Timeline for Is TA (true arithmetic) interpretable in a recursively axiomatizable theory?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2020 at 2:26 | vote | accept | Eric | ||
| Jul 5, 2020 at 19:47 | history | edited | Rodrigo Freire | CC BY-SA 4.0 |
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| Jul 5, 2020 at 19:12 | comment | added | Rodrigo Freire | Sure, there is no interpretation, but my comment was on the issue of the computability of the interpretation. I will edit the answer to eliminate the reference to this issue, thanks. | |
| Jul 5, 2020 at 18:37 | comment | added | Emil Jeřábek | This does not matter. You can’t interpret any extension of TA to any language you want, because you cannot even interpret its finite-language fragment in the usual language of arithmetic. | |
| Jul 5, 2020 at 14:50 | comment | added | Rodrigo Freire | Sure Emil, but the OP may have a different presentation of TA in mind. For example, one with infinitely many primitive constants. | |
| Jul 5, 2020 at 14:29 | comment | added | Emil Jeřábek | Since the language of arithmetic is finite, interpretations in the usual sense are automatically computable (hence arithmetical). | |
| Jul 5, 2020 at 14:09 | comment | added | Eric | I edited the question to address the indeterminacy. But suppose $\tau$ preserves negation, why does this make $T$ impossible? | |
| Jul 5, 2020 at 12:10 | comment | added | Rodrigo Freire | I agree, I have expanded the answer with more information. | |
| Jul 5, 2020 at 12:09 | history | edited | Rodrigo Freire | CC BY-SA 4.0 |
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| Jul 5, 2020 at 12:02 | history | edited | Rodrigo Freire | CC BY-SA 4.0 |
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| Jul 5, 2020 at 11:50 | comment | added | მამუკა ჯიბლაძე | Entirely true, and answers the question in its present form perfectly. Still, it is formulated in such way that it is a comment rather than answer. | |
| Jul 5, 2020 at 11:34 | history | answered | Rodrigo Freire | CC BY-SA 4.0 |