Timeline for A ‘canonical’ bounded lattice with proper de Morgan negation?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2020 at 13:43 | vote | accept | Daniil Kozhemiachenko | ||
| Mar 2, 2020 at 13:43 | |||||
| Feb 26, 2020 at 20:02 | comment | added | Adam Přenosil | You can embed a Boolean algebra into a non-distributive lattice with a De Morgan negation by adding some non-distributive lattice (say, $M_3$) with a De Morgan negation to the top and bottom of your lattice, if you want an example which is not distributive. | |
| Feb 26, 2020 at 19:11 | comment | added | Daniil Kozhemiachenko | I meant lattices with De Morgan, not Boolean negations, and without distributivity. Or will it also fail? | |
| Feb 26, 2020 at 18:06 | comment | added | Adam Přenosil | Not quite: you can find such $x$, $y$, $z$ in a large enough Boolean algebra. | |
| Feb 26, 2020 at 13:23 | comment | added | Daniil Kozhemiachenko | Thanks! Do I get it correctly that $x\wedge(y\vee\neg y)=(x\wedge y)\vee(x\wedge\neg y)$ holds if and only if the lattice does not have $x$, $y$, and $z$ pairwise incomparable w.r.t. $\leq$ such that $\neg y=z$ and $\neg z=y$? | |
| Feb 26, 2020 at 5:29 | history | answered | Adam Přenosil | CC BY-SA 4.0 |