How to prove for that a given non-regular language its given super-set is also non-regular? [duplicate]

Question

Consider the non-regular language $L_1$ = {$a^n$b$a^n$: $n \geq 0$} and its super-set language

L = Language of strings with equal number of trailing and leading a's,

or in other words,

$$L = \{a^n b w b a^n : n \ge 0, w \in \{a,b\}^*\} \cup \{a^n b a^n : n \ge 0\}.$$

How would we prove that L is also non-regular?

I have the idea that I should use the help of $L_2 = L - L_1$, but I'm stuck at the following step:

$L_1 \cap L_2 = \phi $ , thus $L_1 \cap L_2$ is a regular language.

I was trying to prove it using contradiction.

Let, L to be regular which is $L_1 \cup L_2 $ is regular.

And tried to prove, $L_1$ or $L_2$ is equal to a set union or intersection or complement operation over the regular sets $L_1 \cap L_2 $ and $L_1 \cup L_2$

But stuck at

$L_1 = [(L_1 \cup L_2) \cap (L_2)^c] \cup (L_1 \cap L_2)$

With Myhill-Nerode theorem: I might be incorrect please fix me if I am wrong.

Take S =$a^*(b+b.(a+b)^*.b)$
Pick x and y from S such that $x \neq y$.

x = $a^n b$
y = $a^m b$
Take, z = $a^n$

$x.z \in L$ but $y.z \notin L$. Thus, by Myhill_Nerode one can conclude it to be non-regular.

Answer 1

Since you already know that $L_1$ is nonregular, here is a simple proof. Suppose that $L$ is regular. Then since $a^*ba^*$ is regular, the language $L \cap a^*ba^*$ should be regular. But $L \cap a^*ba^* = L_1$, which is not regular. Hence $L$ is nonregular.