 So, far we have discussed DFA and NFA. So, by introducing some non-technical transformations in DFA, we have constructed NFA. Looking at that, that means introducing more flexibility, it looks like that NFA is much more powerful or more general and much more powerful than DFA. That means, it looks like that the class of languages accepted by NFA is different from that of the DFA, but actually that is not the case. We are going to show that DFA and NFA are in fact equivalent. So, when you say equivalence between two automaton say automaton A and automaton A dash, we say that A and A dash are equivalent. However, the language accepted by A and the language accepted by A dash are the same. In that context, we want to prove now that DFA and NFA are equivalent. So, while proving the equivalence, we will see that one direction is quite obvious. That means, every DFA can be treated as an NFA. It is a special case of NFA, but let us prove the other side. That means, the converse. That means, for every NFA, there is an equivalent DFA. So, when we consider an NFA, we will first consider an NFA with epsilon transition. An NFA with epsilon transition, then what we will do? We will construct an NFA, equivalent NFA without epsilon transition. Once you do that, from that NFA, which is equivalent to the original NFA, we will now construct an equivalent DFA. That means, there will not be any multiple next steps or there will not be any move from a state on input symbol to multiple next steps and there has to be a next move or next step from every state on every input symbol. So, NFA without epsilon transition from there, we will construct a DFA. So, we will first start with a multiple example. Let us consider this NFA, which has the three states. Q 0 is the initial state. Q 1 is the next state. Q 0 on A, there is no input symbol, say the input symbols are only A and B. So, Q 0 on B remain the same state or it may also go to the state B, a state Q 1 on input symbol B and Q 0 without taking an input may also go to Q 1. That means, from Q 0 to Q 1, there are two transition on epsilon and on B. Q 1 on A and B, it may remain on the same state, a self flip on A and B and from Q 0, there is a transition to a state Q 2 on epsilon and Q 2 is also a final state. Similarly, from Q 1, there is again a transition on input B to state Q 2. That means, the transition map, say it is delta is given by in a three states Q 0, Q 1, Q 2 on A, B and epsilon. So, Q 0 on A, it is does not go to anywhere. So, it is fine. Q 0 on B goes to two states, one is Q 0 itself and other is Q 1. So, it is a set of states, next states basically Q 0 and Q 1 on input symbol B and similarly, Q 0 on epsilon it will go to only one state, sorry two states Q 1 and Q 2. Similarly, Q 1 on A goes to single state, the same state is Q 1 and Q 1 on B may remain the same state Q 1 or it may go to Q 2 and on epsilon it does not have any transition. So, it is fine. Similarly, Q 2 on A, B and epsilon there is no any transition. So, it is all fine or A, B and epsilon. So, this is a transition map for this particular NFA with epsilon transition. Now, let us see how to remove epsilon transition. That means, how to get an NFA, which is equivalent to this NFA, but there is no any epsilon transition. So, the main point is that, if this NFA where transition map is delta, if we start at Q 0, if on input string acts it leads us to some state there will be multiple next states. Suppose, P is one such state there will be multiple next states in such a case. So, this string acts will be accepted provided from P eventually there is a transition to one of the final states. That means, in the equivalent NFA all those states will be final for which there is an epsilon transition to one of the final states. So, looking at that what we will do is that from Q 0 on epsilon there is an epsilon move to Q 2 or Q 2 is a final state. So, Q 0 will be made a final state in the equivalent NFA without epsilon transition, but from Q 1 there is no any transition on epsilon to the final state Q 2. So, Q 1 will not be final state and hence in the equivalent NFA will have two final states Q 0 and Q 2 and then the transition map state is delta S. So, transition map will be nothing but the extended transition function for the given NFA with epsilon transition and here we will have Q 0 Q 1 and Q 2 the same set of states Q 2 will be a final states. So, here Q 2 was a final state and here in the equivalent NFA without epsilon transition Q 0 will also be a final state here Q 