Proving Equivalence of DFA and NFA

Question

Im trying to learn Equivalence of DFA and NFA.The problem is that in the below explanation Q' is given as the power set of Q.But this statement seems to be contradictory to the previous statement saying NFA is mapped from Q*E to 2^Q,Why is this?

Answer 1

I see nothing at all contradictory. $Q'$ is defined to be $2^Q$, the power set of $Q$. There is no mention of $E$ at all. My best guess is that you mean $\Sigma$, the input alphabet. If that’s the case, the map from $Q\times\Sigma$ to $2^Q$ is $\delta$, the transition function for the NFA $M$: for any state $q\in Q$ and $a\in\Sigma$, $\delta(q,a)$ is a set of states of $M$, i.e., a subset of $Q$. Specifically, $\delta(q,a)$ is the set of states $p$ such that there is an $a$-transition from $q$ to $p$.

This is one of the differences between DFAs and NFAs. If $Q$ is the set of states of an NFA with input alphabet $\Sigma$, the transition function for the NFA is a function from $Q\times\Sigma$ to $2^Q$: for each $a\in\Sigma$ there is a set of $a$-transitions out of each $q\in Q$. If, however, $Q$ is the set of states of a DFA with input alphabet $\Sigma$, the transition function for the DFA is a function from $Q\times\Sigma$ to $Q$: for each $a\in\Sigma$ there is exactly one $a$-transition out of each state $q\in Q$.