Solution
= Solution
Set $p\sim q$ when $\widehat\delta(p,w)\in F\iff\widehat\delta(q,w)\in F$ for every word $w$. Put $\delta'([q],a)=[\delta(q,a)]$ and $F'=\{[q]:q\in F\}$. If $p\sim q$, then testing continuations $aw$ shows $\delta(p,a)\sim\delta(q,a)$, so $\delta'$ is well-defined.