Yes, there is a problem there. In a nonstandard model it would say that G is provable, while it isn't. — alcontali
Yes, there is a problem there. In a nonstandard model it would say that G is provable, while it isn't. — alcontali
Since ~G says that G is provable then, if ~G is true, G is provable. Now ~G is true (in that model) therefore G is provable (in that model). It seems you disagree with this. Which part and why? — Andrew M
Any model in which the Gödel sentence is false must contain some element which satisfies the property within that model. Such a model must be "nonstandard" – it must contain elements that do not correspond to any standard natural number (Raatikainen 2015, Franzén 2005, p. 135). — Truth of the Gödel sentence - Wikipedia (italics mine)
Namely, if there are independent statements such as GF, F must have both models which satisfy GF and models which rather satisfy ¬GF. As ¬GF is equivalent to ∃xPrfF(x, ⌈GF⌉), the latter models must possess entities which satisfy the formula PrfF(x, ⌈GF⌉). And yet we know (because PrfF(x, y) strongly represents the proof relation) that for any numeral n, F can prove ¬PrfF(n, ⌈GF⌉). Therefore, no natural number n can witness the formula. It follows that any such non-standard model must contain, in addition to natural numbers (denotations of the numerals n), “infinite” non-natural numbers after the natural numbers. — 2.6 Incompleteness and Non-standard Models - SEP (italics mine)
On the other hand, I completely agree that in the model M, the sentence G loses its intuitive meaning. That’s the key to this business.
G is a statement that’s all about natural numbers. It’s supposed to encode the English language statement “I am unprovable”. But what it actually says is a bit more like this:
“There is no natural number n that is the number of a proof of this statement”.
(The idea, remember, is that Gödel figured out a way to assign numbers to proofs.)
However, the model M contains nonstandard natural numbers as well as the usual ones. In the model M, one of these nonstandard numbers is the number of a proof of G. So, G does not hold in the model M.
So, G doesn’t hold in M, but that’s because M has nonstandard numbers. We can loosely say that there’s a nonstandard number which is the number of an “infinitely long proof” of G. — John Baez - professor of mathematics at the University of California, Riverside
“There is no natural number n that is the number of a proof of this statement”. So, G doesn’t hold in M, but that’s because M has nonstandard numbers. We can loosely say that there’s a nonstandard number which is the number of an “infinitely long proof” of G. — John Baez - professor of mathematics at the University of California, Riverside
That sounds intriguing and intuitively correct, but unfortunately, also difficult to verify, because these nonstandard numbers are infinite cardinalities. So, yes, if there is a proof it will be encoded in one of these infinite cardinalities, which is indeed not a natural number n. — alcontali
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