...it's decidable - otherwise, it would be impossible to decide whether a proof is valid.
Isn't that the whole issue with PA_2 vs. PA? In PA_2 with "full semantics" there is no effective procedure for determining if a statement is an axiom. In my mind this is what I mean by the incompleteness results not applying to PA_2. They do apply to Z_2 since that is an effective (computable?) system.
But Z2 is usually studied with first-order semantics, and in that context it is an effective theory of arithmetic subject to the incompleteness theorems. In particular, Z2 includes every axiom of PA, and it does include the second-order induction axiom, and it is still incomplete.
Therefore, the well-known categoricity proof must not rely solely on the second-order induction axiom. It also relies on a change to an entirely different semantics, apart from the choice of axioms. It is only in the context of these special "full" semantics that PA with the second-order induction axiom becomes categorical.
...it's decidable - otherwise, it would be impossible to decide whether a proof is valid.
Isn't that the whole issue with PA_2 vs. PA? In PA_2 with "full semantics" there is no effective procedure for determining if a statement is an axiom. In my mind this is what I mean by the incompleteness results not applying to PA_2. They do apply to Z_2 since that is an effective (computable?) system.
But Z2 is usually studied with first-order semantics, and in that context it is an effective theory of arithmetic subject to the incompleteness theorems. In particular, Z2 includes every axiom of PA, and it does include the second-order induction axiom, and it is still incomplete.
Therefore, the well-known categoricity proof must not rely solely on the second-order induction axiom. It also relies on a change to an entirely different semantics, apart from the choice of axioms. It is only in the context of these special "full" semantics that PA with the second-order induction axiom becomes categorical.
https://math.stackexchange.com/questions/617124/peano-arithm...
Thanks for the knoweledge. I'm going to read up more on this stuff.