Wei Dai comments on Decision Theories, Part 3.75: Hang On, I Think This Works After All

Wei Dai 10 Sep 2012 22:54 UTC
3 points
Let me ask a simpler question. Does the following FB1 cooperate with FB2?
```
def FB1(X):
- if "X(FB1)=C" is provable in PA+1:
- - return C
- else return D
```
where FB2 is FB1 with “PA+2” in place of “PA+1″. I think the answer is yes, because PA+1 can prove
1. “FB1(FB2)=C” is provable in PA+1 implies “FB1(FB2)=C” is provable in PA+2
2. “FB1(FB2)=C” is provable in PA+1 implies FB2(FB1)=C (using 1 and looking at FB2)
3. “FB2(FB1)=C” is provable in PA+1 implies FB1(FB2)=C (looking at FB1)
4. FB2(FB1)=C (Löbian cycle)
Does this look right? And I think FB1 would also cooperate with FB-ZFC (FB1 with ZFC in place of PA+1), because PA+1 can prove that ZFC proves a superset of theorems of PA+1?
- orthonormal 15 Sep 2012 22:01 UTC
  0 points
  Parent
  I think that’s right, yes. If one proof system encompasses the other, it looks like it works.
  
  Another question is, what if we had two proof systems, neither of which encompasses the other? (Most horrifyingly, what if one were PA plus the Gödel sentence of PA, and the other were PA plus the negation of the Gödel sentence?)
  - Wei Dai 16 Sep 2012 7:56 UTC
    3 points
    Parent
    Another question is, what if we had two proof systems, neither of which encompasses the other? (Most horrifyingly, what if one were PA plus the Gödel sentence of PA, and the other were PA plus the negation of the Gödel sentence?)
    
    I think that works too. Call these proof systems PA+ and PA- and the FairBots using them FB+ and FB-. PA can prove that any theorem of PA is also a theorem of PA+ and PA-. So PA can prove:
    
    “FB+(FB-)=C” is provable in PA implies “FB+(FB-)=C” is provable in PA-
    “FB+(FB-)=C” is provable in PA implies FB-(FB+)=C (using 1 and looking at FB-)
    
    and similarly for the other half of the Löbian circle. So FB+(FB-)=C and FB-(FB+)=C are theorems of PA and hence theorems of PA- and PA+.