I came up with this proof for the following theorem:

If $L$ is a language produced by an nfa, then there exists a dfa $M$ where $L(M) = L$.

Is it correct? If it's not, what are the flaws of my proof, and if it is, how can I improve it?

My Proof: Let the nfa be $N = (Q, \Sigma, \delta, q_0, F)$. We construct a dfa $M$ where $M = (2^Q, \Sigma, \delta', p_0, F')$ (note that the set operations are defined on the labels of the states of $M$). Let $p_0$ be the state labeled as $\delta^*(q_0, \epsilon)$, and $F'$ be all those states like $p$ where $p \cap F \neq \emptyset$. Finaly, define $\delta'$ in the following manner:

Consider two states of $M$, $u$ and $v$, and a character $a$, we say $\delta'(u, a)=v$ if and only if $v = {\bigcup_{q \in u}} \delta^*(q, a)$.

We now proof by induction on the length of $w$ that $\delta^*(q_0, w)=\delta^{'*}(p_0, w)$.

Induction basis ($|w|=0$): By the definition of $p_0$, $p_0 = \delta^*(q_0, \epsilon)$, and thus $\delta^{'*}(p_0, \epsilon) = p_0 = \delta^*(q_0, \epsilon)$.

Inductive step ($w = w' + a$): We know that $\delta^*(q_0, w) = \bigcup_{q\in\delta^*(q_0, w')} \delta^*(q, a)$. By the induction hypothesis, $\delta^*(q_0, w')=\delta^{'*}(p_0, w) $ (call it as $p$), thus $\delta^*(q_0, w) = \bigcup_{q\in p} \delta^*(q, a)$, and finaly be the definition of $\delta'$, $\delta^*(q_0, w) = \delta'(p, a) = \delta^{'*}(q_0, w)$.

Therefore, $F\cap \delta^*(q_0, w) \neq \emptyset \iff F\cap \delta^{'*}(p_0, w) \neq \emptyset \iff \delta^{'*}(p_0, w) \in F'$, thus $L(N) = L(M)$.


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    $\begingroup$ We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. $\endgroup$
    – Discrete lizard
    Oct 11 '21 at 12:06

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