Is this regular or not

L = {w1w^R | w ∈ {0,1}* (where for any word w ∈ {0,1})*, w^R denotes the reverse of w)

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– D.W.
Oct 1 '20 at 17:27

L is not a regular language.

We can contradict by pumping lemma for regular languages.

Let assume L is regular, so there exists an integer $$k$$ of pummping lemma.

$$w = a^kb$$

$$w^R = ba^k$$

$$ww^R = a^kbba^k \in L$$

$$w' = a^{k-r}bba^k \notin L , r>0$$

Contradiction to $$w' \in L$$.

• Actually in this case you can pump $bb$ and there is no problem. Oct 1 '20 at 17:52