If A is a non-regular language and B is a infinite language such that B ⊆ A, then B must be non-regular.
B is infinite, and B ⊆ A ====> A should be Infinite Language
let B = { $c^{m}$ | m ≥ 0 } and A = { $a^{n}\;b^{n}\;c^{m} \;| n\;\geq 0 \;and \;m \;\geq 0$ } then Given statement is false.
If A is non-regular language which is not Accepted by any PDA and B is a regular language then ( A ∪ B )* may be regular.
if B = ∑* ===> ( A ∪ B )* is RL
Non-regular language is closed under complement but not closed under concatenation.
Non-RL complement should be Non-RL, due to RL is closed under Complement.
i mean, let A is Non-RL, then A' is Non-RL...
if A' is RL, then (A')' is RL due to RL is closed under Complement. ===> (A')' = A is RL, it is contradicting ===> A' should be Non-RL
Non-regular is closed under reversal but not closed under subset operation.
Non-regular is closed under reversal ===> it is true
WHY ?
when you say a language is Non-RL ?
there should be a relation between them, i mean let take L = {$a^n\;b^n\;c^n$}, in this the relation between a and b is no.of a's should be equal to a, and no.of b's should be no.of c's, By reversing the strings, this relation can cancel ? No, even you reverse the relation should be preserved !
i mean, in $ L^{R}$ = { c$^{n}\;b^{n}\;a^{n}$ } , the relation between a and b is no.of a's should be equal to a, and no.of b's should be no.of c's
Non-regular is not closed under subset operation ===> it is true due to
Non-regular language subset may be finite then it is RL.
