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Godel proved that any system expressive enough to produce an arithmetic is incomplete. He initially proved it for the peano axioms but then it got generalized.

Godel proved that any system expressive enough to produce an arithmetic is incomplete. He initially proved it for the peano axioms but then it got generalized. ZFC can produce an arithmetic. Also, before being arrogant and demanding explanations, you should give them first for your claims

andriy_koval 12d ago
> expressive enough to produce you understand that "expressive enough to produce" are not obvious elements of zfc, that's some average consumer napkin math and not strict formalization.
- Almondsetat 12d ago
  why should they be obvious? they are derived and have been thoroughly proven.
  - andriy_koval 12d ago
    looks like we are in disagreement
    - Almondsetat 12d ago
      A quick google search shows different proof assistants have been used to obtain the Peano axioms from ZFC, such as Isabelle/ZF and Metamath. I think you're just wrong
      - cdelsolar 12d ago
        What are you nerds fighting about please explain
        deleted 12d ago
        [deleted]
      - andriy_koval 12d ago
        you are entitled to have your opinion :-)
        Almondsetat 12d ago
        and you are entitled to talk about maths while rejecting maths
        andriy_koval 12d ago
        coming back to your argument about peano being obtained from zfc, you obviously can't prove that it happened using purely zfc, and not some logical framework embedded into those proof assistants. I said I am not expert, I am indeed not expert in zfc and godel theorems, but I am an expert (phd) in actual formalization theory. Formal theory is very simple concept: its alphabet, set of formulas on top of this alphabet, and function which translates one formula to another. ZFC can't "obtain" peano, simply because it doesn't have say * operator defined. You need to do something on top of it. Additionally, zfc itself looks like loosely formalized say in wikipedia (and I am not sure if there is any strict formalization anywhere), we take it as common sense that it can utilize some simple logical rules (e.g. modus ponens), but what are exactly rules, which could be separate topic of research, this detail is skipped.
        Smaug123 12d ago
        Eh? Any first course in set theory will present ZFC as a one-sorted theory with ten axioms (/schemas) in first order logic (inheriting an equality symbol, forall, implies etc) with one binary predicate (namely set membership), or will present a theory that is equiconsistent with a usual ZFC presentation. Honestly I’m not sure how you simultaneously claim to be a PhD in formalisation and also not be aware of the existence of Isabelle/ZF, for example.
        andriy_koval 11d ago
        > Honestly I’m not sure how you simultaneously claim to be a PhD in formalisation and also not be aware of the existence of Isabelle/ZF, for example. I am aware, also I am not sure why you wrote all of this. Your unknown to me "first course" claims to be some authority of formalization purity?
    - jibal 12d ago
      That increases the likelihood that they are right. > support your point with explanation or be ignored :-) Anyone who says "Godel theorems are for systems with basic arithmetic, zfc doesn't include arithmetic, thus are not object of Godel theorems" and isn't joking warrants a permanent ignore. https://math.stackexchange.com/questions/1366560/why-does-g%C3%B6dels-first-incompleteness-theorem-apply-to-zfc https://math.stackexchange.com/questions/1366560/why-does-g%... https://math.stackexchange.com/questions/1090437/how-to-prove-that-g%C3%B6dels-incompleteness-theorems-apply-to-zfc/1090755#1090755 https://math.stackexchange.com/questions/1090437/how-to-prov...
      - andriy_koval 12d ago
        imo, those two links are example of rather low quality weird math discussions, but you can keep your opinion
      - jibal 9d ago
        > imo, those two links are example of rather low quality weird math discussions, but you can keep your opinion I've seen a lot of bad faith on this site, but none exceeding that.

andriy_koval 12d ago

> expressive enough to produce you understand that "expressive enough to produce" are not obvious elements of zfc, that's some average consumer napkin math and not strict formalization.

Almondsetat 12d ago
why should they be obvious? they are derived and have been thoroughly proven.
- andriy_koval 12d ago
  looks like we are in disagreement
  - Almondsetat 12d ago
    A quick google search shows different proof assistants have been used to obtain the Peano axioms from ZFC, such as Isabelle/ZF and Metamath. I think you're just wrong
    - cdelsolar 12d ago
      What are you nerds fighting about please explain
      - deleted 12d ago
        [deleted]
    - andriy_koval 12d ago
      you are entitled to have your opinion :-)
      - Almondsetat 12d ago
        and you are entitled to talk about maths while rejecting maths
        andriy_koval 12d ago
        coming back to your argument about peano being obtained from zfc, you obviously can't prove that it happened using purely zfc, and not some logical framework embedded into those proof assistants. I said I am not expert, I am indeed not expert in zfc and godel theorems, but I am an expert (phd) in actual formalization theory. Formal theory is very simple concept: its alphabet, set of formulas on top of this alphabet, and function which translates one formula to another. ZFC can't "obtain" peano, simply because it doesn't have say * operator defined. You need to do something on top of it. Additionally, zfc itself looks like loosely formalized say in wikipedia (and I am not sure if there is any strict formalization anywhere), we take it as common sense that it can utilize some simple logical rules (e.g. modus ponens), but what are exactly rules, which could be separate topic of research, this detail is skipped.
        Smaug123 12d ago
        Eh? Any first course in set theory will present ZFC as a one-sorted theory with ten axioms (/schemas) in first order logic (inheriting an equality symbol, forall, implies etc) with one binary predicate (namely set membership), or will present a theory that is equiconsistent with a usual ZFC presentation. Honestly I’m not sure how you simultaneously claim to be a PhD in formalisation and also not be aware of the existence of Isabelle/ZF, for example.
        andriy_koval 11d ago
        > Honestly I’m not sure how you simultaneously claim to be a PhD in formalisation and also not be aware of the existence of Isabelle/ZF, for example. I am aware, also I am not sure why you wrote all of this. Your unknown to me "first course" claims to be some authority of formalization purity?
  - jibal 12d ago
    That increases the likelihood that they are right. > support your point with explanation or be ignored :-) Anyone who says "Godel theorems are for systems with basic arithmetic, zfc doesn't include arithmetic, thus are not object of Godel theorems" and isn't joking warrants a permanent ignore. https://math.stackexchange.com/questions/1366560/why-does-g%C3%B6dels-first-incompleteness-theorem-apply-to-zfc https://math.stackexchange.com/questions/1366560/why-does-g%... https://math.stackexchange.com/questions/1090437/how-to-prove-that-g%C3%B6dels-incompleteness-theorems-apply-to-zfc/1090755#1090755 https://math.stackexchange.com/questions/1090437/how-to-prov...
    - andriy_koval 12d ago
      imo, those two links are example of rather low quality weird math discussions, but you can keep your opinion
    - jibal 9d ago
      > imo, those two links are example of rather low quality weird math discussions, but you can keep your opinion I've seen a lot of bad faith on this site, but none exceeding that.