I don’t think there’s a flaw in this argument. I’m pretty sure that ¬□C→C is a theorem of PA for any sentence C. However, even if we consider C to be a sentence like 2=1, that does not mean that we can conclude that C is a theorem of PA, since proving that there doesn’t exist a proof for a given sentence in PA is equivalent to proving that PA is consistent, which is not possible if we assume that PA is consistent.
I could elaborate, but maybe you want to think about this more, so for now I’ll just address your remark about ¬□C→C, where C is refutable. If we assume that ¬□C→C, then, since C is false, ¬□C must be false, so □C must be true. That is, you have proven that PA proves □C, that is, since C is contradictory, PA proves its own inconsistency. You’re right that this is compatible with PA being consistent—PA may be consistent but prove its own inconsistency—but this should still be worrying.
Thanks Sam. I thought about this some more and realized where I went wrong—I was applying the deduction theorem incorrectly (as other comments in this thread have pointed out).
By the way, when you say that PA proves its own inconsistency, do you mean that PA⊢□(¬Con(PA)) or that PA⊢¬Con(PA)? From your reasoning I agree that if we assume ¬□C→C then we can arrive at PA⊢□C and PA⊢□(¬C), from which we can conclude that PA⊢□(¬Con(PA)). If you meant that PA⊢¬Con(PA) though, could you explain how you arrived at that?
I don’t think there’s a flaw in this argument. I’m pretty sure that ¬□C→C is a theorem of PA for any sentence C. However, even if we consider C to be a sentence like 2=1, that does not mean that we can conclude that C is a theorem of PA, since proving that there doesn’t exist a proof for a given sentence in PA is equivalent to proving that PA is consistent, which is not possible if we assume that PA is consistent.
*I* think that there’s a flaw in the argument.
I could elaborate, but maybe you want to think about this more, so for now I’ll just address your remark about ¬□C→C, where C is refutable. If we assume that ¬□C→C, then, since C is false, ¬□C must be false, so □C must be true. That is, you have proven that PA proves □C, that is, since C is contradictory, PA proves its own inconsistency. You’re right that this is compatible with PA being consistent—PA may be consistent but prove its own inconsistency—but this should still be worrying.
Thanks Sam. I thought about this some more and realized where I went wrong—I was applying the deduction theorem incorrectly (as other comments in this thread have pointed out).
By the way, when you say that PA proves its own inconsistency, do you mean that PA⊢□(¬Con(PA)) or that PA⊢¬Con(PA)? From your reasoning I agree that if we assume ¬□C→C then we can arrive at PA⊢□C and PA⊢□(¬C), from which we can conclude that PA⊢□(¬Con(PA)). If you meant that PA⊢¬Con(PA) though, could you explain how you arrived at that?