Modus Tollens} \\ \neg (p\wedge \neg q)&\vdash \neg p\vee q &&1,\textsf{de Morgan's} \\ \neg p\vee q, p &\vdash q&&2,\textsf{Premise 3},\textsf{Disjunctive Syllogism} \\\hline p∧¬q→r,¬r,p &⊢q \end{split}$$

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Avoiding de Morgan's

$$\begin{split} (p\wedge \neg q)\to r, p, \lnot q&\vdash r &\quad&\textsf{Premise 1,Premise 3, Assumption of $q$, Modus Tolens} \\ r, \lnot r &\vdash \bot&&1,\textsf{Premise 2},\textsf{Negation Elimination}\\\hline(p\wedge\neg q)\to r,\lnot r,p,\lnot q&\vdash \bot&&\textsf{Cut}\\\hline (p\wedge\neg q)\to r,\lnot r,p&\vdash \lnot\lnot q&&\textsf{Negation Introduction (discharges the assumtion)}\\\hline (p\wedge\neg q)\to r,\lnot r,p& \vdash q &&\textsf{Double Negation Elimination}\end{split}$$

Answer 2

$$p\land\neg q\to r \iff \neg(p\land\neg q) \vee r \iff (\neg p \vee q \vee r)$$ (ref)

Since $\neg r$ and $p$ are in the premise, $q$ follows.

Answer 3

$$¬r \Rightarrow ¬(p \land ¬q) \mbox{ by modus tollens}$$

$$¬(p \land ¬q) \iff ¬p \lor ¬¬q \iff ¬p \lor q$$

$$( ¬p \lor q) \land p \Rightarrow q \mbox{ by definition of the disjunction operator.}$$

$$\therefore p\land\neg q\to r, \neg r, p ⊢ q$$