An element $S$ is said to be totally ordered set if for all $a, b\in S\implies$ either $a\leq b$ or $b\leq a.$
An algebraic structure $(S, +, \cdot)$ is said to be an idempotent semiring if $x\cdot x=x$ for all $x\in S$. Note that $S$ is addively idempotent semiring if $x+x=x$ for all $x\in S$.
In view of the above definitions, can we confirm that
$\endgroup$ 3a set of elements of an idempotent semirings are totally ordered.
1 Answer
$\begingroup$Your definition of "totally ordered" semiring and your definition of a semiring are wrong. Assuming the standard definitions, the answer is "no": every bounded, distributive lattice is a commutative, idempotent semiring under join and meet with $1=$ the maximal element and $0=$ the minimal element.
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