A nondeterministic Finite State machine computation is a sequence of transitions ending in a halting configuration, $\mathcal{C}{0}=\left\langle q{0}, \tau_{0}\right\rangle \vdash \mathcal{C}{1} \vdash \mathcal{C}{2} \vdash \cdots \mathcal{C}{h}=\langle q, \varepsilon\rangle$. From a given $\mathcal{C}{0}$ there may be many such sequences. If at least one ends with a halting state, having $q \in F$, then the machine accepts the input $\tau_{0}$, otherwise it rejects $\tau_{0}$.
With that, we will modify the definition of the extended transition function $\hat{\Delta}: \Sigma^{} \rightarrow Q$ from section 2. Begin by defining $\hat{\Delta}(\varepsilon)=\hat{E}\left(q_{0}\right)$. Then the rule for going from a string to its extension is that for $\tau \in \Sigma^{}$ and where $\hat{\Delta}(\tau)=\left{q_{i_{0}}, q_{i_{1}}, \ldots q_{i_{k}}\right} .$
$$\hat{\Delta}\left(\tau^{-} t\right)=\hat{E}\left(\Delta\left(q_{i_{0}}, t\right)\right) \cup \cdots \cup \hat{E}\left(\Delta\left(q_{i_{k}}, t\right)\right) \quad \text { for } t \in \Sigma$$
Observe that this nondeterministic machine with $\varepsilon$ transitions accepts a string $\sigma \in \Sigma^{*}$ if any one of the states in $\hat{\Delta}(\sigma)$ is a final state.

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