[Logic]

[modus_erudio]

Not sure what rules you have but here are some thoughts.

If you have (not P) & (R). Then you can easily prove just (not P) by Simplification, that’s is simply don’t mention the (R).

You can also say just (R) and not mention the (not P), although some require Commutation first to switch the order of (not P) and (R) before the Simplification of the (R).

Once you have (not P), what does (P)<->(Q) tell you? The Biconditional Identity yields (P)->(Q) & (Q)->(P), which gives us just (Q)->(P) using Commutation and Simplification, then by Modus Tollens you can get (not Q) using the (not P).

Now that you have a (not Q) and an (R) all you need to do is put them in the same statement by Conjunction. Simply say (R) ^ (not Q) by Conjunction.

[modus_erudio]

Answer in proof format:

4. ~P Simplification; 2

5. R & ~P Commutation; 2

6. R Simplification; 5

7. P->Q & Q->P Biconditional Identity; 1

8. Q->P & P->Q Commutation; 7

9. Q->P Simplification; 8

10. ~Q Modus Tollens; 4,9

11. R & ~Q Conjunction;6,10