Nisan comments on A model of UDT with a halting oracle - Less Wrong

41 Post author: cousin_it 18 December 2011 02:18PM

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Comment author: Nisan 29 January 2012 07:18:00PM 1 point [-]

The conclusions that I think I can draw are

¬(S⊢A()=a)∧(S⊢A()=b) ⇒ Con(S)∧¬Con(S+Con(S)) ⇒ A()=b

So if one player plays chicken on the second inspected action, then all of them do.

Comment author: cousin_it 29 January 2012 09:15:08PM 0 points [-]

I'm still not getting it. Can you explain the proof of the following:

Con(S)∧¬Con(S+Con(S)) ⇒ A()=b

Comment author: Nisan 30 January 2012 12:59:37AM 2 points [-]

¬Con(S+Con(S)) ⇒ S⊢¬Con(S)
⇒ S⊢Prv(A()≠a) (if S is inconsistent, then it proves anything)
⇒ S⊢A()=a (because A plays chicken)
⇒ S⊢A()≠b
Con(S)∧¬Con(S+Con(S)) ⇒ Con(S)∧(S⊢A()≠b) (follows directly from 4)
⇒ (S⊬A()=b)∧(S⊢A()≠b) (a consistent theory can't prove a proposition and its negation)
⇒ (S⊬A()≠a)∧(S⊢A()≠b) (here I'm assuming that S⊢(A()=a∨A()=b). I don't know what to do if A can choose between more than two alternatives.)
⇒ A()=b (A plays chicken on the second inspected action)

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