Solving theorem proof with only primitive rules of logic!

Question

I am having trouble solving the theorem proof of (P-> ~Q)->(Q->~P). I can only use primitive rules and I understand I have to use arrow introduction to introduce my antecedent, but after that I am a bit lost. Any help would be appreciated.

Answer 1

To Prove: $(P\to \neg Q)\to(Q\to \neg P)$

Assume $P\to\neg Q$ and $Q$, and show this implies $\neg P$ via a contradition.

Then discharge the assumptions in the right order, by two applications of implication-introduction

Answer 2

\begin{array}{llll} 1&(1)&P \rightarrow \neg Q&A\\ 2&(2)&Q&A\\ 3&(3)&P&A\\ 1,3&(4)&\neg Q&1,3 \rightarrow E\\ 1,2&(5)&\neg P&2,4 \ RAA \ (3)\\ 1&(6)&Q \rightarrow \neg P&5 \rightarrow I \ (2)\\ &(7)&(P \rightarrow \neg Q) \rightarrow (Q \rightarrow \neg P)&6 \rightarrow I \ (1)\\ \end{array}