Timeline for True by accident (and therefore not amenable to proof)
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2022 at 11:47 | comment | added | Carl Mummert | @Martin Sleziak - I changed the link to rmzoo.math.uconn.edu | |
| Aug 12, 2022 at 11:46 | history | edited | Carl Mummert | CC BY-SA 4.0 |
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| Aug 12, 2022 at 7:53 | comment | added | Martin Sleziak | The first link seems to be dead and I did not find it in the Wayback Machine either. I browsed a bit Damir D. Dzhafarov's new website but did not find it - maybe somebody else might have better luck. | |
| Aug 25, 2011 at 14:50 | history | edited | Carl Mummert | CC BY-SA 3.0 |
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| Aug 25, 2011 at 14:49 | comment | added | Carl Mummert | Is there a common name for what would traditionally have been called the dual of a Lindenbaum algebra? | |
| Aug 25, 2011 at 14:47 | comment | added | Carl Mummert | @Emil: Yes; it's just the order that is of interest here. The algebra itself is often uninteresting as an algebra, as you explained in mathoverflow.net/questions/65851/lindenbaum-algebras-and-models/… . The point of the diagrams I linked at the end of my answer is to show interesting suborders obtained by picking a few recognizable nodes out of the order. | |
| Aug 25, 2011 at 14:40 | comment | added | Emil Jeřábek | (The reason for making falsity the bottom in the standard definition is, of course, that once you really treat it as an algebra rather than just order, it is quite inconvenient to have $[\phi]\land[\psi]=[\phi\lor\psi]$ and $[\phi]\lor[\psi]=[\phi\land\psi]$, where $[\phi]$ denotes the equivalence class of a formula $\phi$.) | |
| Aug 25, 2011 at 14:34 | history | edited | Carl Mummert | CC BY-SA 3.0 |
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| Aug 25, 2011 at 14:23 | comment | added | Carl Mummert | I do mean the algebra which puts $0=1$ at the top. I was only using '$>$' as an arbitrary relation symbol. I changed it to an $R$ to avoid entirely any confusion about whether it should have a horizontal line. | |
| Aug 25, 2011 at 14:15 | history | edited | Carl Mummert | CC BY-SA 3.0 |
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| Aug 25, 2011 at 14:14 | comment | added | Andreas Blass |
For the usual definition of the Lindenbaum algebra, you would mean neither $>$ nor $\geq$ but $\leq$. But your use of "bottom", "top", and "higher" indicates that you really intend $\geq$ and thus the dual of the usual definition.
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| Aug 25, 2011 at 12:41 | comment | added | François G. Dorais | (I think you mean '≥' and not '>'.) | |
| Aug 25, 2011 at 12:26 | history | answered | Carl Mummert | CC BY-SA 3.0 |