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Question 9. Consider the language {ab : n >0}. (i) Is this a regular language? Why or why not? (ii) Is this a recursively e
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Answer #1

9)
given L = {a^nb^n :0<=n}
i)
Given language is not regular
proof using pumping lemma:
let take a string w belongs to L
w = aabb
now divide w into xyz, such that 0<|y|
a|a|bb
x|y|z
x=a,y=a,z=bb
now by pumping lemma, for all 0<i, xy^iz belongs to L
let i=1, xy^1z = aabb belongs to L
i=2, xy^2z = xyyz = aaabb doesn't belongs to L
since, xy^2z doesn't belongs to L which violates pumping lemma, hence L is not regular

ii)
yes it is recursively enumerable
given language is context free,
every context free language is recursive
and every recursive language is recursively enumerable
hence given language is recursively enumerable

for all 970 the pumping limma for regiar lamgelegesi- Let l be a regular lemgriege Then, there exists a constant ac whose val

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