The biconditional is true. (which means it correct both ways)

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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

Which way is it true and which way is it false?

Please explain in detail

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)

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