+1 vote
138 views

retagged | 138 views
+3
$L_1 \cap L_2 = \epsilon$
$\epsilon \cup L_3$  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
$(L_3)^*=\sum ^*$

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