Suppose we have the following automaton:
Find the regular expression.
By inspection we could say that the regular expression is $x(x+y)^*$, but I want to prove it by solving the system of equations: $$\begin{cases}q_0=xq_1\\q_1=xq_1+yq_1+\lambda\end{cases}$$ I am not able to reduce $q_1$.
I know the rule says if $p=ap+bq$ then $p=a^*(bq)=a^*bq$, but here I cannot apply this rule because I have a term (the null word $\lambda$) that is not of the form $bq$ i.e. I cannot do the following: $$q_1=xq_1+yq_1+\lambda\neq x^*(yq_1+\lambda)=x^*yq_1+x^*\lambda=x^*yq_1+x^*.$$ So my question is:
How can we arrive at the following equality?: $$q_1=xq_1+yq_1+\lambda=(x+y)^*$$
Just rewrite $q_1 = xq_1 + yq_1 + \lambda$ as $q_1 = (x+y)q_1 + \lambda$. By Arden's rule, you get $$q_1 = (x+y)^*\lambda = (x+y)^*.$$