# Is the natural order relation on an idempotent semiring total/linear?

We know that on an idempotent semiring $$R$$, the natural order relation is defined as: for all $$x, y\in R$$, $$x\leq y$$ when $$x+y=y$$, which is clearly a partial order relation. I am unable to point out whether this relation is a total order relation too? i.e., does it satisfy Comparability (trichotomy law)?

Take the nonnegative integers with bitwise OR ($$\vee$$) as the sum and bitwise AND ($$\wedge$$) as the product.
$$10\vee 01 = 11\neq 10,01$$ Thus $$10$$ and $$01$$ are incomparable.
The partial order relation in this case is easy to describe. $$x\leq y$$ if and only if whenever $$x$$ has a bit in the $$i$$th position set, so does $$y$$. So this is the product partial order on $$\{0,1\}^\infty$$.
• You write the integers in base $2$, then take OR position by position. So for $10\vee 01$ it's $(1\vee 0)(0\vee 1)$. – Matt Samuel Dec 25 '18 at 14:32
• Thanks i got it now..likewise we can find $10\wedge 01=00$. Right? – gete Dec 25 '18 at 14:42
A distributive lattice is an idempotent semiring (with addition $$\vee$$ and multiplication $$\wedge$$), but most lattices are not totally ordered.