Finding whether given languages are regular or context free

Question

Given

$L_1=\{a^nb^nc^n | n\geq 0\}$
$L_2 =\{a^nb^mc^k|k=n+m  \text{  and  }n,m\geq 0 \}$
$L_3 =\{a^nb^mc^k|n,m,k \geq 0 \}$
Assume 
$L_4=L_1 (L_3)^*$
$L_5=(L_1\cap L_2)\cup L_3 $
Which of the following statement is correct?
A. L4 is regular and L5 is not regular
B. L4 is CFL and L5 is not CFL
C. Both L4, L5 are regular
D. Both L4, L5 are CFL but not regular

Comments

both $L_4$ and $L_5$ are regular languages.

$L_4=(a+b+c)^*$

$L_5=a^*b^*c^*$

hence option C is correct

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sorry I mis-written $L_3$ as $L_2$ in question. Now corrected.
My doubts:

• how $L_4=(a+b+c)^*$?
  Given is $L_4=L_1.(L_3)^*$
  So in each string $s\in L_4$, we will have string from $L_1$ as a prefix, that is we will have string of the form $a^nb^nc^n$ as a prefix, right? (am I understanding the formation of $L_4$ correctly?) If yes, then I guess $a^nb^nc^n$ cannot be part of any regular language right?
• About $L_5$, I can reach till this:
  $L_5=(L_1\cap L_2)\cup L_3=\{\epsilon \}\cup L_3=L_3$
  But then $L_3$ does not seem to be regular right? $L_3$ contains all strings of the form $a^nb^mc^k$. If we look at the definition of $L_1$ (which contains all strings of the form $a^nb^nc^n$), it feels as if $n\neq m \neq k$ and in strings of $L_3$, number of occurrences of $a$, $b$ and $c$ should be different, right? So that does not translate to regular expression $a^*b^*c^*$, right?

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I have just realized that if $n=0$ in $L_1$, then $L_4=L_3^*=(a^*b^*c^*)^*=(a+b+c)^*$, so prefixing any language to it will result in $(a+b+c)^*$ only. But then this presumes that the definition of $L_3={a^nb^mc^k}$ does not imply that $n\neq m\neq k$ and that is why above I wrote $(a^*b^*c^*)^*$ instead of $(a^nb^mc^k)^*$. If we assume that the definition of $L_3={a^nb^mc^k}$ does not imply that $n\neq m\neq k$, $L_5$ also becomes regular as you said. But the assumption is necessary, right?

Answer 1

option c 

Comments

How are you taking intersection as a €