0 was an initial state Q 0 still remains to be an initial state. Q 0 will be final state since we have an epsilon transition from Q 0 to Q 2. Next the transition function for this is defined as suppose on some input symbol A whenever this NFA without epsilon transition goes to some states on any input from any state. Then what we need to consider is that we need to consider all possible transitions inserting epsilon before and after A that means we need to look at the strings epsilon A A epsilon or epsilon A epsilon. So, we need to consider the transition out of every state from out of every state for this NFA on the input strings epsilon A A epsilon and epsilon A epsilon and we need to consider that is the set of next states corresponding to Q 0 on input symbol A. So, Q 0 on A if we consider there is no move only on A from out of Q 0, but taking an epsilon it can go to Q 1 and from Q 1 taking an A it will remain at Q 1. So, Q 1 will be the one of the next states for this. Next taking epsilon from at Q 0 it may go to Q 2, but on Q 2 there is no any transition out of Q 2 on A. So, therefore, Q 2 will not belong to the delta Q 0 A. So, therefore, Q 1 is the only state that will have for delta hat Q 0 A. Similarly, Q 0 on B if we consider the string B. So, Q 0 on B it may taking B it may remain the same state Q 0 or taking a single B it may go to Q 1 or taking epsilon it may go to Q 1 and then it may remain the same state Q 1 taking B. That means, if you consider string epsilon B on that it will again be the same state Q 1. So, Q 1 is already there or taking epsilon at Q 0 it may go to Q 1 and then taking this B it may go to Q 2. So, therefore, Q 2 will also be there in a set of next states for delta hat Q 0 B. Similarly, if you consider Q 0 taking epsilon it may go to Q 2, but there is no transition out of Q 2 on B. So, therefore, Q 2 that will not be included, but Q 2 is already included via this part Q 0, Q 1, Q 0 on epsilon Q 1 and Q 1 taking B it may go to Q 2. So, there are only possibilities that we have Q 0 on B. Similarly, Q 1 on A it may remain the same state Q 1 and there is no epsilon move out of Q 1. So, therefore, there is only possible case for delta hat Q 1 A. Then for delta hat Q 1 B from Q 1 taking B it may go to it may remain the same state Q 1 and there is no epsilon transition and then Q 1 on B it may go to Q 2 also. Therefore, the possible next states are Q 1 and Q 2 and then Q 2 out of Q 2 there is no any transition on any input symbol or epsilon as well. So, there are only possible cases that we have considering this. So, just consider the transition diagram corresponding to this automata that we have that we have just constructed. So, in this case also we have three states Q 0 initial state then you have Q 1 and Q 2 both Q 1 and Q 2 Q 0 and Q 2 are final states as well Q 0 on A goes to Q 1 and Q 0 on B it may remain the same state Q 0 it may go to Q 1 or it may go to B. So, there are three transitions out of Q 0 on input symbol B similarly, from Q 1 on input symbol A it may remain the same state Q 1 on A remain the same state and Q 1 on B may remain the same state or it may go to Q 2. So, this is a transition diagram for this. So, let us see what are these two this NFA and this NFA without epsilon transition this NFA with epsilon transition and we have said that these are NFA without epsilon transition what are they accept the what are they accept the same language or not. If we see carefully that the NFA with epsilon transition can accept the string epsilon because there is an epsilon move from the star state to the final state or if you consider via the other part it may take epsilon transition may go to Q 1 and then take any strings over A B because of this loop self loop on A B and then eventually you will go to Q 2 the final state on input symbol B or beforehand it can take any numbers of B's and this single B and then go to Q 1 then take again any strings over A B and eventually that string terminated by B we accepted by this NFA. That means, eventually any string over A B terminated by symbol B we accepted by this NFA with epsilon transition. Let us see whether this NFA without epsilon transition can also accept the same string same language or not since the star state Q 0 is a final state as well. Therefore, epsilon is automatically accepted by this NFA and whenever from Q 0 it goes to the other final state then it will go on a single input symbol B or if it wants to accept if it accepts any string then the string of the of this form is