4 ms·

I didn’t read that part but the statement is still wrong. Take as your model the standard model of integers. Collects all true statements of this model. Make

by skh 8y ago

I didn’t read that part but the statement is still wrong. Take as your model the standard model of integers. Collects all true statements of this model. Make this your axiom system. It is now complete and contains the first order Peano Axioms.

gnulinux 8y ago
Then that system is not finitely axiomatizable (there are infinitely many truths in that model) so the original claim is still true. It's true for all finitely axiomatized theories.
- skh 8y ago
  The author of the article does not stipulate that the axiom system has to be recursively enumerable. From the article: We assume it to contain the basic Peano axioms of arithmetic. An axiom system is complete, if every true statement can be proven within the system. The theorem in the article is true for recursively enumerable axioms systems that contain first order arithmetic. EDIT: Note that the second order Peano Axioms are categorical but not recursively enumerable. This is also an axiomatic system that has finitely many axiom schema.
- antidesitter 8y ago
  Did the author specify “finitely axiomatizable”? Because neither Peano arithmetic nor ZFC are finitely axiomatizable, for example. Also, “not finitely axiomatizable” doesn’t mean “having infinitely many truths in the (?) model”. Not sure what you meant by that.
  - gnulinux 8y ago
    > Also, “not finitely axiomatizable” doesn’t mean “having infinitely many truths in the (?) model”. Not sure what you meant by that. If you take all truths in the standard model of arithmetic, and make them your axioms you'll get a complete and consistent system with infinitely many axioms, and an extension of Q. If one uses Roser's trick, Godel says a system cannot be all of these at the same time: * a conservative extension of Q ("minimal amount of arithmetic") * finitely axiomatizable * complete * consistent It is ok to have any <=3 of these 4. Note that PA is a conservative extension of Q. So you can replace Q with PA above. Godel's original proof doesn't generalize this much but using Roser's trick Godel immediately implies this.
    - pulisse 8y ago
      Nit: The relevant notion is not finite axiomatizability but recursive axiomatizability.
- andrepd 8y ago
  Not finite, but recursively enumerable, no? Since you can just start from PA and derive conclusions forever, so this is an effective procedure in that any true statement will eventually be thus produced.
  - krusch 8y ago
    No, even if you start with the axioms of PA and enumerate all theorems provable, you'll miss some true sentences, one of them being the Goedel sentence of PA.

by skh 8y ago

gnulinux 8y ago

Then that system is not finitely axiomatizable (there are infinitely many truths in that model) so the original claim is still true. It's true for all finitely axiomatized theories.

skh 8y ago
The author of the article does not stipulate that the axiom system has to be recursively enumerable. From the article: We assume it to contain the basic Peano axioms of arithmetic. An axiom system is complete, if every true statement can be proven within the system. The theorem in the article is true for recursively enumerable axioms systems that contain first order arithmetic. EDIT: Note that the second order Peano Axioms are categorical but not recursively enumerable. This is also an axiomatic system that has finitely many axiom schema.
antidesitter 8y ago
Did the author specify “finitely axiomatizable”? Because neither Peano arithmetic nor ZFC are finitely axiomatizable, for example. Also, “not finitely axiomatizable” doesn’t mean “having infinitely many truths in the (?) model”. Not sure what you meant by that.
- gnulinux 8y ago
  > Also, “not finitely axiomatizable” doesn’t mean “having infinitely many truths in the (?) model”. Not sure what you meant by that. If you take all truths in the standard model of arithmetic, and make them your axioms you'll get a complete and consistent system with infinitely many axioms, and an extension of Q. If one uses Roser's trick, Godel says a system cannot be all of these at the same time: * a conservative extension of Q ("minimal amount of arithmetic") * finitely axiomatizable * complete * consistent It is ok to have any <=3 of these 4. Note that PA is a conservative extension of Q. So you can replace Q with PA above. Godel's original proof doesn't generalize this much but using Roser's trick Godel immediately implies this.
  - pulisse 8y ago
    Nit: The relevant notion is not finite axiomatizability but recursive axiomatizability.
andrepd 8y ago
Not finite, but recursively enumerable, no? Since you can just start from PA and derive conclusions forever, so this is an effective procedure in that any true statement will eventually be thus produced.
- krusch 8y ago
  No, even if you start with the axioms of PA and enumerate all theorems provable, you'll miss some true sentences, one of them being the Goedel sentence of PA.