User:Msiddalingaiah/NFA to CFG

The following context-free_grammar describes the same language as the regular expression (a|b)*abb:

$A_{0}\rightarrow aA_{0}\quad |\quad bA_{0}\quad |\quad aA_{1}$

$A_{1}\rightarrow bA_{2}$

$A_{2}\rightarrow bA_{3}$

$A_{3}\rightarrow \epsilon$

From page 178 of Aho, Sethi, Ullman