B star because of this self loop or otherwise for the other from for the other part from Q 0 it may go to Q 1 on taking any A single A or B and then any numbers of A's and B's because of this self loop and eventually it will be terminated by B. So, here also if you look carefully is that any string over A B terminated by B is also accepted. So, therefore epsilon plus A B is a language of this NFA as well that means this NFA is equivalent this NFA without epsilon transition is equivalent to this NFA with epsilon transition. In fact, how have you constructed this NFA without epsilon transition from the original NFA with epsilon transition. So, it is quite simple that means given an NFA then say this Q sigma delta Q 0 F suppose NFA with some epsilon transition then we can construct NFA say and this without epsilon transition then we will retain the same set of states Q over the same input alphabet we are going to modify the transition map which is nothing but the extended transition function of the original NFA n then the set of start set remains same and the set of fun sets we modified to say F this is an equivalent NFA without epsilon transition. What we have done is that the transition function for this NFA without the transition there is delta this defined like this. So, delta this Q A is basically delta hat Q A for all A belonging to sigma and Q belonging to capital Q set of states. So, for any state Q and input symbol A delta this Q A is defined as delta hat Q A for the original NFA and F dash is nothing but as you have you observe that is nothing but we have retained the same final sets F union any state Q in the original NFA such that epsilon closure that means if we can if we take epsilon move from any state and the set of next sets we see if it contains any one final states then we say that that state is also in the final state that epsilon closure of Q and F if this intersection is not equal to phi then we consider this Q to be included in the set of final states in the modified NFA. So, this is a construction that we have observed and we have seen that the corresponding NFA is equivalent to the original NFA the original NFA had epsilon transition but this NFA does not have any epsilon transition but they except the seven language. So, this is a construction that we have got but let us now prove we have shown it by an example and intuitively it looks like that this construction works but let us now prove that this construction really works. So, we are going to give a formal proof that these two NFA are equivalent that means they except the same language we claim that the language of the original NFA with epsilon transition is equivalent or identical to the language of the NFA without epsilon transition that we have constructed. So, first we will see that if epsilon belongs to the language of the original NFA with epsilon transition. So, epsilon belongs to l n if and only if delta hat that means there is a transition from Q 0 on epsilon from a star state Q 0 on epsilon and the set of next states that we have on epsilon has at least one state that belongs to the set of final state that means delta hat Q 0 epsilon and f has some elements come on that means interaction of these two sets not equal to 5. So, there is a condition according to the acceptance when NFA. So, this is a case if and only if so this is nothing but epsilon closure of Q 0 delta hat Q 0 epsilon is nothing but epsilon closure of Q 0. So, this interaction f is not equal to the null set this is a case if and only if Q 0 belongs to f just because there is a way we have defined the set of final states. So, f there is is f union all those states Q we are into the set of states such that epsilon closure Q and f the interaction is not equal to 5. So, accordingly this will this is nothing but Q 0 must belong to F dash, but what it says that since Q 0 belongs to F dash epsilon is automatically in the language of n s. So, we have seen that if epsilon belong to the language of this NFA without which epsilon transition then epsilon mass also belong to the language of this NFA without epsilon transition. Now, for any string other than epsilon belong to sigma that means for any string x belong to sigma plus we proved that delta dash hat Q 0 x equal to delta hat Q 0 x. So, that this eventually says that l of n equal to l of n s this because if delta dash hat Q 0 x. So, if delta dash hat Q 0 x equal to delta hat Q 0 x then x belongs to l of n x belongs to l of n if and only if delta hat Q 0 x intersection f