¬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)
¬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)