1 votes 1 votes Show that the language $L=$ {$a^nb^n:n\geq0$} $\cup$ {$a^nb^{n+1}:n\geq0$} $\cup$ {$a^nb^{n+2}:n\geq0$} is not regular. Theory of Computation peter-linz peter-linz-edition4 theory-of-computation pumping-lemma + – Naveen Kumar 3 asked Apr 11, 2019 Naveen Kumar 3 149 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.