# Proving that $(a \cup b)^* = (b^*(a\cup\lambda)b^*)^*$?

How would I prove that these two regexes are equal to one another? $$(a \cup b)^* = (b^*(a\cup\lambda)b^*)^*$$

I'm permitted to use the following regular expression identities.

• Please do not delete your questions after receiving answers. That's against our policies. – Raphael Sep 25 '17 at 7:54

One of the equalities 12 tell us that now $(u\cup v)^* = u^*(vu^*)^*$. Applying this to the last expression we obtain $(b\cup (a\cup\lambda))^*$.
Also $(u\cup \lambda)^* = (u^*\lambda^*)^* = (u^* \lambda)^* = (u^*)^* = u^*$.
Do you have associativity of $\cup$ so that $(b\cup (a\cup\lambda))^* = ((b\cup a)\cup\lambda))^*$? It should be a basic property.
• Also, re-reading your answer I don't see how your second line for $u^*$ is used? But, I can derive what we would get from associativy by starting with $(b \cup a)^*$ and using the second line in reverse. – Skeletor Sep 25 '17 at 0:02