Consider the following three statements:
Which of the following gives the correct true/false evaluation of the above?
Correct Answer: $D$
Not getting last one...
@arjun
$L1 \cap L2 \cap L3 \cap L4 \cap \dots$
if $L1 \cap L2 \cap L3 \cap L4 \cap \dots$ is regular then it means $(L1' \cup L2' \cup L3' \cup L4' \cup \dots)'$ is regular, then its complement $L1' \cup L2' \cup L3' \cup L4' \cup \dots$ must be regular, but it is not. then it means $L1 \cap L2 \cap L3 \cap L4 \cap \dots$ is not regular.
is that correct ?
@praveen saini, in the third one,
L = {} is regular and M be non-regular like {0n1n | n>0}
then, can L.M becomes {0n1n}???????
concatenation means appending the second string to the end of the first string right??? in that case L.M is not regular. right???
please Clarify this????
yes sir. so in the above case, L.M becomes {0n1n} which is not regular. is it correct??
@Praveen sir concatenation of L=(a+b)* and M={a^nb^n | n>=0} two strings L and M will be
L.M = (a+b)*
because L = (ϵ, a, b, aa, ab, ba, bb, .....) * M=(ϵ,ab, aabb, aaabbb)
result will be (a+b)* may i right?
I did the same mistake. It’s necessarily not regular, and not, not necessarily regular.
(i) False.Infinite union or intersection both are not closed under regular language.
(ii)False . a^n b^n is subset of a*b* , but a^n b^n not regular
(ii)True . Let L=regular ,M=CFL ,then L.M=CFL
So, ans (B)
Answer is B .
and the reason for (iii) L.M necessarily regular is ,
consider ..L=(a+b)* which is regular ,
M=anbn which is CFL and not regular .
so concatinating these two strings , the resulting string will be regular which is nothing but (a+b)*.
So the word necessarily is important in this question.
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