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.
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.
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.
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.
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.
> 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.
