Prove $p \wedge \neg p \vdash q$ for any propositional variables $p$ and $q$ without using disjunctive syllogism or excluded middle or $\neg$-elimination.
I can prove this easily using $\neg$-elimination: assume $p\wedge \neg p$ and $\neg q$. Then by $\wedge$-elimination, we have $p$ and again by $\wedge$-elimination, we have $\neg p.$ But then by $\neg$-elimination, we have $q$. However I'm not sure how to do it without using $\neg$-elimination. Will Peirce's law (i.e. $((A \to B)\to A) \to A$)) be useful?
Clarification: $\neg$-elimination is defined as follows: Let $\sum, A, B$ be formulas. Then if $\sum, \neg A \vdash B$ and $\sum, \neg A \vdash \neg B,$ then $\sum \vdash A$. Informally, it resembles the "proof by contradiction" method.
By the rule of conjunctive simplification, we have $p$ and $\neg p$. Now, let $q$ be some proposition. Thus, we can use disjunctive addition to derive $\neg q \lor p$. By conditional exchange, we have $q \to p$. Finally, by modus tollens, we have $\neg q$