not equal to 5 that is by definition of acceptance of a string x by the original automata original NFA. So, if and only if delta dash hat Q 0 x intersection f not equal to 5 because that is how we have defined the function of the new NFA delta dash. Now, what we know is that this f dash contains. So, f is a subset of f dash. So, f dash may contain some element which is not there in f. So, therefore, if this is the scenario say delta dash hat Q 0 x intersection f not equal to 5 then this implies that delta dash hat Q 0 x intersection f dash not equal to 5. But the converse may not be true that means if delta dash hat Q 0 x intersection f dash intersection f dash not equal to 5 then it may not be the case that delta dash hat Q 0 x intersection f not equal to 5 because f is a subset of f dash. If this is the case then it implies that x belongs to l of n dash according to the definition and if x belongs to l of n dash then delta dash hat Q 0 x intersection f dash must not equal to m t. That means there is at least one state is common to f dash and the set of next is from Q 0 delta dash hat Q 0 x. So, that is by definition. So, therefore, it says it is clear that l of n is a subset of l of n dash. So, it is quite clear that l of n is a subset of l of n dash. Conversely suppose that x belong to l of n dash. That means delta dash hat Q 0 x intersection f dash not equal to 5. So, according to definition of acceptance of string x by this n f n dash delta dash hat Q 0 x intersection f dash must not be equal to 5. So, if that is the case. So, in this case if we can show that delta dash Q 0 x intersection f delta dash hat Q 0 x intersection f is also not equal to 5. So, that was the case for f dash and you can show that this is intersection f is also not equal to 5 then we are true. Otherwise then we are true because that will also belong to x must will also belong to l of n. Otherwise there exists some state Q that belongs to delta dash hat Q 0 x such that there exist Q belong to this set of next steps for this transition on string x such that epsilon closure of Q intersection f not equal to 5. This implies that Q belongs to delta hat Q 0 x and E Q intersection f not equal to 5 which in turn implies that delta dash Q 0 x intersection f not equal to 5 that is x belongs to l of n. Therefore, l of n dash is a subset of l of n. So, considering the previous result that l of n is a subset of l of n dash and l of n dash is subset of l of n we have now we now concluded l of n is equal to l of n dash. So, it is enough to prove that delta dash hat Q 0 x equal to delta hat Q 0 x for all x belonging to sigma plus that means for any string over sigma other than epsilon. So, we prove this by induction on the length of the string x let us prove it by induction on the length of string x. So, if x belongs to sigma that means if x is a single symbol that means length of x equal to 1 that is what we have said length of x equal to 1 that means x is a single symbol that means x belongs to the input of habit. Then by definition of delta dash delta dash hat Q 0 x equal to delta dash Q 0 x with nothing but delta hat Q 0 x. So, for any string of length 1 it is true and there is a base case. Now, we will assume that this result is true for all strings of length less than or equal to n that means for any length of a string x less than or equal to n say the statement is true there is an inductive hypothesis. That means for string x of length n delta dash Q 0 x equal to delta dash hat Q 0 x equal to delta hat Q 0 x. Now, consider the string x where a is a single symbol the length of this string is n plus 1 since length of x is n and by inductive hypothesis this is the case this is true that is the statement is true. Now, delta dash hat Q 0 x a equal to so this can be written as first we apply delta dash hat Q 0 on the string x and then whatever the state of states we get apply here the transition function delta dash. So, this is how we define the x and transition function but this is nothing but since this delta dash hat Q 0 x is a set of states and for every such state we apply this delta dash a and eventually we take the union of all those that means union of all those for every p belonging to delta dash hat Q 0 x if we apply delta dash this delta dash on for every p on symbol for all a input symbol for input symbol a that we have over here. So, by definition so this is the case by definition now according to inductive hypothesis this delta dash hat Q 0 x can be written as p belonging to delta hat Q 0 x