Ullman 2nd edition Theorem 2.22 (Section 2.5)

Question

Here why they are saying we must convert DFA transition into NFA transition? 

 

Comments

It is simply saying that we can write the DFA transition of $\delta _{D}(q,a) = p$  into $\epsilon$-NFA transition of $\delta _{E}(q,a) = \{p\}$ because in DFA, on any input symbol, we can go from one single state to another single state because it is deterministic machine model unlike one state to multiple states in case of $\epsilon$-NFA. So, $p$ of DFA is same as $\{p\}$ of $\epsilon$-NFA that's why we say that every DFA is already NFA or $\epsilon$-NFA.