Different application of Arden's theorem leads to different answers

Question

So, I have to solve for the following set of equations

$q_1$ = $q_1$a + $q_2$b + $\epsilon$

$q_2$ = $q_1$a + $q_2$b + $q_3$a

$q_3$ = $q_2$a

There are two ways to do this.

I did this

$q_1$ = $q_2$b + $\epsilon$ + $q_1$a

$q_1$ = ($q_2$b + $\epsilon$)a* Applying Arden's theorem

Now substituting in $q_2$ the values of $q_1$ and $q_3$

$q_2$ = $q_2$ba*a + a*a + $q_2$b + $q_2$aa

$q_2$ = a*a + $q_2$(ba*a+b+aa)

$q_2$ = a*a(ba*a+b+aa)* Applying Arden's theorem

$q_2$ = a*a(ba*+aa)*

Now substituting in $q_3$, the answer should be

$q_3$ = a*a(ba*+aa)*a

However, the correct answer is

(a + a(b+aa)*b)*a(b+aa)*a

which can be obtained by first substituting $q_3$ in $q_2$, and then substituting $q_2$ in $q_1$, and finally solving $q_2$, $q_3$ from the obtained regular expression for $q_1$.

Can someone tell where I have gone wrong in the above method, or am I applying ardens theorem in the wrong way ?

Answer 1

The two regular expressions describe the same language.

Answer 2

Using various Kleene algebraic properties, such as the following $$ (u + v)^* = v^* (u v^*)^* = (v^* u)^* v^* = u^* (v u^*)^* = (u^* v)^* u^*,\\ u u^* = u^* u,\quad (u v)^* u = u (v u)^*,\quad u^* φ(u)^* = u^*,\\ u + v = v + u,\quad (u + v) w = u w + v w,\quad u 1 = u,\quad 1 + u u^* = u^*, $$ where $φ(u)$ denotes an arbitrary Kleene-algebraic polynomial in $u$, we can write $$\begin{align} a (a + b)^* a &= a a^* (b a^*)^* a\\ &= a a^* (a a)^* (b a^* (a a)^*)^* a\\ &= a^* a (a a)^* (b a^* (a a)^*)^* a\\ &= a^* a (b a^* + a a)^* a,\\ \end{align}$$ and $$\begin{align} a (a + b)^* a &= a a^* (b a^*)^* a\\ &= a a^* (a a)^* (b a^* (a a)^*)^* a\\ &= a a^* (b a^* + a a)^* a\\ &= a a^* (b (1 + a a^*) + a a)^* a\\ &= a a^* (b + a a + b a a^*)^* a\\ &= a a^* ((b + a a)^* b a a^*)^* (b + a a)^* a\\ &= (a a^* (b + a a)^* b)^* a a^* (b + a a)^* a\\ &= (a^* a (b + a a)^* b)^* a^* a (b + a a)^* a\\ &= (a + a (b + a a)^* b)^* a (b + a a)^* a. \end{align}$$

So, actually, you could have said it was $a (a + b)^* a$.