I want to prove that $L$ is not regular: $$L = \{ww^Rv \mid |w|\ge1 , |v|\ge 0\},$$ where the alphabet contains at least two symbols.
Can someone prove it?
I prefer to use "Pumping Lemma for Regular Languages" to prove it, but I think that's somehow impossible!
Would someone help me, please?
If the alphabet has only one symbol then $L$ is regular, so assume the alphabet contains at least two symbols, which we'll call $0$ and $1$.
For $n>0$, let $w_n = (01)^n$ and $s_n = w_n w_n^R\in L$. Let $\sim$ be the equivalence relation of the Myhill-Nerode theorem: $$ x\sim y \iff \forall z\,(xz\in L \leftrightarrow yz\in L). $$ By the theorem, $L$ is regular iff $\sim$ has only finitely many equivalence classes.
Suppose $m < n$, and let $z=w_m^R$. We have $w_m z = s_m \in L$. However, if $w_n z = (01)^n(10)^m \in L$, then by definition of $L$ there are $s,t$ with $|s|>0$ such that $$ w_n z = (01)^n(10)^m = (01)^{n-m}(01)^m(10)^m = ss^Rt $$ But because $n>m$, there are no such $s$ and $t$, so $w_n z\notin L$ after all. So this $z$ distinguishes $w_m$ and $w_m$.
It follows that $$ m\ne n \implies w_n \not\sim w_m, $$ so $\sim$ has infinitely many equivalence classes, hence $L$ is not regular.