Proving equality of two regular languages using operations closed under regular languages Is there any way to prove that two regular languages A and B are equal using only closed operations under regular languages?
(The languages can be expressed as regular expressions,NFAs, eNFAs or DFAs)
 A: Consider regular languages $L_1, L_2,$ and $L_3$. For $\textbf{some}$ closed unary operations $*$ and binary operations $\diamond$ on regular languages, $$
(L_1)^* = (L_2)^* \Longrightarrow L_1=L_2 \\
L_1\diamond L_3 = L_2\diamond L_3 \Longrightarrow L_1=L_2
$$
So showing that applying the same operation to two regular languages $L_1$ and $L_2$ produces equal regular languages may be enough to show that $L_1$ and $L_2$ are equal. However, this is not true for every operations. For this to be true, the operation $\star:\mathscr{R}\rightarrow\mathscr{R}$ must be one-to-one on $\mathscr{R}$, the set of all regular languages. Here is a breakdown of the most common closed operations on regular languages:
$\textbf{One-to-one}$:


*

*Complement of regular languages on same alphabet $\Sigma$ $$
\Sigma^*-L_1=\Sigma^*-L_2 \implies L_1=L_2
$$

*Reversal $(R)$ $$
(L_1)^R=(L_2)^R \implies L_1=L_2
$$

*One-to-one homomorphism $(h)$ $$
h(L_1)=h(L_2) \implies L_1=L_2
$$
$\textbf{Not one-to-one}$:


*

*Union $(\cup)$ with another regular language $$
L(a^*)\cup L(a^*b^*) = L(b^*)\cup L(a^*b^*) \\
L(a^*)\ne L(b^*)
$$

*Intersection $(\cap)$ with another regular language $$
L(b^*)\cap L(b^*) = L(a^*b^*)\cap L(b^*) \\
L(b^*)\ne L(a^*b^*)
$$

*Concatenation $(\circ)$ with another regular language $$
L(ab^*)\circ L(b^*) = L(a)\circ L(b^*) \\
L(ab^*)\ne L(a)
$$

*Kleene closure $(*)$ $$
L(b)^* = L(bb^*)^* \\
L(b) \ne L(bb^*)
$$

*Difference $(-)$ with another regular language $$
L(a^*)- L(a^*b^*) = L(b^*)-L(a^*b^*) \\
L(a^*)\ne L(b^*)
$$
