Prove the following closure properties for regular languages. Prove that for any

regular language $L\frac{,}{b}ar(L)$ is also regular.

You can break the proof into following steps:

Closure Under Intersection: Intersection of $L_1$ and $L_2$ is $L_1{L}_2=$

$\{x|xin{L}_1^{{?}^{?}}xin{L}_2\}.$

Hint: Use the ideas from "closure under union of DFAs" $($closure of DFAs

directly, and not through NFAs$).$ From DFA $D_1=\{,Q_1,s_1,A_1,{}_1\}$

for $L_1,$ and DFA $D_2=\{,Q_2,s_2,A_2,{}_2\}$ for $L_2,$ construct DFA $D=$

$\{,Q_1{Q}_2,(s_1,s_2),A,\}$ for $L_1{L}_2.$ Clearly define A and $$ in terms of

$D_1$ and $D_2\text{'}$s components.

Closure Under Complement: Complement of $L$ is $\frac{?}{\text{b}\text{a}\text{r}}(L)=\{x|x!$inL$\}.$

Hint: From DFA $D=\{,Q,s,A,\}$ for $L,$ construct DFA $D^{\text{'}}=\{,Q,s,A^{\text{'}},{}^{\text{'}}\}$

for $\frac{?}{\text{b}\text{a}\text{r}}(L).$ Clearly define $A^{\text{'}}$ in terms of $,Q,s,A,.$

Closer Under Difference: Difference of $L_1$ and $L_2$ is

$\{$:$L_1^{{?}^{?}}x!in{L}_2\}.$

Hint: For $L_1$ and $L_2,$ represent $L_1-L_2$ as an expression over $L_1$ and $L_2$

that only uses "closure under intersection and complement".