because the length of x is equal to n and according inductive hypothesis delta dash hat Q 0 x nothing but delta hat Q 0 x. So, this is delta p a again according to definition this is nothing but delta hat Q 0 x a by the definition of x and transition function delta hence by induction we have shown that delta dash hat Q 0 x a is nothing but delta hat Q 0 x a and hence this completes the proof that means the language except by the original n f a n with some epsilon transition can be played is equal identical to the language except by the n f a that we constructed which does not have any epsilon transition. Next once we have constructed this n f a from the original n f a without any epsilon transition and this epsilon transition does not have any this n f a does not have any epsilon transition. Now, we will construct from this n f a which does not have any epsilon transition but may have multiple next states on input symbol on some input symbol from out of any state we will constructed the f a that means what we will show is that for every n f a say n s without epsilon transition but it might have some other non deterministic transitions like transition to multiple next states on some input symbol or no transition out of any state on some input symbol. Then there exist a d f a say a such that the language of these two are same. Now, we will give a proof for this just consider that the original n f a without epsilon transition has a set of states Q sigma is the input alphabet delta s is the transition function Q 0 is the star state and as this is a set of final states. So, this is a n f a without any without epsilon transition. Now, we construct a d f a a which contains say p set of states the input alphabet is same say transition function is suppose mu. Then we will have say p 0 is a star state and say e is a set of final states where the set of states p is nothing but the power set of Q that means every subset of Q will be considered as a state in the d f a it also written as sometimes 2 power Q. So, it means that if the original n f a has n states the the equivalent d f a may have 2 power n states total is a maximum force of states. And the set of states p can be written as Q i 1 Q i 2 say Q i k within square bracket. So, this is a state such that this Q i 1 Q i 2 Q i k is a subset of the set of states of the n f a. So, we indicate a state in the d f a within square bracket for a subset of the original set of states in the n f a. And the star state p 0 is nothing but the original star state which is considered as subset containing the single element the original state of the n f a. Then this is the star state of the d f a let us make it more clear. So, this indicates that the subset containing Q 0 is basically considered as the star state of the d f a. And then the set of final states e is nothing but say Q i 1 up to say Q i k. So, this is considered it belong to p is considered as the set of final states in the d f a provided Q i 1 up to say Q i k this is a final state this intersection f f dash not equal to 5. That means, this contains this subset contains one only final states in the original n f a then this that corresponding to that subset be considered as a final set in the d f a. And this transition function mu which is from the set of states of p then sigma to p is defined by is defined by mu of say Q i 1 Q i 2 up to say Q i k. So, this mu on this state the set of states of the original n f a on input symbol a is nothing but say Q j 1 Q j 2 up to some say Q j m. If and only if we have the transition in the original n f a delta dash Q i l on a say it is Q j 1 Q j 2 up to say Q j m for every l i l belonging to the set of states every l i l belonging to i 1 i 2 up to say i k. And you take the union of all those and eventually we get the set Q j 1 Q j 2 Q j m in such a case we provided provide this move mu Q i 1 Q i 2 up to Q i k on a it will go to Q j 1 Q j 2 Q j m. So, clearly the automaton that we have defined by this construction is a d f a because it satisfies all the properties to being to be a d f a. Now, what we want to show is that l n dash is equal to l of a to prove this what we need to prove is that for all string acts belonging to C master mu P 0 where P 0 is a star state mu P 0 x for mu dash extended transition function for this d f a. So, mu dash P 0 x equal to Q i 1 up to say Q i k if and only if delta dash hat Q 0 x equal to Q i 1 Q i 2 up to say Q i k. So, if you can show this then you can prove that l of n equal to l of n dash equal to l of a