0 votes 0 votes Here why they are saying we must convert DFA transition into NFA transition? Theory of Computation theory-of-computation finite-automata non-determinism + – Bhaskar Singh asked Mar 3, 2019 • reopened Mar 3, 2019 by Bhaskar Singh Bhaskar Singh 597 views answer comment Share Follow See 1 comment See all 1 1 comment reply ankitgupta.1729 commented Mar 3, 2019 reply Follow Share 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. 0 votes 0 votes Please log in or register to add a comment.