A DFA embeds as an epsilon-free NFA

For each letter pair $(q,a)$ the element $d(q,a)$ exists in $Q$, so its singleton is a subset of $Q$. Each epsilon value is also a subset of $Q$, and the tagged epsilon domain is disjoint from the letter domain. Thus the prescription is total into $\mathcal{P}(Q)$; the other tuple requirements are inherited from $D$. It defines $N_D$.

For any $S\subseteq Q$, the first saturation stage is $S\cup {\bigcup }_{q\in S}\mathrm{\varnothing }=S$. Whenever a stage equals $S$, its next stage has the same value. Induction gives $C_k(S)=S$ for all $k$, hence $E(S)=S$, including $S=\mathrm{\varnothing }$.

Fix $q\in Q$. For the empty word, $\widehat{\delta }(q,\epsilon )=E(\{q\})=\{q\}=\{d^{\ast }(q,\epsilon )\}$.

Assume $\widehat{\delta }(q,u)=\{d^{\ast }(q,u)\}$ for all words of length $k$. For a word $ua$ of length $k+1$, the union defining its next NFA state set has the single index $d^{\ast }(q,u)$. Therefore $\widehat{\delta }(q,ua)=E(\delta (d^{\ast }(q,u),a))=E(\{d(d^{\ast }(q,u),a)\})=\{d^{\ast }(q,ua)\}$. These equalities use the displayed singleton prescription, the identity closure, and the DFA successor equation.

Induction proves the singleton identity for every word and every initial state $q$. This remains valid when the alphabet is empty, since then there is only the empty word, and when $Q$ has one state.

At $q=q_0$, nonempty intersection $\widehat{\delta }(q_0,w)\cap F\mathrm{\ne }\mathrm{\varnothing }$ implies its sole possible element $d^{\ast }(q_0,w)$ lies in $F$. Conversely membership of that element in $F$ witnesses the nonempty intersection. The NFA and DFA acceptance tests therefore agree for every word, giving $L(N_D)=L(D)$; in particular empty $F$ rejects all words and $F=Q$ accepts all words.