Reply
Mon 23 May, 2011 10:26 pm
1. (B > P) > S

2. P > S

3. (S v B) > C / ~C v B

I've tried to come at this with both indirect and conditional proof methods (and both together), but I can't make any headway at all. Any help or suggestions would be appreciated!

@whatisinevidence,

Here is an example of one of my failed attempts...

1. (B > P) > S

2. P > S

3. (S v B) > C / ~C v B

4. ~(S v B) v C (3 Impl)

5. (~S • ~ B) v C (4 DM)

6. C v (~S • ~ B) (5 Comm)

7. ~C > (~S • ~ B)

. 8. ~C (ACP)

. 9. ~S • ~ B (7, 8 MP)

. 10. ~S (9, Simpl)

. 11. ~P (2, 9 MT)

12. ~C > ~P (8-11 CP)

13. P > C (12, Trans)

. 14. P (13 ACP)

. 15. C (13, 14 MP)

. 16. S (2, 14 MP)

17. C > S (CP)

(8-11 and 14-16 should be indented)

@whatisinevidence,

This is not a tautology ,hence not provable

Here is a proof that the above is not a tautology:

Put B=false , P=true ,S=true ,C= true and we have :

for .......1) (F=>T)=>T ................................which is T

..............2) T=> T ........................................Which is T

..............3) (TvF) => T.................................Which is T

Hence : [(B=>P) =>S] & (P=>S)&[(SvB) =>C] .......................IS T

But ~CvB = FvF .....Hence F. So we have T=>F.

Thus the above is not a tautology and therefor not provable