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Let M be a proof-checking Turing machine which takes two inputs, A and B. :

M(A,B) = 0 if A codes a valid proof of the sentence coded by B in ZFC.

M(A,B) = 1 if A does not code a valid proof of the sentence coded by B in ZFC.


Let M' be a provability-checking Turing machine which takes input B. :

M'(B) = 0 if B is provable in ZFC.

M'(B) does not terminate if B is not provable in ZFC.


My question: If M(X,B) = 0 then, M'(X) = 0 always?

It means, a valid proof is always provable?

share|cite|improve this question
By the definition of formal proof, yes. One can assume that (among other things) the "machine" $M'$, which for many theories cannot be a Turing machine, generates all sequences that are formal proofs, and verifies for each squence whether our sentence is the last entry in the proof. – André Nicolas Mar 21 '13 at 21:56
I edited that it is on ZFC. Now, the M' can be Turing machine? – HoCheol SHIN Mar 21 '13 at 22:08
On the understanding that on many inputs (refutable sentences, and more) $M'$ does not terminate, sure, since ZFC is recurisvely axiomatizable. – André Nicolas Mar 21 '13 at 22:14

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