Star-continuity of a partially ordered, idempotent semiring implies Kleene Algebra?

Question

I've been looking into Kleene algebras for an upcoming presentation I'm giving on regular expressions. I've read that (in an idempotent semiring with partial order $a\le b \iff a+b=b$) star-continuity, i.e.

$$xy*z = \sup(\underset{0\le i\le n}{\{xy^i z\}})$$

implies the four Kleene algebra axioms related to Kleene stars.

I'm stuck on how star-continuity implies this axiom:

$$1+xx^*\le x^*$$

I can show that it's true for $n\ge 1$:

$$ \begin{eqnarray*} 1+xx^* & = & 1 + \sup(\underset{0\le i\le n-1}{\{xx^i\}}) \\ & = & x^0 + \sup(\underset{1\le i\le n}{\{x^i\}}) \\ & = & \sup({x^0}) + \sup(\underset{1\le i\le n}{\{x^i\}}) \\ & = & \sup(\underset{0\le i\le n}{\{x^i\}}) \\ & = & x^* \end{eqnarray*} $$

(and if $1+xx^*=x^*$ then certainly $1+xx^*\le x^*$.)

However, this clearly won't work for $n=0$ so now I'm starting to question whether this is even a correct way to prove this implication. Any hints?

Edit: I should add that I'm getting a lot of this out of the lecture notes from Prof. Dexter Kozen's course. My hangup is trying to interpret and understand the arguments therein.

Answer 1

The thing I failed to realize was that star-continuity holds for all $n\ge 0$.

$$xy*z = \sup(\underset{0\le i\le n}{\{xy^i z\}}) \forall n\ge 0$$

So when doing the computation, we don't care that the $n$ at the start doesn't line up with $n+1$ at the end. This isn't an inductive step, it's just a computation. Hence we can do this:

$$ \begin{eqnarray*} 1+xx^* & = & 1 + \sup(\underset{0\le i\le n}{\{xx^i\}}) \\ & = & x^0 + \sup(\underset{1\le i\le n+1}{\{x^i\}}) \\ & = & \sup({x^0}) + \sup(\underset{1\le i\le n+1}{\{x^i\}}) \\ & = & \sup(\underset{0\le i\le n+1}{\{x^i\}}) \\ & = & x^* \end{eqnarray*} $$

(and if $1+xx^*=x^*$ then certainly $1+xx^*\le x^*$.)