A complete determinisation table for a three-state epsilon-NFA

There are nine domain pairs; the three prescribed nonempty entries and the six empty entries all lie in $\mathcal{P}(Q)$. This makes a total epsilon-NFA with the stated start and finals. Its only epsilon edge is from $p$ to $q$, so saturation adds $q$ precisely when $p$ is present and $q$ absent. Hence $E(\{p\})=A$, and $A,B,C,Z$ are each closed.

From $A$, the $a$-union is $\mathrm{\varnothing }\cup \{q,r\}=B$ and the $b$-union is $\mathrm{\varnothing }\cup \mathrm{\varnothing }=Z$. From $B$, the $a$-union is $\{q,r\}\cup \mathrm{\varnothing }=B$ and the $b$-union is $\mathrm{\varnothing }\cup \{r\}=C$. Closing these unions leaves them unchanged. This computes the first two rows.

From $C$, the $a$-union is $\mathrm{\varnothing }=Z$ and the $b$-union is $\{r\}=C$. From $Z$, both unions are empty, so both transitions return $Z$. Again each result is closed. This computes the last two rows.

The start is $A$, and the words $\epsilon ,a,ab,b$ reach $A,B,C,Z$, respectively: the middle two follow the transitions $A\stackrel{a}{\to }B\stackrel{b}{\to }C$, and $A\stackrel{b}{\to }Z$. Thus each of the four states is reachable.

Every word reaches one of these four states. Indeed at length zero the state is $A$. If a prefix reaches one of the four, then appending $a$ or $b$ reaches one of the four by the computed rows. Induction on word length therefore gives the assertion for every word; the induction applies to the property that every word of a given length reaches the four-state set.

The eight subsets are $Z,\{p\},\{q\},C,A,\{p,r\},B,\{p,q,r\}$. By the two reachability implications, precisely $\{p\},\{q\},\{p,r\},\{p,q,r\}$ are unreachable. Of the reachable subsets, $B\cap F=C\cap F=\{r\}$ and $A\cap F=Z\cap F=\mathrm{\varnothing }$, giving exactly the displayed accepting states. The full DFA also designates its unreachable subsets containing $r$ as final; the table states the finals of its reachable part.