Prove that every NFA can be converted to an equivalent one that has a single accept state.

Each NFA may be convertible to a DFA of the same size. DFAs, NFAs, as well as conditionals are all the same thing. In above question through we can see proper correction or we can in technical language output in step 2.

Below we can see the above question related answer.

Make $\mathrm{M}$ a NFA (non deterministic finite automatic)

Let role="math" localid="1663185624203" $\mathrm{N}$be a second NFA with only one approve state,$q_{fina{l}_{-}}$

We go over each accept state $\mathrm{M}$and $\mathrm{I}$.

1. create it a semi state
2. add a -transition from such a stage to$q_{final}$

After that, we'll receive NFA $\mathrm{N}$.

There will be no transfers into $q_{final}$if $\mathrm{M}$does not have any accept states.

From Now, we can communicate on this topic further.

Then,

$$\begin{gathered} M=\left(Q,\Sigma ,\delta ,q_0,F\right)then \\ N=\left(Q\nu \right\{q_{final}\},\sum ,\delta ,q_0,\{q_{final}\}foranyq\sum Qanda\in \sum \end{gathered}$$

$$\begin{gathered} \delta \text{'}(q,a)=\left\{\delta (q,a)ifa\ne \in orq\notin F \\ \delta (q,a)\cup \left\{q_{final}\right\}ifa=\in and q\in F\right\} \end{gathered}$$

And$q\text{'}(q_{final},a)=\varnothing$

Thus, we obtain N by converting each accept state ofrole="math" localid="1663186183860" $\mathrm{M}$ to a non-accepting state and adding a crossover from such a state to qfinal.

As a result, $M$equals$N$.

As a result, each NFA is transformed into an equivalent with a single accept state.