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Did the author specify “finitely axiomatizable”? Because neither Peano arithmetic nor ZFC are finitely axiomatizable, for example. Also, “not finitely axiomati
by 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 agoNit: The relevant notion is not finite axiomatizability but recursive axiomatizability.