non-deterministic finite automaton state transition table language question I have the nfa problem below which I dont quite understand. I have been reading many slides on this but I'm having a really hard time understanding them enough to be able to complete the question. I can see that on the $q_0 $ state when we have an $a$ it should stay in the $q_0$ state and for the $q$ state when we have an $a$ it should go to the $q_0$ state but still dont understand the other information given well enough to construct the diagram. 

If anyone could help me understand the problem below it would be much appreciated.
Starting from the $M$ you gave in a comment (which you should really include in the question), we have the transition table
$$\begin{array}{r|cc}
\delta &amp; 0  &amp; 1 \\ \hline
q_0    &amp; q_1 &amp; q_2\\
q_1    &amp; q_0 &amp; q_2\\
q_2    &amp; q_2 &amp; q_2
\end{array}$$
with $q_0$ the start state and $q_1$ the final state. Now we modify this to a new NFA $M'$ by adding another final state $q'_0$ and construct the transitions $\delta'$:


$\delta'(q'_0, 0) = \{\delta(q_0, 0), q_0\}=\{q_1,q_0\}$, since $\delta(q_0,0)=q_1\in F$
$\delta'(q'_0, 1) = \{\delta(q_0, 1)\}=\{q_2\}$, since $\delta(q_0,1)=q_2\notin F$
$\delta'(q_0, 0) = \{\delta(q_0, 0), q_0\}=\{q_1,q_0\}$, since $\delta(q_0,0)=q_1\in F$
$\delta'(q_0, 1) = \{\delta(q_0, 1)\}=\{q_2\}$, since $\delta(q_0,1)=q_2\notin F$
$\delta'(q_1, 0) = \{\delta(q_1, 0)\}=\{q_0\}$, since $\delta(q_1,0)=q_0\notin F$
$\delta'(q_1, 1) = \{\delta(q_1, 1)\}=\{q_2\}$, since $\delta(q_1,0)=q_2\notin F$
$\delta'(q_2, 0) = \{\delta(q_2, 0)\}=\{q_2\}$, since $\delta(q_2,0)=q_2\notin F$
$\delta'(q_2, 1) = \{\delta(q_2, 1)\}=\{q_2\}$, since $\delta(q_2,0)=q_2\notin F$


giving us this transition table for $M'$
$$\begin{array}{l|cc}
\delta'     &amp; 0           &amp; 1 \\ \hline
\{q'_0\}    &amp; \{q_0,q_1\} &amp; \{q_2\}\\
\{q_0\}     &amp; \{q_0,q_1\} &amp; \{q_2\}\\
\{q_1\}     &amp; \{q_0\}     &amp; \{q_2\}\\
\{q_2\}     &amp; \{q_2\}     &amp; \{q_2\}\\
\{q_0,q_1\} &amp; \{q_0,q_1\} &amp; \{q_2\}
\end{array}$$
with $\{q'_0\}$ the start state and $\{q'_0\}, \{q_1\}, \{q_0,q_1\}$ the final states. Notice that $\{q_0\}$ and $\{q_1\}$ are unreachable so can be eliminated from the NFA. Notice also that in both FAs, state $q_2$ is a "trap" state: once you enter it (on input $1$), you'll never leave. This means that neither FA will accept any string containing a $1$ and from there it's not too hard to see what $L(M)$ and $L(M')$ are.