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Consider the languages $L_1 = \phi$ and $L_2 = \{a\}$. Which one of the following represents $L_1 {L_2}^* \cup {L_1}^*$ ?

  1. $\{\epsilon\}$
  2. $\phi$
  3. $a^*$
  4. $\{\epsilon, a\}$
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Best answer
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Concatenation of empty language with any language will give the empty language and ${L_1}^ * = \phi^* = \epsilon$.
 

Therefore,

$L_1L_2^* \cup L_1^*   $
$=\phi.(L_2)^* \cup \phi^ *$
$= \phi \cup \{\epsilon\} \left(\because \phi \text{ concatenated with anything is } \phi \text{ and }\phi^* = \{\epsilon\} \right)  $
$= \{\epsilon \} $.

Hence, option (A) is true.

PS: $\phi^* = \epsilon$, where $\epsilon$ is the regular expression and the language it generates is $\{\epsilon\}$. 

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L1.anything is empty language and empty union empty* is epsilon hence a
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5 votes

L_1L_2^* = \phi \because \text{ concantenation of any language with } \phi \text{ gives } \phi. \phi^* = \epsilon

Answer is A

Answer:

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