5 ms·
What most ppl don't notice is that Gödels incompleteness theorems are themselves expressible only in a logic system capable of expressing Peano arithmetic. Now
by erichahn 5y ago
What most ppl don't notice is that Gödels incompleteness theorems are themselves expressible only in a logic system capable of expressing Peano arithmetic. Now this means that they apply to themselves which means that we cannot know if they are made up from a axiomatic system that can prove anything.
- kaba0 5y agoI’m not too knowledgeable on the topic, but Gödel’s proof uses a smaller logic system — on which a meta-language can be used to prove consistency/completeness. It is precisely about not being able to prove these properties “from within”.
- erichahn 5y agoWhich one? I don't think so tbh because his proof uses PA.
- kyberias 5y agoIt's very easy for you to check this from his 1931 article.
- erichahn 5y agoIf you figure this out correctly you can claim the third incompleteness theorem. (If I am right)
- ProfHewitt 5y agoResults in [Gödel 1931] depend on existence of proposition I'mUnprovable. Since, the proposition doesn't exist in foundations, the results in [Gödel 1931] do not hold for foundations.
- ProfHewitt 5y agoYes, it is very easy to check that [Gödel 1931] was for a system for the foundation of mathematics.
- ProfHewitt 5y agoThe reason that [Gödel 1931] was influential was that it claimed to prove incompleteness for a system for the foundations of mathematics. 1st-order systems such a PA were introduced later and quickly shown to be inadequate for the foundations of mathematics.
- ProfHewitt 5y agoAs explicitly stated in the title [Gödel 1931] for a system for the foundation of mathematics.
- dandanua 5y agoIt's ok to use logic to prove theorems in logic. We also use thinking to deduce some facts about our thinking. Self-reference is not always paradoxical. In a higher-order logic some self-referential definitions are legitimate.
- erichahn 5y agoYes and then you can do self-referential statements again which leads you to the conclusion that Gödels theorems are provable only in a system that cannot prove its own consistency.
- dandanua 5y agoTrue, but it's not a logical paradox. The fact that a system can't prove its own consistency doesn't imply that this system is inconsistent. The fact that Godel's theorems are provable only in such systems also doesn't imply that they are wrong. From the common sense the whole situation looks paradoxical, indeed. To prove consistency of some theory we have to use some stronger theory, to prove consistency of that theory we need even stronger theory and so on. Perhaps, the best way to realize why there is no paradox is the following: Our memory is finite. As well as our thinking. But the total number of true facts about mathematics is infinite. By constructing theories we are trying to compress the infinite number of true facts into some finite form. Godel's theorem says it's impossible. And it looks quite natural from this perspective.
- erichahn 5y agoI never said it is a paradox. I never said that PA or Q is inconsistent. I said that they cannot prove their own consistency by Gödels theorem. Hence we don't know if Gödels theorem was formalized in an inconsistent system. Honestly it would be weird if the incompleteness theorems don't apply to themselves.
- ProfHewitt 5y agoThe logical contradiction is that allowing the [Gödel 1931] proposition I'mUnprovable into foundations makes the foundations inconsistent.
- bidirectional 5y agoIt's not Peano arithmetic, it is basically 'enough of' arithmetic for Goedel's methods to apply. Robinson arithmetic is weaker than PA but Goedel still applies. Goedel's argument is basically a meta-argument about any mathematical system which is rich enough to describe useful mathematics, it does not rely on any particular axiomatisation, rather it applies to all axiomatisations with a few simple features.
- erichahn 5y ago"it does not rely on any particular axiomatisation" -- OK, what does it rely on then?
- carnitine 5y agoThe axioms allowing one to express enough of arithmetic for Goedel’s methods to apply. As the comment says.
- erichahn 5y agoI think if Gödels proof needs Q then that is OK. Q cannot prove its own consistency. Which means there is no way of telling that Gödels theorems are proved in a theory that is inconsistent (where everything is true).
- rssoconnor 5y agoWhile it is true that Goedel's theorem applies to weak systems such as Robinson Arithmetic (and any decidable extensions there of), The proof of Goedel's result itself requires at least some amount of induction. As a consequence the minimum system that Goedel's second incompleteness applies to is stronger than the minimum system that the first incompleteness theorem applies to.
- IngoBlechschmid 5y agoThis is not quite true, Gödel's incompleteness theorems can luckily be formalized in extremely weak fragments of Peano arithmetic such as primitive recursive arithmetic (PRA) with its very limited induction principle. :-) The only position on the philosophy of mathematics I know which does not accept PRA is ultrafinitism.
- erichahn 5y agoCan you provide a paper or book on this?
- ProfHewitt 5y agoProblem with 1st-order PRA is that it is too weak to serve as foundation of mathematics.