I am reading Hopcroft Ullman - Automata Theory(2nd Edition).
In page Number 65 (Red Underline part in the given Image); I understand when i=1 but unable to understand when i>1 then how the Accepting and Non-accepting State are same.
My Point is "if i > 1 then a1 or b1 may be 1. So, then How it must be both Accepting and Non-Accepting; it may be Accepting and Non-Accepting ". I uploaded the Respective Image also
Please help me............ THIS IS THE PAGE WHERE I STUCK
At that point you know that the string
$$a_1\ldots a_n\underbrace{00\ldots 00}_{i-1\text{ zeroes}}\tag{1}$$
is a string of length $n+i-1$ that has a $1$ in position $i$, while
$$b_1\ldots b_n\underbrace{00\ldots 00}_{i-1\text{ zeroes}}\tag{2}$$
is a string of length $n+i-1$ that has a $0$ in position $i$. There are
$$(n+i-1)-i=n-1$$
bits after $a_i$ in string $(1)$, so $a_i=1$ is the $n$-th symbol from the end of string $(1)$. Similarly, $b_i=0$ is the $n$-th symbol from the end of string $(2)$. Thus, string $(1)$ should be accepted, and string $(2)$ should not.
Recall that $D$ is a DFA that accepts $L(N)$ and has fewer than $2^n$ states, and that $q$ is a state of $D$ such that $D$ is in state $q$ after reading either $a_1\ldots a_n$ or $b_1\ldots b_n$. Thus, $D$ is in state $q$ after reading the first $n$ bits of string $(1)$ or the first $n$ bits of string $(2)$. Let $p$ be the state reached by $D$ when it starts in state $q$ and reads $i-1$ zeroes; $D$ is deterministic, so $p$ is well-defined. Then $D$ ends up in state $p$ after reading either string $(1)$ or string $(2)$. The first of these strings is in $L(N)$, so $p$ should be an acceptor state; the second is not in $L(N)$, so $p$ should not be an acceptor state. This is the desired contradiction.
There’s nothing magical about a string of $i-1$ zeroes, by the way: the same argument would work equally well if we let $x$ be any string of $i-1$ bits and looked at the strings $a_1\ldots a_nx$ and $b_1\ldots b_nx$. The first of these has $a_i=1$ in the $n$-th position from the end, so it should be accepted by $D$, and the second has $b_i=0$ in the $n$-th position from the end, so it should not be accepted by $D$. But the inputs $a_1\ldots a_n$ and $b_1\ldots b_n$ both put $D$ in state $q$, and since $D$ is deterministic, the input $x$ starting from state $q$ will always take $D$ to the same state (corresponding to $p$ above). That state has to be an acceptor in order for $D$ to handle the input $a_1\ldots a_nx$ correctly, but it cannot be an acceptor if $D$ is to handle the input $b_1\ldots b_nx$ correctly.