prove Certainly! Let's go through the proofs for $**$complement$**,**$union$**,$ and $**$intersection$**$ in more detail, thoroughly explaining each step and concept. ### $1.**$Complementation of a Regular Language$**$ #### Theorem: If a language $\backslash ($ L $\backslash )$ is regular, then its complement $\backslash (\backslash$overline$\{$L$\}\backslash )$ is also regular. #### Proof: $1.**$Given$**$: A regular language $\backslash ($ L $\backslash ),$ meaning there exists a Deterministic Finite Automaton $($DFA$)\backslash ($ M $=($Q$,\backslash$Sigma$,\backslash$delta$,$ q$\_0,$ F$)\backslash )$ that recognizes $\backslash ($ L $\backslash ).-\backslash ($ Q $\backslash )$ is the set of states. $-\backslash (\backslash$Sigma $\backslash )$ is the alphabet. $-\backslash (\backslash$delta $\backslash )$ is the transition function $\backslash (\backslash$delta: Q $\backslash$times $\backslash$Sigma $\backslash$to Q $\backslash ),$ determining the next state. $-\backslash ($ q$\_0\backslash )$ is the initial $($start$)$ state. $-\backslash ($ F $\backslash )$ is the set of final $($accepting$)$ states where, for any input string $\backslash ($ w $\backslash ),$ if $\backslash ($ M $\backslash )$ ends in a state in $\backslash ($ F $\backslash ),\backslash ($ M $\backslash )$ accepts $\backslash ($ w $\backslash ).2.**$Goal$**$: To construct a DFA that recognizes the complement of $\backslash ($ L $\backslash ),\backslash (\backslash$overline$\{$L$\}\backslash ),$ where: $-\backslash (\backslash$overline$\{$L$\}=\backslash$Sigma$^*\backslash$setminus L $\backslash ),$ meaning it contains all strings that are not in $\backslash ($ L $\backslash ).3.**$Method$**$: $-$ The idea is simple: to recognize the complement language $\backslash (\backslash$overline$\{$L$\}\backslash ),$ we need a DFA that accepts exactly those strings that the DFA for $\backslash ($ L $\backslash )$ rejects. $-$ The only change required is to $**$flip the accepting states$**$ and the $**$non$-$accepting states$**$ of the original DFA. $-$ Specifically, the new DFA $\backslash ($ M$\text{'}\backslash )$ will have the same states and transitions as $\backslash ($ M $\backslash ),$ but its accepting states will be $\backslash ($ Q $\backslash$setminus F $\backslash )($the complement of the final states in the original DFA$).4.**$Construction$**$: $-$ Let the new DFA $\backslash ($ M$\text{'}=($Q$,\backslash$Sigma$,\backslash$delta$,$ q$\_0,$ Q $\backslash$setminus F$)\backslash ),$ where: $-$ The state set $\backslash ($ Q $\backslash ),$ transition function $\backslash (\backslash$delta $\backslash ),$ and start state $\backslash ($ q$\_0\backslash )$ remain unchanged. $-$ The final states are now $\backslash ($ Q $\backslash$setminus F $\backslash )($states that were not final in $\backslash ($ M $\backslash )).5.**$Conclusion$**$: $-$ This new DFA $\backslash ($ M$\text{'}\backslash )$ recognizes the complement of $\backslash ($ L $\backslash ),\backslash (\backslash$overline$\{$L$\}\backslash ),$ since it accepts exactly the strings that $\backslash ($ M $\backslash )$ rejects and rejects the strings that $\backslash ($ M $\backslash )$ accepts. $-$ Thus, the complement of a regular language is also regular. ### $2.**$Union of Two Regular Languages$**$ #### Theorem: If two languages $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash )$ are regular, then their union $\backslash ($ L$\_1\backslash$cup L$\_2\backslash )$ is also regular. #### Proof: $1.**$Given$**$: Two regular languages $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash ),$ with corresponding DFAs $\backslash ($ M$\_1=($Q$\_1,\backslash$Sigma$,\backslash$delta$\_1,$ q$\_1,$ F$\_1)\backslash )$ and $\backslash ($ M$\_2=($Q$\_2,\backslash$Sigma$,\backslash$delta$\_2,$ q$\_2,$ F$\_2)\backslash )$ that recognize $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash ),$ respectively. $-\backslash ($ Q$\_1\backslash ),\backslash ($ Q$\_2\backslash )$: States of DFAs $\backslash ($ M$\_1\backslash )$ and $\backslash ($ M$\_2\backslash ).-\backslash ($ F$\_1\backslash ),\backslash ($ F$\_2\backslash )$: Final states of $\backslash ($ M$\_1\backslash )$ and $\backslash ($ M$\_2\backslash ).