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1. • CommentRowNumber1.
   • CommentAuthorJ-B Vienney
   • CommentTimeDec 19th 2022
   • (edited Dec 19th 2022)
   • PermaLink
   Author: J-B Vienney
   Format: MarkdownItexAdded as examples: $0$, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{B}=\{0,1\}$ with $1+1=0$. Proved that they are exactly the boolean rigs of cardinal less or equal than $2$. I don't know if boolean rigs in the sense of this entry are always commutative. In [The variety of Boolean semirings](https://core.ac.uk/download/pdf/81147572.pdf), they show that they are commutative assuming that $1+x+x=1$. If some noncommutative boolean rig exists, it must be of cardinal $\ge 3$, not be a ring (because boolean rings are commutative) and not verify this equation. <a href="https://ncatlab.org/nlab/revision/diff/Boolean+rig/6">diff</a>, <a href="https://ncatlab.org/nlab/revision/Boolean+rig/6">v6</a>, <a href="https://ncatlab.org/nlab/show/Boolean+rig">current</a>

   Added as examples: $0$, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{B}=\{0,1\}$ with $1+1=0$. Proved that they are exactly the boolean rigs of cardinal less or equal than $2$.

   I don’t know if boolean rigs in the sense of this entry are always commutative. In The variety of Boolean semirings, they show that they are commutative assuming that $1+x+x=1$. If some noncommutative boolean rig exists, it must be of cardinal $\ge 3$, not be a ring (because boolean rings are commutative) and not verify this equation.

   diff, v6, current

2. • CommentRowNumber2.
   • CommentAuthorGuest
   • CommentTimeMay 22nd 2023
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   Author: Guest
   Format: MarkdownItexAdded related concepts section <a href="https://ncatlab.org/nlab/revision/diff/Boolean+rig/9">diff</a>, <a href="https://ncatlab.org/nlab/revision/Boolean+rig/9">v9</a>, <a href="https://ncatlab.org/nlab/show/Boolean+rig">current</a>

   Added related concepts section

   diff, v9, current