0 votes 0 votes L is regular <=> there exists a linear grammar for L. Which way is it true and which way is it false? Please explain in detail Rhythm asked Feb 10, 2019 Rhythm 406 views answer comment Share Follow See 1 comment See all 1 1 comment reply smsubham commented Feb 10, 2019 reply Follow Share The biconditional is true. (which means it correct both ways) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes https://gateoverflow.in/303591/cfg-doubt and more over if Let L is: A-> a (RG) and also Linear G and S-> aSb | epsilon (which is linear. But for it we can't drive any RG) Aks9639 answered Feb 10, 2019 Aks9639 comment Share Follow See all 0 reply Please log in or register to add a comment.