$L_1 \cap L_2 = \epsilon $

$\epsilon \cup L_3$ is regular

so $L_5$ is regular

so only option C matches

$\epsilon \cup L_3$ is regular

so $L_5$ is regular

so only option C matches

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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 ?

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 ?

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