No that’s not right. The theorem says that if PA proves “◻C → C” then PA proves C. so that’s ◻(◻C → C) → ◻C.
The flaw is that the deduction theorem does not cross meta levels. Eliezer says “Löb’s Theorem shows that, whenever we have ((◻C)->C), we can prove C.” and goes on to claim that (◻C->C)->C. But he’s intentionally failed to use quotes and mixed up the meta levels here. Lob’s Theorem does not give us a proof in first order logic from ((◻C)->C) to C. It gives us a proof that if there is a proof of ((◻C)->C) then there is a proof of C. Which is an entirely diffirent thing altogether.
Robert: “You can only say (((◻C)->C)->(◻C))”
No that’s not right. The theorem says that if PA proves “◻C → C” then PA proves C. so that’s ◻(◻C → C) → ◻C.
The flaw is that the deduction theorem does not cross meta levels. Eliezer says “Löb’s Theorem shows that, whenever we have ((◻C)->C), we can prove C.” and goes on to claim that (◻C->C)->C. But he’s intentionally failed to use quotes and mixed up the meta levels here. Lob’s Theorem does not give us a proof in first order logic from ((◻C)->C) to C. It gives us a proof that if there is a proof of ((◻C)->C) then there is a proof of C. Which is an entirely diffirent thing altogether.