DFA 'M'
$M = (Q,\Sigma, \delta,q_0,F)$
Let, $Q = \{q_0,q_1\}, F= {q_1}$ and the transition function be defined as -
$\delta:$
$L = \{\sigma,\sigma_3,\sigma_5\ldots \}$ (odd number of $\sigma s$)
NFA $'M_2'$
$M_2 = (Q_2,\Sigma,\delta_2,q_{00},F_2)$
$Q_2 = \{q_{00}, $
- $\langle q_0,q_0,q_0 \rangle$
- $\langle q_0,q_0,q_1\rangle$
- $\langle q_0,q_1,q_0\rangle$
- $\langle q_0,q_1,q_1\rangle$
- $\langle q_1,q_0,q_0\rangle$
- $\langle q_1,q_0,q_1\rangle$
- $\langle q_1,q_1,q_0\rangle$
- $\langle q_1,q_1,q_1\rangle$}
Now, (A) $F_2 = A = \{\langle p,q,r \rangle\mid p \in F; q,r \in Q\} \forall p,q,r \in Q$ & $\sigma \in \Sigma$
Now NFA $M_2$ will be
$L' = \{\sigma,\sigma_3,\sigma_5\ldots \}$
so $L' = L$ so $M_2 \equiv M$
$\Rightarrow$ we can take other transition rules for $'M'$ like,
etc or $Q= \{q_0,q_1,q_2\ldots \}$
In all cases, $L' = L$ because, we have to start with triplet $\langle q_0, , \rangle$ & then follow the path as in $'M'$ and first part of the triplet is important here because it shows the final state of NFA $'M_{2}’$ since $\Sigma = \{\sigma\}$ only, we have to trace first part of the triplet of $M_2$ which is same as tracing of DFA $'M'$.