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Consider the languages $L_{1}= \phi$ and $L_{2}=\{1\}$. Which one of the following represents $L_{1}^{\ast}\cup L_{2}^{\ast} L_{1}^{\ast}$?

1. $\{\in \}$
2. $\{\in,1\}$
3. $\phi$
4. $1^{\ast}$

### 1 comment

1* ∪ L 2* L 1*

φ* ∪ (1)*.φ*

* > . >  ∪ (priority)

ε ∪ (1)*.ε

ε ∪ (1)*

1*

## 7 Answers

Correct option is (4) 1*

Given $L_1 = \phi$, so, $L_1^* = \{ \epsilon \}$. Remember, Kleene's star operator means to repeat the strings any number of times (including 0 times).

$L_2 = \{ 1 \}$, therefore, $L_2^* = \{ \epsilon, 1, 11, 111, ... \}$, i.e $= 1^*$.

Now, $L_1^* \cup L_2^* . L_1^*$

$= \{ \epsilon \} \cup \{ 1^* \} . \{ \epsilon \}$

$= \{ \epsilon \} \cup \{ 1^* \}$

$= \{ 1^* \}$
L1=ϕ,

$L1^{*}$= {$\epsilon$,$\phi$,$\phi$ $\phi$,$\phi$$\phi$$\phi$.......}={$\epsilon$}

$L2={1}$

$L2^{*}$={$\epsilon$,1,11,111,1111............}=$1^{*}$

$L1^{*}\cup L2^{*} L1^{*}$

$\epsilon^{*}\cup1^{*}{\epsilon}$

$1^{*}$
option D) is correct 1*,

As, L1* ∪  L2*.L1*

={ϵ} ∪ {1∗}.{ϵ}

={ϵ} ∪ {1∗}

={1∗}
by
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