-\backslash (\backslash$delta$\_1\backslash ),\backslash (\backslash$delta$\_2\backslash )$: Transition functions of $\backslash ($ M$\_1\backslash )$ and $\backslash ($ M$\_2\backslash ).2.**$Goal$**$: Construct a DFA that recognizes the union $\backslash ($ L$\_1\backslash$cup L$\_2\backslash ).3.**$Method$**$: $-$ The key technique is to construct a $**$product automaton$**,$ a DFA that simulates both $\backslash ($ M$\_1\backslash )$ and $\backslash ($ M$\_2\backslash )$ in parallel. The idea is to track both automata's state simultaneously and accept if either one of them is in a final state. $4.**$Construction$**$: $-$ The state set $\backslash ($ Q $\backslash )$ of the new DFA $\backslash ($ M $\backslash )$ is the $**$Cartesian product$**$ of the states of $\backslash ($ M$\_1\backslash )$ and $\backslash ($ M$\_2\backslash ),$ i$.$e$.,\backslash ($ Q $=$ Q$\_1\backslash$times Q$\_2\backslash ).$ A state in the new DFA represents a pair of states, one from $\backslash ($ M$\_1\backslash )$ and one from $\backslash ($ M$\_2\backslash ).-$ The transition function $\backslash (\backslash$delta $\backslash )$ for $\backslash ($ M $\backslash )$ is defined as follows: for any symbol $\backslash ($ a $\backslash$in $\backslash$Sigma $\backslash )$ and any pair of states $\backslash (($q$\_1,$ q$\_2)\backslash$in Q$\_1\backslash$times Q$\_2\backslash ),\backslash [\backslash$delta$(($q$\_1,$ q$\_2),$ a$)=(\backslash$delta$\_1($q$\_1,$ a$),\backslash$delta$\_2($q$\_2,$ a$)).\backslash ]-$ The start state of $\backslash ($ M $\backslash )$ is $\backslash (($q$\_1,$ q$\_2)\backslash ),$ where $\backslash ($ q$\_1\backslash )$ is the start state of $\backslash ($ M$\_1\backslash )$ and $\backslash ($ q$\_2\backslash )$ is the start state of $\backslash ($ M$\_2\backslash ).-$ The set of final states is $\backslash ($ F $=\backslash \{($p$\_1,$ p$\_2)\backslash$in Q$\_1\backslash$times Q$\_2$ : p$\_1\backslash$in F$\_1\backslash$text$\{$ or $\}$ p$\_2\backslash$in F$\_2\backslash \}\backslash ).$ In other words, a string is accepted if $**$either$**\backslash ($ M$\_1\backslash )$ or $\backslash ($ M$\_2\backslash )$ is in a final state after processing the string. $5.**$Conclusion$**$: $-$ The resulting DFA $\backslash ($ M $\backslash )$ accepts exactly the strings that are accepted by $\backslash ($ M$\_1\backslash )$ or $\backslash ($ M$\_2\backslash ),$ which means it recognizes $\backslash ($ L$\_1\backslash$cup L$\_2\backslash ).-$ Therefore, the union of two regular languages is regular. ### $3.**$Intersection of Two Regular Languages$**$ #### Theorem: If two languages $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash )$ are regular, then their intersection $\backslash ($ L$\_1\backslash$cap L$\_2\backslash )$ is also regular. #### Proof: $1.**$Given$**$: Two regular languages $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash ),$ with corresponding DFAs $\backslash ($ M$\_1=($Q$\_1,\backslash$Sigma$,\backslash$delta$\_1,$ q$\_1,$ F$\_1)\backslash )$ and $\backslash ($ M$\_2=($Q$\_2,\backslash$Sigma$,\backslash$delta$\_2,$ q$\_2,$ F$\_2)\backslash )$ that recognize $\backslash ($ L$\_1\backslash )$ and $\backslash ($ L$\_2\backslash ),$ respectively. $2.**$Goal$**$: Construct a DFA that recognizes the intersection $\backslash ($ L$\_1\backslash$cap L$\_2\backslash ).3.**$Method$**$: $-$ Similar to the union case, we construct a $**$product automaton$**.$ This time, however, we want the DFA to accept a string if $**$both$**$ DFAs are in a final state after processing the string. $4.**$Construction$**$: $-$ The state set $\backslash ($ Q $\backslash )$ is again the Cartesian product $\backslash ($ Q$\_1\backslash$times Q$\_2\backslash ).-$ The transition function $\backslash (\backslash$delta $\backslash )$ is the same as in the union case: for any symbol $\backslash ($ a $\backslash$in $\backslash$Sigma $\backslash ),\backslash [\backslash$delta$(($q$\_1,$ q$\_2),$ a$)=(\backslash$delta$\_1($q$\_1,$ a$),\backslash$delta$\_2($q$\_2,$ a$)).\backslash ]-$ The start state is $\backslash (($q$\_1,$ q