simon:
To argue that a proof is being made concluding ?C using the assumption ?(◻C -> C) given the theory PA, to which proof we can apply the deduction theorem to get (PA |- "?(◻C -> C) -> ?C") (i.e. my interpretation of Löb's Theorem)
OK so the question marks are boxes right? In that case then yes, PA |- "?(?C -> C) -> ?C". This is OK. The contradiction comes if PA |- "(?C->C)->C". But morally this doesn't have anything to do with the deduction theorem. PA proves Lob because everything in the proof of Lob is expressible inside of PA.
Like I said before, the deduction theorem is really just a technical lemma. If I'm doing ordinary mathematics (not logic), and I assume X, and prove Y, and then say "ok well now I've proved X -> Y", then I have not used the deduction theorem, because I'm writing a proof, not explicitly reasoning about proofs. The deduction theorem lies a meta level up, where we have a explicit, specific, technical definition of what constitutes a proof, and we are trying to prove theorems from that definition.
But the proof uses an additional assumption which is the antecedent of an implication, and comes to a conclusion which is the consequent of the implication. To get the implication, we must use the deduction theorem or something like it, right?
Nope, we are using an ordinary principle of mathematical reasoning. The deduction theorem says that if you have a proof that uses this principle and is otherwise first-order, you can convert it into a pure first order proof.
Is this fact a theorem of first order logic without any additional assumptions, or is it merely a theorem of PA? I admit I don't know, as I'm not very familiar with first order logic, but it intuitively seems to me that if first order logic were powerful enough on its own to express concepts like "PA proves X" it would probably be powerful enough to express arithmetic, in which case the qualification in Gödel's theorem that it only applies to theories that express arithmetic would be superfluous.
First order logic without any additional assumptions can't even express concepts like like PA. So, yea; that's why Gödel's theorem has that qualification, because there are plenty first order theories that are simple enough that they can't express integers.
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?