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# i confirm answer plz check???

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Given answer is 100% correct.

1. False : DCFL's are not closed under intersection.

Example:L1={anbncm} intersection L2={ambncn}={anbncn}

2.False: RE is not closed under complement. What does this mean?? It means that the complement may or may not be RE. It is not certain that it will not be RE.

3.False: All languages are countable. Set of languages may be uncountable.

4.False: No language is closed under subset// this reason is correct.

Someone is commented that reguar languages are closed under subset. No, they are not. L=sigma * is regular right? L1={anbncm} is a subset of this language is which is not regular.

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let sigma=(a1,a2,a3,......) as input symbols ,


1.then sigma* represent the set of all the possible combinations of input symbols (i.e set of all the strings) and we know that sigma* is a countable set(proven with Diagonalization method).

2. we know that any language is the set of strings, means it is the subset of sigma* (subset of a countable set is also countable) therefor, all the languages are countable.  but the set of Not RE languages is uncountable.

and for 3rd option....  it is talking about the language not set.

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