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Apr 6, 2020 at 18:37 history edited Tim Campion CC BY-SA 4.0
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Apr 6, 2020 at 18:11 history edited Tim Campion CC BY-SA 4.0
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Apr 6, 2020 at 15:59 comment added Emil Jeřábek No, being consistent in $\omega$-logic is a much stronger property than $\omega$-consistency. (The terminology is fairly confusing.) “Sound” is simply overloaded; in the context of theories of arithmetic (and perhaps other theories with a clearly distinguished “standard” model), it means “true in the standard model”, this is unrelated to the notion of a derivation system being sound with respect to a semantics.
Apr 6, 2020 at 15:52 comment added Tim Campion @EmilJeřábek Thanks so much! I suppose the terminology "$\omega$-consistent" makes some sense -- it looks like a theory is $\omega$-consistent iff it is consistent in $\omega$-logic. But the terminology "sound" for "weakening of true arithmetic" strikes me as strange -- I'm used to "soundness" being a property of a logic rather than a property of a theory.
Apr 6, 2020 at 15:30 comment added Emil Jeřábek See e.g. en.wikipedia.org/wiki/ω-consistent_theory for more background.
Apr 6, 2020 at 15:23 comment added Emil Jeřábek The $\omega$-rule generates all of true arithmetic, yes, but you need repeated applications of the $\omega$-rule. Therefore your conclusion that any theory of arithmetic that is not sound (weakening of true arithmetic, as you call it) must prove $\exists x\,\neg\phi(x)$ and all instances $\phi(n)$ , $n\in\omega$, for some formula $\phi(x)$, is incorrect. Theories such that no such $\phi$ exists are known as $\omega$-consistent, and there exist unsound $\omega$-consistent theories.
Apr 6, 2020 at 14:27 history edited Tim Campion CC BY-SA 4.0
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S Mar 31, 2020 at 18:51 history answered Tim Campion CC BY-SA 4.0
S Mar 31, 2020 at 18:51 history made wiki Post Made Community Wiki by Tim Campion