Timeline for Is $\mathsf{R}$ axiomatizable by finitely many schemes?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2022 at 12:45 | comment | added | Fedor Pakhomov | @EmilJeřábek Yes, indeed it should have been $1<p\le u+1$, thank you. | |
| Feb 28, 2022 at 10:33 | comment | added | Emil Jeřábek | Oh I see; it didn’t occur to me that one can force the universe to be a power of $2$ with finitely many axioms like this. Presumably you want $1<p\le u+1$ rather than $1<p<u+1$, as otherwise the axioms also hold in $\mathbb Z_{2p}$ for odd primes $p$. | |
| Feb 28, 2022 at 8:50 | history | edited | Fedor Pakhomov | CC BY-SA 4.0 |
fixed minor mistake
|
| Feb 28, 2022 at 8:49 | comment | added | Fedor Pakhomov | @EmilJeřábek Of course I meant that we should fix $\mathsf{U}_0$ axiomatized by finitely many schemes (I have edited the answer to clarify this). I have in mind the following axiomatization of $\mathsf{U}_0$: 1. < is a linear order, where 0 is the least element and there is the greatest element $m$; 2. $0+1=1$; 3. $m+1=0$; 4. $x<x+1$ and $(x,x+1)=\emptyset$, for $0\le x <m$; 5. $x+0=x$; 6. $x+(y+1)=(x+y)+1$; 7. $x0=0$; 8. $x(y+1)=xy+x$; 9. either $0=m$ or there is $u<m$ s.t. $(u+1)+u=m$ and for any $0<x<u+1$ and $1<p<u+1$ if $xp=u+1$ and $p$ isn't $zw$ for any $1<z,w<p$, then $p=1+1$. | |
| Feb 28, 2022 at 7:39 | comment | added | Emil Jeřábek | Why is $U_0$ axiomatizable by finitely many axiom schemata? | |
| Feb 28, 2022 at 7:06 | history | edited | Fedor Pakhomov | CC BY-SA 4.0 |
added some context for the main construction
|
| Feb 28, 2022 at 0:02 | comment | added | Fedor Pakhomov | @NoahSchweber Yes, sure. | |
| Feb 27, 2022 at 23:59 | history | bounty ended | Noah Schweber | ||
| Feb 27, 2022 at 23:58 | history | answered | Fedor Pakhomov | CC BY-SA 4.0 |