5 ms·

I think that's another theorem.

by fabiospampinato 5y ago

scotty79 5y ago
Dunno, I thought you were refering to Godel's theorem.
- fabiospampinato 5y ago
  Yes I think you were referring to the first incompleteness theorem while I was referring to the second one: https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_... > The second incompleteness theorem, an extension of the first, shows that the system cannot demonstrate its own consistency.
  - scotty79 5y ago
    This one just says that any non-trivial consistent set of math doesn't contain the proof of its consistency. Where consistency means it doesn't contain proofs of falsehoods. It doesn't seem like a bug to me. Just a way to ensure that any non-trivial math has no self sufficient bubbles. That no set of math is somehow better than others. That math is free and unlimited with it's choice of axioms and rules and no subset is better than others.
    - fabiospampinato 5y ago
      I wouldn't consider it a bug either, I didn't mention it because I thought it was a bug, I mentioned it following this line of thought: "You don't have bugs in math." -> if math were inconsistent it would be buggy -> if math is inconsistent it's buggy -> if it's not inconsistent you can't prove that it's not inconsistent -> you can't tell whether math has bugs in it (unless you can find some), which is a contradiction from your statement that there are just no bugs. It's basically impossible to prove that, and if you can't prove something you are left with nothing more than doubt.
      - scotty79 5y ago
        Ah I see. But it's only the case if you limit yourself to a subset of math. Subset of math is not provably consistent if you are trying to use just itself for the proof, but this subset is provably consistent if you extend your proving toolset with something else. But the new extended subset has new, other troubles that make it impossible to prove its consistency by using just itself, so to prove its consistency you need to extend it again. You can still have fully provable (i.e. bugless) subset of math, you just need other math to prove that it is so.
        fabiospampinato 5y ago
        Other math that you can't prove it's consistent, you end up just moving the problem one step away, but it's not going away.
        scotty79 5y ago
        Yes. But you don't need the whole math to run the universe. Just a subset of your choosing. And you can go beyond that subset (to a wider god-math) to prove that this subset is consistent. And the fact that your extended god-math can't prove that it is itself fully consistent (doesn't generate any proofs of falsehoods) is not necessarily a problem. It's enough if no falsehoods were involved in the proof of the universe-math consistency. And that's entirely possible. Theorem you mentioned doesn't say anything about that.

I think that's another theorem.

scotty79 5y ago

Dunno, I thought you were refering to Godel's theorem.

fabiospampinato 5y ago
Yes I think you were referring to the first incompleteness theorem while I was referring to the second one: https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_... > The second incompleteness theorem, an extension of the first, shows that the system cannot demonstrate its own consistency.
- scotty79 5y ago
  This one just says that any non-trivial consistent set of math doesn't contain the proof of its consistency. Where consistency means it doesn't contain proofs of falsehoods. It doesn't seem like a bug to me. Just a way to ensure that any non-trivial math has no self sufficient bubbles. That no set of math is somehow better than others. That math is free and unlimited with it's choice of axioms and rules and no subset is better than others.
  - fabiospampinato 5y ago
    I wouldn't consider it a bug either, I didn't mention it because I thought it was a bug, I mentioned it following this line of thought: "You don't have bugs in math." -> if math were inconsistent it would be buggy -> if math is inconsistent it's buggy -> if it's not inconsistent you can't prove that it's not inconsistent -> you can't tell whether math has bugs in it (unless you can find some), which is a contradiction from your statement that there are just no bugs. It's basically impossible to prove that, and if you can't prove something you are left with nothing more than doubt.
    - scotty79 5y ago
      Ah I see. But it's only the case if you limit yourself to a subset of math. Subset of math is not provably consistent if you are trying to use just itself for the proof, but this subset is provably consistent if you extend your proving toolset with something else. But the new extended subset has new, other troubles that make it impossible to prove its consistency by using just itself, so to prove its consistency you need to extend it again. You can still have fully provable (i.e. bugless) subset of math, you just need other math to prove that it is so.
      - fabiospampinato 5y ago
        Other math that you can't prove it's consistent, you end up just moving the problem one step away, but it's not going away.
        scotty79 5y ago
        Yes. But you don't need the whole math to run the universe. Just a subset of your choosing. And you can go beyond that subset (to a wider god-math) to prove that this subset is consistent. And the fact that your extended god-math can't prove that it is itself fully consistent (doesn't generate any proofs of falsehoods) is not necessarily a problem. It's enough if no falsehoods were involved in the proof of the universe-math consistency. And that's entirely possible. Theorem you mentioned doesn't say anything about that.