Thread: a "basis" in {0,1}^n with operations induced from the boolean algebra {0,1}

i think you can for , especially since you know that spanning is the hard part. it might be more challenging to prove the more general result (an arbitrary idempotent semiring).

OK. It's actually very easy when you know what you want to prove. Unless I'm mistaken of course.

Let . Suppose X spans the space. Then there must be a combination of vectors in X that is equal to Suppose Then since every vector in X has 1 on some coordinate different from . No combination of such vectors can be equal to . Therefore for any must be in X. If we suppose that X is independent, there can be no other vectors in it.