this suffices the result because x belongs to l of n dash if and only if delta dash hat Q 0 x interaction f dash not equal to 5 according to the definition of acceptance of a string by this n f a n dash. In this case if and only if according to the construction of this d f a from this n f a we have seen that mu hat P 0 x belongs to if and only if x belongs to l of a. Now, we prove this statement that for all x belonging to sigma star we have that delta dash hat P 0 x equal to Q i 1 up to Q i k if and only if delta dash hat Q 0 x equal to the set Q i 1 up to Q i k. Now, the base case is that we can again prove this result on the length of a string x. So, the base case is that the length of a string is 0 in such a case delta dash hat Q 0 x equal to Q 0 also mu dash P 0 x equal to P 0 this is nothing but Q 0. So, this is proved order statement holes for length for strings of length 0 this will also be the case for strings of length equal to 1 that means for every symbol belonging to the input alphabet. It is very clear from the definition that you have given for the transition function mu it holds for strings of length 1 as well now assume that it is true for all strings of length n or less for all strings of length for all strings x of length n or less say this is to be true this is true now consider say length of string x equal to n and a single symbol a. Now, if you consider string x a what will have delta hat P 0 x a can be written as mu of mu hat P 0 x then a. So, first apply this x n transition function on x and then whatever set of states we get there we apply the transition function mu on each of that. Now, by inductive hypothesis we have found that mu hat mu hat P 0 x is nothing but Q 1 up to say Q i 1 up to say Q i k if and only if delta this hat Q 0 x equal to Q i 1 say Q i k. So, this is why inductive hypothesis because the length first string x is equal to n. Now, by definition of mu we know that mu Q i 1 up to say Q i k on a is nothing but Q j 1 up to say Q j m if and only if union of i l belonging to i 1 up to say i k delta this Q i l a equal to Q j 1 up to say Q j m does mu hat P 0 x a equal to Q j 1 up to say Q j m if and only if delta this hat Q 0 x a equal to Q j 1 up to say Q j m. Therefore, by induction on the length for the string x the statement that we have already said is true and hence the theorem follows. Now, from the above two theorem that is first one is that for an NFA if there is an NFA with some epsilon transition then you can have an equivalent NFA without a transition and then if we have an NFA without a transition then you will have there exists DFA which is equivalent to this NFA. Therefore, for every NFA for every NFA there exists an equivalent DFA. So, this is a theorem that we have proved from the above. Now, let us construct an example or given example. So, how to construct an equivalent DFA from any given NFA without epsilon transition. So, let us start with the same example that we have already constructed let us redraw the equivalent NFA that we have drawn from the original NFA with epsilon transition where we have Q 0 as the start state as well as the final state on input B there is self-loop on input A and B it goes to Q 1, Q 1 on A B remains in the same state and Q 2 is a final state again Q 0 on B it goes to Q 2 and Q 1 on B as well it goes to Q 2. So, this was the equivalent NFA without epsilon transition for a given example. So, the transition map for this was simply Q 0, Q 1, Q 2 on A and B goes to Q 1 and Q 0, Q 1, Q 2, Q 1 on A goes to Q 1 again and on B it goes to Q 1 and Q 2 and Q 2 on A and B goes to 5. So, this is the transition map for this given NFA without epsilon transition. So, let us construct the equivalent DFA for this NFA. So, according to our construction we have seen that since there are three states in this NFA there may be 2 power Q that means the set of states that we have P which is equal to the power set of Q there may be 2 power Q states that means there may be total 2 power 3 H states, but all those H states are given an N state NFA in the equivalent DFA even though there may be possible 2 power N states all those states may not be accessible from the start state or all the final states may not be accessible from the remaining states of the DFA. So, therefore, in the construction what we will do we will start with only the accessible states and then we will proceed in such a way that only the states which are accessible will be there in the DFA. So, in the construction we will start like this let