Let's now look at A: . It must hold true for every sentence s. — alcontali
Where have you used the assumption that f is computable? — fdrake
It is easy to show via the construction of Prov('s') that s -> Prov('s') when s is a theorem. — sime
Do you think it works? — alcontali
for every computable function f that takes a number as an argument and returns false or true, — alcontali
there exists some computable function f with the special property that it takes the godel number of any 'gappy' sentence and returns the godel number of the same sentence eating itself,
When you consider A, I think you can omit the s1 discussion and just note that A is an instance of the diagonal lemma that is (true,true) for function ~f. So, per the negated lemma, A can't be true. — Andrew M
But that still leaves disjunct B that could be true. So you would also need to show a case where B fails. — Andrew M
returns the godel number of the same sentence eating itself
Is your f really Γ, the "graph" predicate assumed available to "represent" (presumably like the way points on a 2d coordinate graph represent a relation of ordered pairs) any computable function and therefore f? — bongo fury
this somewhat bewildering and disorienting critique of wikipedia and everyone else — bongo fury
Because the Gödel numbering function is a function in the meta-language to the system T, there cannot be any such relation δ in the system T that contains the information of the definition of the Gödel numbering function and which can unambiguously reference formulas of its own formal system T.
What you just wrote, is indeed the gist of it. I wonder if the proof can just be phrased like that? Would it still be considered a proof? — alcontali
Concerning the expression in B, instead of being true for all true sentences, it will be true for all false sentences. So, B is then an instance of the diagonal lemma that is (false,false) for function ~f.
The whole point of the proof is that the practice of requiring that a sentence stays off the diagonal for f will always force the sentence onto the diagonal for ~f. This is the case for both true and false sentences. — alcontali
I think that the main issue left now, is to come up with a succinct way of phrasing this principle. — alcontali
The proposition that there exists a Γ for which Γ(x,y=f(x)) -> true and Γ(x,y≠f(x)) -> false, is tautological for any (computable) function f. But then again, this proposition is needed just for the original proof, and not for this one. — alcontali
Then, the diagonal lemma says that:
∀f ∈ F:N→{false,true} ∃s,t ∈ S: s ↔ f(⌜s⌝) ∧ ¬t ↔ f(⌜t⌝) — alcontali
[/quote]I don't see how you are addressing anything like the same claim, e.g.,
Any symbol system that can prove all arithmetic proves at least one liar sentence. — bongo fury
Let T be a first-order theory in the language of arithmetic and capable of representing all computable functions. Let F be a formula in the language with one free variable, then: There is a sentence ψ such that ψ ↔ F(°#(ψ)) is provable in T.
Why shouldn't we assume that at least one valuation (call it f and even interpret it as the predicate "is false") creates a complete set of mis-matches (true-false and false-true) of s to f(s) while its opposite (~f = g) makes a complete set of matches? — bongo fury
Why shouldn't we assume that at least one valuation (call it f and even interpret it as the predicate "is false") creates a complete set of mis-matches (true-false and false-true) of s to f(s) while its opposite (~f = g) makes a complete set of matches? — bongo fury
Toward a Big Surprise.Tarski’s Theorem is equivalent with the Semantic Diagonal Lemma. — Saeed Salehi
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