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in Theory of Computation by Active (1.3k points)
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$L_1 \cap L_2 = \epsilon $
$\epsilon \cup L_3$  is regular

so $L_5$ is regular

so only option C matches
+1
L1 intersection L2 = $\epsilon$
0
Lokesh, can you comment on how L4 is regular?
0
$(L_3)^*=\sum ^*$

1 Answer

0 votes
By elemination we can conclude that answer is C

But.

Although I understand that L5 is regular. But i dont inderstand how L4 is regular.

CSL is closed under concatenation and since

L1.L3*

CSL. REG*

CSL. REG

Since every regular language is also a CSL and its a strict subset. Therefore

CSL.(REG union $)

CSL. CSL

wiz. CSL only.

How L4 can be regular ?
by Active (2.3k points)
+1
If you think in this way you will never get it regular
see,
do you agree $(L_3)^* =\sum ^*$ which is regular

now $L_2\cdot (L_3)^* = \{ a^nb^nc^n|n\geq0\}\cdot \sum ^*$

put n=0 then you can generate any string$\in \sum ^*$

which makes $L_4$ regular
0
And if n=1,2,3,4... then ?
It will be CSL, right ?
0
every language which is regular is CSL also

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