J Thomas:
Once more through the mill. If PA proves that 6 is a prime number, then 6 is really a prime number. Can you prove that if PA proved 6 was a prime number then 6 would be a prime number? How would you do it?
If PA |- "forall x y . x y = 6 => |x|=1 || |y|=1" then N |= "forall x y . x y = 6 => |x|=1 || |y|=1" (N = the natural numbers equiped with + and ) so for all x and y in N, N |= ",x ,y = 6 => |,x|=1 || |,y|=1" (where ,x means a constant symbol for x) if xy = 6 then N |= ",x ,y = 6" so therefore N |= "|,x|=1 || |,y|=1" thus either N |= "|,x| = 1" or N |= "|,y| = 1" thus either |x|=1 or |y|=1 therefore we have that if x*y = 6 then either |x| = 1 or |y| = 1 therefore 6 is prime therefore if PA |- "6 is prime" then 6 is actually prime
Of course it is also a meta-theorem that for any sentence phi in the language of PA that
ZF |- "PA |- phi => phi_omega"
where phi_omega is phi relativeized to the finite ordinals.
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?