A $($Q$,\backslash$Sigma $,\backslash$delta $,$ q$0,$ F$),$ the complement of L$($M$)$ is equal to the set $\{$w in $\backslash$Sigma $$ : $\backslash$delta $($q$0,$ w$)\backslash$cap $($Q $$ F$)=\}?$ If so$,$ prove it; if not, give a counterexample.

$1.$ If we L$($M$)\text{'},$ then w & L$($M$),$ meaning that the NFA M does not accept the string w$.$ This implies that dx $($q$0,$ w$)$ does not lead to any accepting state, i$.$e$.,$ d x $($q$0,$ w$)\backslash$cap F $=0.$

$2.$ If w & L$($M$)\text{'},$ then w in L$($M$).$ This means that the NFA M accepts the string w$,$ and there exists some path in M from the initial state qo to an accepting state using the transitions specified by d$.$ Therefore, dx $($q$0,$ w$)$ contains at least one state from F$,$ and d x $($q$0,$ w$)\backslash$cap F $!=0.$ Thus, the equivalence between w being in L$($M$)\text{'}$ and d x $($q$0,$ w$)$F $!=0.$ In summary, the correct definition for the complement of the language accepted by an NFA is L$($M$)\text{'}=\{$w in $\backslash$times : w$$L$($M$)\},$ and it is not equivalent to the set $\{$w $\backslash$Sigma $\backslash$Chi : $\backslash$delta $\backslash$times $($q$0,$ w$)\backslash$cap F $=0\}.$

The correct definition for the complement of the language accepted by an NFA is

L$($M$)\text{'}=\{$w in $\backslash$Sigma $\backslash$Chi : w & L$($M$)\},$ not$\{$w in $\backslash$Sigma x : $\backslash$delta $\backslash$times $($q$0,$ w$)$ F $=\}.$ W X: X

$.$ The proof establishes the equivalence between a string being in L$($M$)\text{'}$ and the absence of an intersection between the computed states and the set of accepting states. In summary, the provided set is not equivalent to the complement of the NFA language, which is accurately defined as the set of strings not in the language accepted by the NFA.

$9.$ Prove that for every nfa with an arbitrary number of final states there is an equivalent nfa with only one final state. Can we make a similar claim for dfa$$s$?$

Let there is an NFA N$1$ with n final states we can create another NFA N$2$ which is exactly same as N$1$ with exactly one more state Q$,$ such that all the final states of N$1$ have a transition $\backslash$lambda which goes to the Q$.$

Now make Q the final state and all the previou