How to Apply Elementary Axioms from Kleene Star to an Inequality

Question

Axioms For *

\begin{align} 1 + aa^* &\leq a^* \\ 1 + a^*a &\leq a^* \\ b + ax &\leq x \to a^*b \leq x \\ b + xa &\leq x \to ba^* \leq x \\ \end{align}

Elementary Results

\begin{align} a \leq b &\to a + c \leq b + c \\ a \leq b &\to ac \leq bc\, \wedge\, ca \leq cb \\ a \leq b &\to a^* \leq b^* \end{align}

Problem

Prove the following identity in a Kleene algebra using only the axioms and elementary results. $$(a + ab + b)^* = (a + b)^*$$

Solution: \begin{align} (a + b)^* &= (a + ab + b)^* \\ (a + b)^* &\leq (a + ab + b)^* \\ 1 + (a + b)(a + b)^* &\leq 1 + (a + ab + b)(a + ab + b)^* \\ (a + b)(a + b)^* &\leq (a + ab + b)(a + ab + b)^* \\ \end{align}

Quesiton

1. So for them to be equal the sets should be contained in each other. At which point do I transition to an inequality?

2. Is it right to say a $*$ cannot be removed since it has no inverse?

3. Can I distribute into a $*$? Say $(a +b)(a + b)^*$ or do they need to have the same $*$ height?

Some hints to get further would be greatly appreciated.

Answer 1

I assume that in your definition $a \leq b$ iff $a + b = b$.

First, note that if $a \leq b$ and $b \leq a$, then $a = a + b = b + a = b$. Therefore, in order to show that $a = b$, it is sufficient to show that $a \leq b$ and $b \leq a$.

Now, you want to show that $(a + b)^\ast = (a + ab + b)^\ast$. As explained in the previous paragraph, it is sufficient to show that inequalities with $\leq$ in both directions hold ("[...] for them to be equal the sets should be contained in each other").

---

The inequality $(a + b)^\ast \leq (a + ab + b)^\ast$ is almost trivial:

$$(a + b) + ab = a + ab + b $$

therefore $(a + b) \leq (a + ab + b)$, and by monotonicity (your ER-3), it thus holds

$$(a + b)^\ast \leq (a + ab + b)^\ast.$$

---

The other direction is a bit trickier.

By monotonicity (ER-2), (def-$\leq$), (Ax-1):

$$a(a + b)^\ast \leq (a + b)(a + b)^\ast \leq 1 + (a + b)(a + b)^\ast \leq (a+b)^\ast$$

Analogously, $b(a + b)^\ast \leq (a+b)^\ast$.

From both previous statements, again using monotonicity (ER-2):

$$ab(a+b)^\ast \leq a(a+b)^\ast \leq (a+b)^\ast.$$

To summarize: so far, we know that $a$,$b$ and $ab$ are all $\leq (a+b)^\ast$.

Now, observe that from (ER-1), it follows that if $x \leq w$ and $y \leq w$, then $x + y \leq w + y \leq w + w = w$. Thus, from $a\leq (a + b)^\ast$, $ab \leq (a + b)^\ast$ and $b \leq (a + b)^\ast$, you obtain:

$$(a + ab + b)(a+b)^\ast = a(a+b)^\ast + ab(a+b)^\ast + b(a+b)^\ast \leq (a + b)^\ast.$$

Since additionally $1 \leq (a+b)^\ast$ (directly from Ax-1), you obtain:

$$1 + (a + ab + b)(a+b)^\ast \leq (a+b)^\ast.$$

With (Ax-3) it follows:

$$(a + ab + b)^\ast = 1\cdot(a + ab + b)^\ast \leq (a + b)^\ast$$

---

Thus, both inequalities hold, hence we get the equality.

You cannot simply "remove" $^\ast$, nothing is told about "inverse" in the definition of a Kleene algebra. I also don't know what you mean by "$\ast$-height".

Answer 2

You need a full set of axioms. For this, you can use the axioms for

1. a monoid: associativity $x(yz) = (xy)z$ and identity $x1 = x = 1x$;

2. a partial ordering possessing least upper bounds $⋁F$ of finite subsets $F$: defining $0 = ⋁∅$ and $a + b = ⋁\{a,b\}$: an equivalent set of identities is captured by another instance of monoid axioms $x + (y + z) = (x + y) + z$, $x + 0 = x = 0 + x$, along with upper semi-lattice axioms $x + y = y + x$ and $x + x = x$; and

3. finite distributivity: $x\left(⋁A\right)y = ⋁\left(\{x\}A\{y\}\right)$, which is equivalently captured by the zero property $y0z = 0$ and distributivity property $w(x + y)z = wxz + wyz$.

So, yes, you can always distribute and the sum operator, $+$, is the least upper bound operator. The ordering relation is equivalently given in terms of the least upper bound operator by way of the following equivalences $x ≤ y ⇔ y ≥ x ⇔ ∃z (y = x + z) ⇔ y = x + y$. In addition, the sum and product operator are both order-preserving, as a consequence ... so are all polynomials.

The resulting algebraic structure is an idempotent semiring, which is also known as a dioid. Adding in the *-operator, along with your axioms (or any of a number of other equivalent axiomatizations), make this a set of axioms suitable for Kleene algebra.

The * operator is also order-preserving. If $a ≤ b$, then $1 + a b^* ≤ 1 + b b^*$, therefore $[a^* = ]1 a^* ≤ b^*$, by your *-axioms. So, all Kleene-algebraic operators, and functions composed therefrom, are order-preserving.

Since $a + b ≤ a + ab + b$, then $\left(a + b\right)^* ≤ \left(a + ab + b\right)^*$. The proof of $\left(a + ab + b\right)^* ≤ \left(a + b\right)^*$ was already given in another reply.

It generalizes to $f(a,b) ≤ (a + b)^*$, for all Kleene-algebraic functions $f(a,b)$ of $a$ and $b$. In fact, the only values a Kleene-algebraic function of $(a + b)^*$ can take, are $0$, $1$ and $(a + b)^*$, which, by virtue of $f(a,b) ≤ f((a + b)^*, (a + b)^*)$, renders the proof of $f(a,b) ≤ (a + b)^*$ trivial. Another way of saying this is that the Kleene-subalgebra generated by $(a + b)^*$ is just $\{0, 1, (a + b)^*\}$, which is either a 3-element Kleene algebra or else is the unique 2-element Kleene algebra $\{0,1\}$, with $(a + b)^* = 1$, if $a + b ≤ 1$.

Thus, $(a + b)^*$ is the maximum element of the Kleene-subalgebra generated by $\{a,b\}$.