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

A word $w$ is accepted by the Epsilon-NFA precisely whenBy part (b), this reached subset is exactly ${\hat{\delta }}_D(q_D,w)$, and by the definition of $F_D$ the intersection condition is equivalent to ${\hat{\delta }}_D(q_D,w)\in F_D$. Hence