ab L w w ssR s L w w qstsRq q s and t L is not a r

=(a,b}

L = { w : w = ss^R, s *}
L = { w: w = qsts^Rq : q, s, and t *}

L is not a regular language, but L is. Prove L is a regular language by exhibiting a concise regular expression that generates every string in the language.

Solution

Answer:

L2 is regular. Assume Q and S lemda

By assuming them lemda we don\'t need to make any effort for their equivalence. Thus L2 is regular and can draw DFA to it.