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by Tainnor
547 days ago
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> Take the collection of all true statements and make that your axiomatic system. A complete proof system needs to be able to derive Γ |- φ for every pair Γ, φ such that Γ |= φ. Not just when Γ is the complete theory of some structure. Completeness of first-order logic (and its failure for second-order logic) is about the logical system itself, while the incompleteness theorems are about specific theories - people often mix these up, but they talk about very different things. > Andreas Blass in the comments says that Incompleteness does not apply to PA_2. He says something rather different, namely that its "meaningless". That's a value judgement. Incomplete proof calculi for second order logic do exist (e.g. any first-order proof calculus) and for those, what I wrote is true. Andreas Blass would probably just think of this as an empty or obvious statement. |
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However, my understanding is that the incompleteness results apply to only recursively enumerable axiomatic systems. I can find references for this. If I take the standard model of ZFC and collect all true statements in the one model of PA_2 and make that my axiomatic system then I have an axiomatic system that is not recursively enumerable and contains PA_1. It’s not a nice set of axioms. It’s not computable. But it shows that one can have an axiomatic system that contains PA_1 for which the Incompleteness theorems don’t apply.
Andreas wrote “meaningless” not “nonsensical”. I’m not a pedant but the former term evokes in me the idea of “does not apply in this situation becausethe hypotheses of the incompleteness theorem are not satisfied”.
From a mathematical logic book is the following. It’s the set up for the Incompleteness theorems.
Suppose that A is a collection of axioms in the language of number theory such that A is consistent and is simple enough so that we can decide whether or not a given formula is an element of A.
PA_2 is not such a system and as such the Incompleteness Theorems don’t apply. Maybe we are talking past each other. You know more than me.