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Let $\def\N{\mathcal{N}}$ $\N$ be some (powerful enough) theory of arithmetic. Is there an effective procedure to find a Gödel sentence $A$ in $\N$? That is, a sentence for which $$\vdash A \iff \neg P(A).$$ To be clear, this means building a computer program that runs through all sentences in $\N$, expressed as finite strings of symbols in the language of $\N$, until the program halts at $A$.

Naively it seems that any standard proof of Gödel's First Incompleteness Theorem could give us such a procedure:

  • Step 1: use Gödel numbers to define an effectively computable predicate $P$ for which $P(X)$ is true if and only if $X$ is provable in $\N$.
  • Step 2: let $A_1(n), A_2(n), \ldots$ be an effectively computable enumeration of all formulas having exactly one free variable in $\N$.
  • Step 3: search through the above list until you find the sentence $A_k(n) = \neg P(A_n(n))$. This is possible as both $P$ and $n\mapsto A_n$ are effectively computable.

Then it is easy to show that $A_k(k)$ is a Gödel sentence. Am I missing something, or is this indeed true?

Sam
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    You don't even need to search for it. The incompleteness theorem gives an explicit construction of the Gödel sentence. – spaceisdarkgreen Oct 08 '24 at 20:56
  • While technically your definition of 'effective' is equivalent to any other, I would say 'program that runs through all sentences in $\mathcal{N}$ until it halts at $A$' is not the usual way of doing such a thing. Rather, I would expect something that given an encoding of a theory $T$ produces a number / string (they are of course equivalent) representing the sentence $A$. – Steven Stadnicki Oct 08 '24 at 20:58
  • I think I'm repeating the comment by @StevenStadnicki in other words, but "run through all sentences in $\mathcal{N}$" is not necessary. The algorithm or computer program just needs to "result in" / "output" / "halt at" $A$, by any means. – aschepler Oct 08 '24 at 21:06
  • @spaceisdarkgreen do you mean the sentence $P(0 = 1)$? – Sam Oct 08 '24 at 21:17
  • @TankutBeygu I will look into it! – Sam Oct 08 '24 at 21:18
  • @Sam No, I mean the regular Gödel sentence that says "I'm not provable", although it's equivalent to the negation of that over strong enough theories. – spaceisdarkgreen Oct 08 '24 at 21:24
  • @spaceisdarkgreen I worry I'm not understanding then. You say "the incompleteness theorem gives an explicit construction of the Gödel sentence". Do you mean that the proof of the theorem gives an explicit construction? Otherwise it seems to me that if the proof of the theorem was non-constructive we would merely know there exists some Gödel sentence $A$, but know which of the infinitely many sentences in $\mathcal{N}$ is $A$. – Sam Oct 08 '24 at 21:31
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    @Sam Yes, I mean the proof is fully constructive. By the way, someone appears to have actually attempted to write it out long-hand though I'm of course completely unable to discern its correctness https://math.stackexchange.com/a/1484929 (and there are various subtleties... e.g. I expect just by the quantifier depth, this sentence wouldn't work for very weak theories like $\sf Q$... it's probably for $\sf PA$. ) – spaceisdarkgreen Oct 08 '24 at 21:32

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