us see construct the transition map for this DFA there are 2 symbols A and B. So, according to definition Q 0 is the only start state that we have this is a start state and Q 0 on A it goes to Q 1. So, it is a only state where it goes to. So, the subset containing Q 1 is a state. So, we started Q 0 because the start state because it is a only accessible state initially in the NFA or in the DFA equivalent DFA. So, on A it goes to Q 1. So, the subset containing Q 1 is the state where the DFA goes on input A and this state is accessible. Similarly, on B Q 0 on B may remain in the state Q 0 or it may go to Q 2 or it may go to it may go to Q 1 or it may go to Q 2. So, there three possibilities. So, the subset containing Q 0, Q 1, Q 2 is the other state that is accessible from the start state Q 0. Now, we will start constructing a transition or complete transition for all those states which are accessible that means which are introduced newly in the table. So, Q 1 is the new state introduced in the table. So, we will start with Q 1 the subset containing Q 1. So, Q 1 on A it may remain the same state Q 1 and there is no other possibility. So, this is the only possible state which is accessible from Q 1 and Q 1 on B it may remain on Q 1 or it may go to Q 2. So, these are only two states where Q 1 goes on B. So, therefore, this state contains the two states Q 1 and Q 2. So, initially we introduced this Q 1 is a new state and then Q 0, Q 1, Q 2 the subset containing Q 0, Q 1, Q 2. So, therefore, we will introduce this construct the transition function for Q 0, Q 1, Q 2 the subset containing Q 0, Q 1, Q 2. So, we will consider the transition from this state on A which is nothing but the union of the first column Q 0 on of the first column this NFA, Q 0 on A goes to Q 1, Q 1 on A goes to Q 1 and Q 2 on A goes to phi and we will take the union of all those. So, that is the only possibility this Q 1 is the only possibility. Similarly, Q 0, Q 1, Q 2 individually on B goes to Q 0 goes to Q 0, Q 1, Q 2, Q 1 goes to Q 1, Q 2 and Q 2 goes to phi taking the union we will get Q 0, Q 1, Q 2, Q 1, Q 2. Next Q 1 is already done in this second line, second row and Q 1, Q 2 is the only other state that we have. So, for which is new for which we have not constructed the transition map, transfer function. So, Q 1, Q 2, Q 1, Q 2 on Q 1 may go to only one state this Q 1. Similarly, Q 1, Q 2 on B may go to Q 1 and Q 2. So, it is the only subset that we have correspond to this and we have found that no more states new states are introduced in the table. So, therefore, we need not compute further or we need not proceed further. That means, we have 4 states only Q 0, Q 1, Q 0, Q 1, Q 2 and Q 1, Q 2. So, in this case Q 0 is a star state and we construct a final state by considering if any subset contains a final state of the previous NFA. For example, say this Q 0 this state star state Q 0 will be a final state because original NFA contains Q 0 is a star state and final state. Similarly, this Q 0, Q 1, Q 2 both Q 0, Q 2 is a final state and Q 1, Q 2 this must also be a final state because Q 2 is a final state, but Q 1 is not a final state because in the original NFA Q 1 was not a final state. So, this is the equivalent DFA that we have got from the given NFA without any absent relation. So, if we can we can rename or relabel the states for example, this state we can call it as P 0, this may be called as P 1, this may be called as P 2 and this may be called as P 3. And then we can construct the transition function corresponding to this transition map which is the DFA, equivalent DFA for the given NFA. So, let us quickly draw it. So, P 1 is a star state as well as final state, P 1 on A goes to P 2 and P 1 on B goes to P 3 which is also a final state. Then P 2 on A remains on P 2 because P 2 on A remains on this is P 0, this is P 1 and this is P 2. So, P 2 P 1 on A remains the same state and P 1 on A remains on the same state Q 1 and on B it goes to on B it goes to Q 1 Q 2, but Q 1 Q 2 is a new state, Q 1 Q 2 this is P 3 on B it goes to that state. And then P 2 on A, P 2 on A goes to Q 1 and P 2 on B remains on same state, P 2 Q 0 Q 1 Q 2 that means Q 0 Q 1 Q 2 on B it remains on same state. And finally, P 3 on A, P 3 on A remains goes to state Q 1 that means it is P 1 and P 3 on B remains on same state with Q 1 Q 2. So, this is the equivalent DFA, this is the equivalent DFA for the given NFA. And we have proved that this DFA is equivalent to this NFA which are multiple next states on some input symbol.