A re-definition of {analytic} that seems to overcome ALL objections that anyone can possibly have Before the reply to my post, I deleted "To see that, you just need to read the article that you yourself say is "clear and accurate"", as I thought it would be better not to invite referencing that article again.
But to address that article, here are the proofs without ""liar"" (scare quotes in original), "ask", "truth bearer" or anything else extraneous to the mathematical proofs:
In this context, 'formula' and 'sentence' mean 'formula in the language of first order arithmetic' and 'sentence in the language of first order arithmetic'.
In this context, 'true' and 'false' mean 'true in the standard model for the language of first order arithmetic' and 'false in in the standard model for the language of first order arithmetic'.
For every formula M, let g(M) be the numeral for the Godel number of M.
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Theorem: There is no formula T(x) such that for every sentence S, T(g(S)) is true if and only if S is true.
Proof:
Toward a contradiction, suppose there is such a T(x).
So, there is a formula D(x) such that for every numeral m, D(m) is true if and only if m is the numeral for the Godel number of a formula P(x) such that P(m) is false. (The steps in obtaining this line from the previous line are not included in the article.)
D(g(D(x))) is true
if and only if
g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.
Toward a contradiction, suppose D(g(D(x))) is true.
So g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.
g(D(x)) is g(P(x)), so D(x) is P(x), so D(g(S(x))) is P(g(S(x))), so D(g(S(x))) is false. Contradiction.
Toward a contradiction, suppose D(g(D(x))) is false.
So it is not the case that g(D(x)) is the numeral for the Godel number of a formula P(x) such that P(g(D(x))) is false.
So D(g(D(x))) is true. Contradiction.
So there is no formula T(x) such that for every sentence S, T(g(S)) is true if and only if S is true.
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Theorem: There is no formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.
Proof:
Lemma: For every formula P(x) there is a sentence D such that D <-> P(g(D)) is true.
Toward a contradiction, suppose there is a formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.
So, for every sentence S, S <-> T(g(S)) is true.
By the lemma, there is a sentence D such that D <-> ~T(g(D)) is true. But also, D <-> T(g(D)) is true. Contradiction.
So there is no formula T(x) such that for every sentence S, S is true if and only if T(g(S)) is true.