A word $w$ is accepted by the <Epsilon-NFA> precisely when
$$
\widehat\delta_E(q_0,w)\cap F_E\ne\varnothing.
$$
By part (b), this reached subset is exactly $\widehat\delta_D(q_D,w)$, and by the definition of $F_D$ the intersection condition is equivalent to $\widehat\delta_D(q_D,w)\in F_D$. Hence
$$
\boxed{\mathcal L(D)=\mathcal L(E).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 4G / c / Solution

{c}

An epsilon-NFA is a <nondeterministic finite automaton> that may traverse transitions labelled $\epsilon$ without consuming an input symbol.

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 Incoming links: Epsilon-NFA