It would kind of use assumption 3 inside step 1, but inside the syntax, rather than in the metalanguage. That is, step 1 involves checking that the number encoding “this proof” does in fact encode a proof of C. This can’t be done if you never end up proving C.
One thing that might help make clear what’s going on is that you can follow the same proof strategy, but replace “this proof” with “the usual proof of Lob’s theorem”, and get another valid proof of Lob’s theorem, that goes like this: Suppose you can prove that []C->C, and let n be the number encoding a proof of C via the usual proof of Lob’s theorem. Now we can prove C a different way like so:
n encodes a proof of C.
Therefore []C.
By assumption, []C->C.
Therefore C.
Step 1 can’t be correctly made precise if it isn’t true that n encodes a proof of C.
If that’s how it works, it doesn’t lead to a simplified cartoon guide for readers who’ll notice missing steps or circular premises; they’d have to first walk through Lob’s Theorem in order to follow this “simplified” proof of Lob’s Theorem.
Yes to Alex that (I think) you can use an already-in-hand proof of Löb to make the self-referential proof work, and
Yes to Eliezer that that would be cheating wouldn’t actually ground out all of the intuitions, because then the “santa clause”-like sentence is still in use in already-in-hand proof of Löb.
(I’ll write a separate comment on Eliezer’s original question.)
löb = □ (□ A → A) → □ A
□löb = □ (□ (□ A → A) → □ A)
□löb -> löb:
löb premises □ (□ A → A).
By internal necessitation, □ (□ (□ A → A)).
By □löb, □ (□ A).
By löb's premise, □ A.
When disambiguating as far as possible, löb becomes □(□B → A) → □A, but □löb becomes □(□(□B → A) → □B). Perhaps Ψ has a universal property related to this?
It would kind of use assumption 3 inside step 1, but inside the syntax, rather than in the metalanguage. That is, step 1 involves checking that the number encoding “this proof” does in fact encode a proof of C. This can’t be done if you never end up proving C.
One thing that might help make clear what’s going on is that you can follow the same proof strategy, but replace “this proof” with “the usual proof of Lob’s theorem”, and get another valid proof of Lob’s theorem, that goes like this: Suppose you can prove that []C->C, and let n be the number encoding a proof of C via the usual proof of Lob’s theorem. Now we can prove C a different way like so:
n encodes a proof of C.
Therefore []C.
By assumption, []C->C.
Therefore C.
Step 1 can’t be correctly made precise if it isn’t true that n encodes a proof of C.
If that’s how it works, it doesn’t lead to a simplified cartoon guide for readers who’ll notice missing steps or circular premises; they’d have to first walk through Lob’s Theorem in order to follow this “simplified” proof of Lob’s Theorem.
Yes to both of you on these points:
Yes to Alex that (I think) you can use an already-in-hand proof of Löb to make the self-referential proof work, and
Yes to Eliezer that that would be cheating wouldn’t actually ground out all of the intuitions, because then the “santa clause”-like sentence is still in use in already-in-hand proof of Löb.
(I’ll write a separate comment on Eliezer’s original question.)
Similarly:
Noice :)
When disambiguating as far as possible, löb becomes □(□B → A) → □A, but □löb becomes □(□(□B → A) → □B). Perhaps Ψ has a universal property related to this?