+1 vote
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Is it c?
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+1
We know that Regular grammars can only be either right linear or left linear. If we can express a language in one of the form then we can convert it to the other form as well. So we can say that there exists both right and left linear grammar for regular languages. option a and b are true.

If we have an Epsilon-NFA there is always a way out to remove Epsilon transition and get an NFA which is Epsilon free. So e is also true.

The condition in d is not at all necessary for a language to be regular. We can have multiple final states in a DFA as the formal definition states so. So d is False.

Suppose you obtain an NFA having multiple final states. Then you can add Epsilon transition from each of them to another state and call it the new final state and make the previous final ones as non-final. Option c can be proved true in this way. (you can always remove epsilon moves to get epsilon free nfa)

• ## if a language is regular then there must exist a left linear and right linear grammar .so option a and b is true .

Example : L= a*

Left linear grammar is : s--->aS/null

Right linear grammar is: s---->Sa/null

• ## If a language is regular then there exist a NFA . AND NFA WITH NULL IS EQUIVALENT TO NFA . BECZ CONVERSION IS POSSIBLE . SO OPTION e is also true .

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what is the significance of c and d statements?
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While conversion of finite autometa to reguler expresion then we need to make single final state in NFA .

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Thanks a lot!
it's b.
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no

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