Prove $(P\to (Q \to R)) \to (Q \to (P \to R))$ (derivation)...

Question

The problem is as stated in the title. With this problem and I am restricted to modus tollens (MT), modus ponens (MP), repetition (R), and double negation (DN). I'm just getting used to logic derivation, so I've really no idea where to proceed. The issue I'm having is just not knowing how to properly start this. This is all I've got, and I think it's the wrong direction:

1. Show $(P\to (Q \to R)) \to (Q \to (P \to R))$
   2. $\;\;\;\;\;\;\;$ $P\to (Q \to R)$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ Assume CD
   3. $\;\;\;\;\;\;\;$ Show $Q\to R$
   4. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\neg(Q\to R)$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$Assume ID
   5. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\neg P$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ 2,4 MT

Anyone able to give me a hand? I don't need a full answer, just a step in the right direction.

Answer 1

Suppose $P\to (Q\to R)$. (1)

We want to show $Q\to (P\to R)$.

Well, suppose $Q$. (2)

We want to show $P\to R$.

Suppose $P$. (3)

Then $Q\to R$ by 1 and 3. (4)

Then $R$ by 4 and 2.

Answer 2

you are correct in line 3 you go in the wrong direction:

Start with :

1 | Show $ (P \to (Q \to R)) \to (Q \to (P \to R))$

2 || $ (P \to (Q \to R)) $ Assume CD

3 || Show $ (Q \to (P \to R))$

4 ||| $ Q $ Assume CD

5 ||| Show $ P \to R $

and the rest you can do yourself (just a lot of MP)

GOOD LUCK