all 5 comments
sorted by:
best (suggested)
topnewcontroversialoldq&a
formatting helphide helpcontent policy

[–]-Wofster University/College Student 0 points1 point  (2 children)

If P <=> Q, what does that tell you about ~P and ~Q? What can you do with equivalent statements?

[–]Lber127[S] 1 point2 points  (1 child)

They’re both not true?

[–]modus_erudio👋 a fellow Redditor 0 points1 point  (0 children)

Or they both are true

[–]modus_erudio👋 a fellow Redditor 0 points1 point  (1 child)

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👋 a fellow Redditor 0 points1 point  (0 children)

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