Sorry in advance, the following is quite messy, I didn’t find a way to express myself more clearly and rigorously. I would appreciate suggestions on how to make the following better and try to act upon them as soon as possible. I am by no means claiming my argument is correct, in fact my question is: in what ways is it flawed?

The following notation and terminology is taken from Boolos’s book: “Logic, logic and logic” (chapter 26: a new proof of the Gödel incompleteness theorem, page 383). A sketch is also found in the wikipedia page “Proof sketch for Gödel’s first incompleteness theorem". Here I'll outline some definitions used by Boolos in his proof.

A formula F names the number n iff the following is provable: $$\forall x(F(x)\leftrightarrow x=[n])$$ The predicate $Cxz$ is defined to be true iff an arithmetic formula containing z symbols names the number x. The following predicates are based on $Cxz$: $$Bxy ↔ ∃z(z < y ∧ Cxz)$$$$ Axy ↔ ¬Bxy ∧ ∀a(a < x→Bay)$$$$Fx ↔ ∃y((y = [10] × [k]) ∧ Axy)$$

I would like to make a premise. In my understanding, you can introduce new functions/symbols to create a new formal system (that still behaves exactly like the old one), but with them come new axioms: if you introduce the function/symbol “$+$” and integrate it in your Gödel numbering, you need to introduce some more axioms, such as the couple “$x + 0 = x, x + S(y) = S(x + y)$”, so that you have a function/algorithm to “move” from those axioms and a Gödel number encoding a formula containing just old symbols to another equivalent formula where the symbol “$+$” is encoded (and vice versa). Those axioms have to have a Gödel number. I think at least one axiom needed to define the symbol $A$ would have a Gödel number with an infinite number of digits, therefore not a natural number. 

The problem I have with Boolos’s arguments is it seems to me that the formula $F$ is going to instantiate itself infinite times, and therefore its Gödel number would have an infinite number of digits and not be a natural number. I think it is implicitly assumed that the symbol/predicate $A$ can be constructed by using a finite number of the other symbols, and I think this is not true. It’s value seems not definable because of infinite recursion. If so, the proof would be a circular argument: it is assumed the decidability of $A$ (which leads to contradiction) to deduce the decidability of F (which leads to contradiction), but then is deduced F as true and not provable instead of deducing the truth value of $A$ to be not decidable (because $A$ has to be defined by an infinite number of symbols).

You can't say to have a general definition of $A$ until you are able to determine the truth value of any of its instantiations $Axy$. For some $x$ and $y$, the truth value of $Axy$ has to be defined somehow, but its definition/algorithm is necessarily self referential, and falls into infinite recursion. It has to check for number namings (using “name” as defined by Boolos) for every possible formula of less than $y$ symbols, including the symbol $A$, including the possible instantiations of $F$ you can have in less than $y$ symbols. A necessary condition for the truth of $Axy$ is that every single possible formula with less than y symbols does not name $x$, and that has to be verified by the algorithm so that a truth value of $Axy$ can be computed. If the algorithm did not check for formulas containing the symbol $A$, the truth value of the resulting formula would say something about a system of axioms where $A$ is not defined, not about the current one. And it cannot simply be assumed that the old and new system behave the same way anyways: this is something to be achieved as a consequence of a suitable definition of the new symbols. So, $A$ instantiates itself when it checks for instantiations of $F$.

Now, you need to be able to “move” from these formulas containing the symbol $A$ to equivalent ones containing just old symbols, but substituting $A$ in its definition again obviously leads to infinite recursion. Therefore, $A$ cannot be defined by a finite number of symbols that are not $A$, and there is no set of axioms that can move an infinite formula to a finite one in a finite number of deduction steps, therefore $A$ cannot be defined and so the truth value of $A$ cannot be determined, and it seems to me the contrary is assumed in Boolos’s proof.

I feel like this is something along the lines of asking whether the infinite formula “$0=0+0+0+…$” is true. There is no way to exclude a “$1$” will pop up at some point if the number of symbols is not finite. I would say that truth is fundamentally tied to computability in a finite time.

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    $\begingroup$ "You can't say to have a general definition of $𝐴$ until you are able to determine the truth value of any of its instantiations $𝐴𝑥𝑦$." No, that has nothing to do with it. $A$ is a formula, defined completely syntactically. (Fine: Boolos doesn't give an explicit syntactic definition of $C$, which is part of the definition of $A.$ But he sketches out a construction of the formula.) $\endgroup$ Oct 17 '20 at 1:32
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    $\begingroup$ Same misunderstanding of your previous post $\endgroup$ Oct 17 '20 at 7:41
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    $\begingroup$ @AlessandroM.Agostinelli It does have more symbols than $A,$ but I don't see how that's a problem. Boolos only claims it has less than 10 times more, and demonstrates this by counting the symbols. $\endgroup$ Oct 17 '20 at 15:05
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    $\begingroup$ Formula $F$ is not a new symbol of the language of arithmetic. It is an "abbreviation" used in the discussion about the language of arithmetic. In the language of arithmetic the "name" for the number five is not $5$ but $sssss0$. In order to simplify the discussion, the author introduce the abbreviation $[5]$ that must be read "$sssss0$. $\endgroup$ Oct 19 '20 at 12:11
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    $\begingroup$ Thus, Boolos introduces the definition: a formula of the language of arithmetic names number $[n]$ iff $\forall x (F(x) \leftrightarrow x= [n])$. Due to the fact that formula $x+x=ssss0$, we have that formula $(x+x=ssss0)$ names the number four. $\endgroup$ Oct 19 '20 at 12:18

These predicates $A,B,C,$ and $F$ are not new symbols that are being added to the language. Instead, they are merely abbreviations: when we talk about $Axy$, that is merely a shorthand for $¬Bxy ∧ ∀a(a < x→Bay)$, which is in turn itself a shorthand using the definition of $B$. So, there are no formulas including the symbol $A$ that you have to worry about when defining $A$; the language that is used for the formulas referred to in $C$ is just the ordinary language of arithmetic.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Oct 28 '20 at 12:16

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