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Consider a language $L$ that is recognized by a machine $M$. Which of the following statements might not be true?

  1. If $M$ is a deterministic finite automaton, then $L$ can be represented by a regular expression.
  2. If $M$ is a Turing machine, then $L$ is recursive.
  3. If $M$ is a deterministic pushdown automaton, then $L$ can be represented by a context-free grammar.
  4. If $M$ is a non-deterministic pushdown automaton, then $L$ is recursively enumerable.
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The Chomsky hierarchy of language classes and the corresponding machines that decide them.

Regular languages are the simplest. L is regular if and only if there exists a deterministic finite state automaton M that decides L.

Also, L is regular if and only if there exists a nondeterministic finite state automaton M that decides L. Finally, L is regular if and only if there exists a regular expression E such that E generates a string w if and only if w is an element of L.

All regular languages are context free. L is context free if and only if there exists a nondeterministic pushdown automaton M that decides L. (Note: The class of languages decided by deterministic pushdown automata is a strict subset of the class of context free languages.)
L is context free if and only if there exists a context free grammar G such that G generates a string w if and only if w is an element of L.

All context free languages are recursive. L is recursive if and only if there exists a Turing machine M that decides L. (It appears that the decidable languages are not technically a member of the Chomsky hierarchy, but they fit in there nicely.)

All recursive languages are recursively enumerable. L is recursively enumerable if and only if there exists a Turing machine M that recognizes L.

Choice B is not true because if L is recognized by Turing machine M, then this only implies that L is recursively enumerable. It does not imply that L is recursive, since M might not be able to decide L.
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how?

in option D,

it is CFl then it RE also.

so why not option D?
Answer:

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