Timeline for Logical completeness of Hilbert system of axioms
Current License: CC BY-SA 4.0
22 events
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| Feb 5, 2020 at 14:24 | history | edited | Matthé van der Lee | CC BY-SA 4.0 |
Added some comments.
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| Feb 5, 2020 at 13:38 | comment | added | Sergei Akbarov | I don't want to continue this discussion in chat. What you explain here, is clear without your explanations. What is not clear (and sounds strange) are declarations like this: $$ \text{> Hilbert system is equivalent to Tarski, as explained above.} $$ You should give a definition of this equivalence before writing this, @R.Matveev. And giving advices like this: $$ \text{> For more details, see the answer and the discussion.} $$ | |
| Feb 5, 2020 at 12:46 | comment | added | R. Matveev | Let us continue this discussion in chat. | |
| Feb 5, 2020 at 11:22 | comment | added | R. Matveev | @SergeiAkbarov. I am not sure what your question/concern is. Let me summarize the answer/discussion. 1. Hilbert's system is not a formal system in a modern sense of the word, consequently no mathematically/logically rigorous statement can be made about it. 2. Tarski devised another system of axioms based on (guided by, resembling, similar to, -- not a mathematical notion) Hilbert's. 3. Tarski proved that his system is complete. For more details, see the answer and the discussion. | |
| Feb 5, 2020 at 10:02 | comment | added | Sergei Akbarov | @R.Matveev, if there is no definition (of equivalence) is this rigor in declarations apt? | |
| Feb 5, 2020 at 9:04 | comment | added | R. Matveev | @SergeiAkbarov. Hilbert system is equivalent to Tarski, as explained above. This is somewhat informal, because Hilbert's system is slopy by modern standards and contains two statements which are not in the first order logic language. This is explained in the answer by Matthé van der Lee, above. By Tarski's theorem his and Hilberts systems are complete. ZFC is not complete and has nothing to do with this discussion. | |
| Feb 5, 2020 at 8:51 | comment | added | Sergei Akbarov | @R.Matveev this is another puzzle for me. If one theory is complete while another incomplete, then what is meant by "equivalence"? What are you arguing about? | |
| Feb 5, 2020 at 8:29 | comment | added | R. Matveev | @SergeiAkbarov. Any extension of ZFC or any other set theory capable of modelling $\mathbb{N}$ will be incomplete by Gödel's Theorem, while Tarski's system, which is based on Hilbert's as explained above, is complete. | |
| Feb 5, 2020 at 8:13 | comment | added | Sergei Akbarov | @R.Matveev here is the meaning of the term: en.wikipedia.org/wiki/Extension_by_definitions I would say, Hilbert axioms can easily be presented as an extension by definition of ZFC (or NBG, MK, etc.). As, actually, everything in mathematics described in terms of set theory. | |
| Feb 5, 2020 at 8:05 | comment | added | R. Matveev | @SergeiAkbarov. Hilbert's system is not an "extension" of ZFC, whatever "extension" means. As we all know, there is a model in ZFC for the Hilbert's system. Hilbert's system is not first order, because two of the axioms, Archimedean and completeness use the wrong language (the range of quantification is wrong, in the first case it is natural numbers, which are not inherently present, in the second -- models) | |
| Feb 5, 2020 at 7:58 | vote | accept | R. Matveev | ||
| Feb 5, 2020 at 7:23 | comment | added | Sergei Akbarov | I do not understand this discussion. Is it not possible to say that the Hilbert system is an extension by definitions of some axiomatic set theory, say, ZFC, and therefore can be considered as a first-order theory? The only vagueness, as far as I understand, lies in the meaning of the expression "equivalent first-order theories". Do people use this term? | |
| Feb 5, 2020 at 2:47 | comment | added | LSpice | After agreeing that it shouldn't remain, your edit left the assertion that Tarski's system is "different (but equivalent) to Hilbert's". | |
| Feb 5, 2020 at 0:42 | history | edited | Matthé van der Lee | CC BY-SA 4.0 |
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| Feb 4, 2020 at 22:18 | comment | added | Matthé van der Lee | Hilbert's completeness axiom ("the model cannot be extended") contradicts the upward Löwenheim-Skolem theorem of first order model theory. (The second order axiom of full continuity ensures that $K=\mathbb{R}$, though.) And archimedeanity comes down to the field $K$ mentioned above being an Archimedean field. | |
| Feb 4, 2020 at 22:10 | comment | added | Matthé van der Lee | @ToddTrimble I agree. Hilbert's archimedean and completeness axioms are not first order statements. Rather, they would be requirements on models of a first order formalization of geometry. So, as suggested by Matt, it would indeed be better to speak of Tarski's formalization as "based on Hilbert's system" rather than "equivalent to it". | |
| Feb 4, 2020 at 20:40 | comment | added | Todd Trimble | @MatthévanderLee How is your comment responsive to Matt's point that Hilbert's system is not first-order? The archimedean and completeness axioms in Hilbert's system fall outside Tarski's system (and so cannot be deduced therein). | |
| Feb 4, 2020 at 20:14 | history | edited | Matthé van der Lee | CC BY-SA 4.0 |
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| Feb 4, 2020 at 20:05 | comment | added | Matthé van der Lee | @Matt F. Hilbert's axioms, as set forth in his Grundlagen, are, although informal, no less precise. Formalizations of his principles can be deduced in Tarski's system. | |
| Feb 4, 2020 at 19:26 | history | edited | Matthé van der Lee | CC BY-SA 4.0 |
typo
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| Feb 4, 2020 at 19:22 | comment | added | user44143 | Why do you say that Tarski's setup is equivalent to Hilbert's? Tarski's is first-order and formalized, Hilbert's is not first-order and by 21st-century standards rather informal. I wouldn't say "different (but equivalent) to Hilbert's", when "based on Hilbert's ideas" seems more accurate. | |
| Feb 4, 2020 at 19:00 | history | answered | Matthé van der Lee | CC BY-SA 4.0 |