Psy-Kosh: There are two points of view of where the flaw is.
My point view is that the flaw is here:
"Löb's Theorem shows that, whenever we have ((◻C)->C), we can prove C. Applying the Deduction Theorem to Löb's Theorem gives us, for all C: ((◻C)->C)->C"
Because, in fact Lob's Theorem is: (PA |- "◻C -> C") => (PA |- "C") and the Deduction theorem says (PA + X |- Y) => (PA |- "X->Y"). We don't have PA + X proving anything for any X. The deduction theorem just doesn't apply. The flaw is that the informal prose just does not accurately reflect the actual math.
Eliezer's point of view (correct me if I'm wrong) is that in the cartoon, we have 10 steps all of the form "PA proves ...." They each follow logically from the ones that came before. So they look like a proof inside of PA. And if they were a proof inside of PA then the deduction theorem would apply, and his contradiction would go through. So the flaw is that while all of the steps are justified, one of them is only justified from outside of PA. And that one is step 8.
Both of these views are correct.
Brian Jaress:
I think if you used Lob's Theorem correctly, you'd have something like: if PA |- []C -> C then PA |- C [Lob] PA |- ([]C -> C) -> C [Deduction]
This is incorrect because the if-then is outside of PA. The deduction theorem does not apply.
Lo! A cartoon proof of Löb's Theorem!
Löb's Theorem shows that a mathematical system cannot assert its own soundness without becoming inconsistent. Marcello and I wanted to be able to see the truth of Löb's Theorem at a glance, so we doodled it out in the form of a cartoon. (An inability to trust assertions made by a proof system isomorphic to yourself, may be an issue for self-modifying AIs.)
It was while learning mathematical logic that I first learned to rigorously distinguish between X, the truth of X, the quotation of X, a proof of X, and a proof that X's quotation was provable.
The cartoon guide follows as an embedded Scribd document after the jump, or you can download as a PDF file. Afterward I offer a medium-hard puzzle to test your skill at drawing logical distinctions.
Cartoon Guide to Löb's ... by on Scribd
Cartoon Guide to Löb's Theorem - Upload a Document to Scribd
And now for your medium-hard puzzle:
The Deduction Theorem (look it up) states that whenever assuming a hypothesis H enables us to prove a formula F in classical logic, then (H->F) is a theorem in classical logic.
Let ◻Z stand for the proposition "Z is provable". Löb's Theorem shows that, whenever we have ((◻C)->C), we can prove C.
Applying the Deduction Theorem to Löb's Theorem gives us, for all C:
However, those familiar with the logic of material implication will realize that:
Applied to the above, this yields (not ◻C)->C.
That is, all statements which lack proofs are true.
I cannot prove that 2 = 1.
Therefore 2 = 1.
Can you exactly pinpoint the flaw?