CSL and Regular language

Question

if L1 = { aⁿbⁿcⁿ   | n>= 0 } 

and L2 = { aⁿb^mc^k | k,n,m>=0}

L1 is CSL and L2 is regular.

Now  L3 = L1.(L2)*.

Is L3 is regualar or CSL?

Comments

Yes there exists concatenation but no cross product in toc

---

so sorry,i mixed them..:-P

---

Is L2 is regular?

Answer 1

Regular

L1 = { $\epsilon$, abc, aabbcc, ... }

L2 = a*b*c*

L3 is L1.(L2)*, means $a^nb^nc^n(a^*b^*c^*)^*$

The important thing to notice is that n can be 0. So L3 will be  $(a^*b^*c^*)^*$. Which is regular.

Comments

@Archies09 Oh I see it now. Sorry. They are same :)

---

So , shouldn't the answer be both ?

Every Regular Language is a CSL right.

---

@Harsh That's correct. Every regular language is CSL. So both are correct. Though regular language answer is more specific.

Answer 2

R1(R2)* is simply (a+b+c)*....and hence regular

Answer 3

L1 is CSL

and L2 is regular.

then (L₂)* will be reguler (by using closure properties)

and now L₁.(L₂)*=CSL.Reguler(push up) 

CSL.CSL=CSL

Comments

correct i have also same explanation..