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Jul 3, 2022 at 4:55 comment added Double Knot @NoahSchweber thx for your critique for my above comment made 5 months ago! In Q ∀x(x+0=x) is axiom 4, but a famous result is Q⊬∀x(0+x=x) by simply adding 2 nonstandard members satisfying certain relations which can quickly demo incompleteness of Q and surprising for people not familiar with 1st order arithmetic incompleteness (generic group theory incompleteness is easier to understand for Abelian/non-Abelian models). Thus for natural numbers satisfying commutativity I said it's easy to prove ∀x∈N(0+x=x). The point is in Q/PA it's more than natural numbers thus UG needs applied with care...
Jul 3, 2022 at 2:54 comment added Noah Schweber being able to prove $\forall xP(x)$. I'm not really sure how what you've written is related to UG at all.
Jul 3, 2022 at 2:53 comment added Noah Schweber @DoubleKnot "Robinson Q can easily prove ∀x∈N(0+x=x)" First of all, the expression "∀x∈N(0+x=x)" isn't a first-order sentence at all, so it can't even be stated in Robinson arithmetic as usually construed. (And Godel's incompleteness theorem doesn't apply to logics strong enough to state "∀x∈N(0+x=x)" in the first place.) So while what you've written isn't technically correct, in my opinion it's rather misleading. More substantively, you're not describing a failure of UG at all. A failure of UG would consist of Q being able to prove, in some sense, a formula $P(x)$ but not (cont'd)
Jul 2, 2022 at 2:50 answer added emesupap timeline score: 1
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Jun 1, 2022 at 21:00 history tweeted twitter.com/StackPhilosophy/status/1532104705282908160
Jun 1, 2022 at 20:02 history edited J D CC BY-SA 4.0
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Feb 2, 2022 at 5:19 comment added Double Knot Your suspect of UG's applicability in non-trivial abstract theories such as the famous Robinson arithmetic Q is not wrong. Surprisingly due to nonstandard models per Godel's incompleteness Q⊬∀x(0+x=x), though Robinson Q can easily prove ∀x∈N(0+x=x)! Indeed for such inductively axiomatized abstract system, for UG to work we need to add an infinitary ω-rule of ω-logic, then per model theory's omitting type theorem a theory has an ω-model iff it is consistent in ω-logic...
Feb 2, 2022 at 4:18 comment added RodolfoAP Because in final terms, all objects are abstract; objects only exist in reason, not necessarily physically. Descartes' cogito ergo sum means not that he's a physical object (in fact, he doubts about the existence of an external world and its things), but that he perceives himself as an abstract object, by means of thinking, therefore he exists... for himself.
Feb 1, 2022 at 19:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jan 2, 2022 at 17:41 answer added David Gudeman timeline score: 0
Dec 29, 2021 at 23:14 comment added user1578232 @Conifold Thank you for your comment, I would appreciate if you would expand it to an answer, since I think I would benefit from a more detailed version of your comment. Especially, if you could mention what you mean by manipulated symbols, explain why it makes no difference whatsoever and giving more insight into your last sentence, I would be very greatful.
Dec 29, 2021 at 22:31 comment added Conifold To avoid this difficulty, think of it in terms of manipulated symbols, not what they represent. Universal generalization works because arguing with not otherwise specified symbol simply unfolds the rules for manipulating formulas with ∀ into a more parsed form. What the symbols happen to represent, concrete, abstract or nothing at all, makes no difference whatsoever. For justifying the analogy of numbers to concrete objects and applying ∀ to them see abstract objects, but that has little to do with universal generalization in particular.
Dec 29, 2021 at 15:06 history edited user1578232
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Jan 2, 2022 at 17:42
S Dec 29, 2021 at 14:47 history asked user1578232 CC BY-SA 4.0