# My proof for equivalence of DFA and NFA

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 $$L(N) = L(M)$$.

Q.E.D.

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