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:
((◻C)->C)->C
However, those familiar with the logic of material implication will realize that:
(X->Y)->Y
implies
(not X)->Y
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?
Tiiba: "(Smoke implies fire, therefore fire) implies (no smoke means fire)?" "therefore fire" should be left out. i) (Smoke implies fire) implies fire. ii) No smoke implies fire.
You're confused because i) is false.
Medium puzzle:
"Applying the Deduction Theorem to Löb's Theorem gives us, for all C:
Wouldn't it be: ((?C) -> C) -> ?C If the provability of C implies C's truth, then C is provable.
In other words, where you write "(X->Y)->Y", it should be "(X -> Y) -> X".