 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/4g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

w

is accepted by the Epsilon-NFA precisely when

δ_{E} (q_{0}, w) \cap F_{E} \neq = \emptyset .

By part (b), this reached subset is exactly

δ_{D} (q_{D}, w)

, and by the definition of

F_{D}

the intersection condition is equivalent to

δ_{D} (q_{D}, w) \in F_{D}

. Hence

L (D) = L (E) .

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