 Let us first see what is the present situation with this course. So, far we have introduced the following. First we introduce the notion of formal languages and then we look for the finite representation. In that context, we have introduced the notion called regular expressions and the respective languages called regular languages. So, after introducing these regular languages, given a language whether it is regular or not, it is not that easy to understand. We understood that point because giving regular expression is not that easy. So, in that context, we took the help of grammars and we specialized the two so called right linear grammars that we are calling regular grammars because our intention is to understand that these regular grammars characterizes regular languages. Of course, we have not proved that we are going to prove all these things. So, through regular expressions, whatever that we have defined so called regular languages, our intention is to show that regular grammars characterizes that means given any regular language, we can understand that through a regular grammar. Still there we have certain examples of regular languages, we are unable to give regular grammar for that. In that context, we gave a tool called finite automaton and through finite automaton, we are understanding regular languages and as I hope you understood clearly that giving regular expression is not that easy and regular grammar is relatively better, but for many of the things I hope now you could have given this finite automaton. In that we have variants of course, so called DFA and NFA and we have observed that these two are equivalent. So, here the notion of finite automaton that may be DFA or NFA is essentially to capture the information related to regular languages, this characterizes that. So, in this context when we are proving DFA is equivalent to NFA, the number of states when you are. So, the DFA is introduced and whenever you are given in NFA, if you want to convert it to your DFA, what is happening the situation here, if you have n states here in this conversion, you are getting 2 power n states that is exponential number of states. Now, in order to understand that what is a possible minimal DFA or you know when you are considering a particular language corresponding to that if you are if you are constructing a NFA and converting to DFA, is it the situation that you will always have that many number of states or whether there are any states which are useless. So, that kind of aspects now in this module we are going to discuss. So, here our concentration is to understand that what is a minimal DFA to accept a regular language that is our concentration now. Anyways, so in this cycle of course, once again when to conclude here this particular cycle, this finite automata and we are going to observe that in the following lectures that all these are equivalent. So, the present concentration of course, at present we are concentrating to understand that what is the minimal DFA to accept a regular language. So, the present lecture is essentially present module essentially concentrating on that 1 or 2 lectures in this direction. In this direction, we require one important tool called Mahill Neurot theorem that I will be discussing in this lecture and in fact, this is a very important characterization for the languages acceptable DFA. Of course, eventually what we are understanding the language acceptable DFA is regular languages. So, this is very important characterization for regular languages. So, in this lecture I will concentrate on so called Mahill Neurot theorem and to introduce that theorem I require some basic definitions in the direction. First let me introduce this an equivalence relation tilde on sigma star like strings over an alphabet sigma is said to be right invariant. If you take any two strings x y if they are related with respect to tilde then any string you take that and concatenate it on right side of both the strings then they should also be related with respect to tilde. So, if this property is satisfied there for all strings x y which are related if you concatenate any string z on the right hand side the resultant string should also be equivalent with respect to the tilde. So, such an equivalence relation we call it as right invariant equivalence relation. Let us look at an example so that you can understand this concept better. So, this example I am defining tilde l of course that I will be using throughout this lecture. So, please remember now you consider a language over sigma that means it is a subset of sigma star define the relation tilde l on sigma star with this condition that x related if you take any two strings x and y in sigma star you relate them if and only if for any string z x z is in l if and only if y z is in l. So, whenever x z is in l y z should also be in l. So, this advice ours of course. So, with this condition we are defining this tilde l and of course you can quickly understand that this is an equivalence relation because to understand a relation is an equivalence you have to understand reflexivity symmetry and transitive. So, any string x is related to itself because if you concatenate any string z the resultant string on both the situations here are same. And therefore, both are in l or not in l and thus you can quickly understand that it is reflexive symmetry follows very quickly because of this if and only if condition here and a transitive to also you can verify to understand that this tilde is an equivalence relation on sigma star. Now, our point here of course important thing you have to cross check here in this context is right invariant property. That means x related to y implies x z related to y z for all z that is what we have to understand for right invariant relation. So, take x y in sigma star and choose take x y in sigma star and assume they are related and pick up any z arbitrary in from sigma star. Now, we have to show that x z related to y z with respect to this relation that is we have to show that for all w in sigma star x z w is in l if and only if y z w is in l this is the condition we have to cross check. Now, if you write for any w take an arbitrary w in sigma star and write u to be z w and now you see that since x related to y with respect to tilde l what we have x u related x u is in l if and only if y u is in l this is the definition of tilde l. So, x u is in l if and only if y u is in l and what is u we have chosen that is z w that means what are the property that we are aiming to that is this we have got it that is this x z w is in l if and only y z w is in l and hence this tilde l is a right invariant equivalence relation. Now, I give you one more example this is also very important in this lecture that is I am defining it as tilde a for a d f a a consider a d f a q sigma delta q naught f take a d f a now define the relation tilde a on sigma star as below x and y are related with respect to tilde a if and only if you put those strings in the initial state you should essentially reach to the same state if you are reaching to the same state then you say they are related and you can understand again that this is a equivalence relation because if you take any string that is related itself because what are the state that you are reaching to with respect to x of course, that is fixed in a d f a and hence x related to x for all x and symmetry is very straight forward because here we have we are putting equal to. So, x is reaching to a particular state and you know y is reaching to a particular state and they are same then of course, vice versa and hence x related to y implies y related to x and when you are looking for a trans duty because of the equal to here you are getting trans duty also very quickly and hence this is an equivalence relation tilde a is an equivalence relation. So, here what we have to understand that it is a right invariant equivalence relation let us see take x related to y that is this property satisfied. Now, you pick up any arbitrary z in sigma star what I have to observe that delta cap q naught x z is equal to delta cap of q naught y z that is what we have to observe to show that x z is related to y z. Now, consider this delta cap q naught x z I hope by now you have proved that this is equal to this you can use induction to prove this property and now since this two are equal you can take in place of delta cap of x naught sorry q naught at x you can replace it with delta cap q naught at y and again using the property you can have this and thus x z is related to y z for all z x here that is an arbitrary thing and thus you can understand this is also a right invariant equivalence relation on sigma star of course, on sigma star tilde a is defined on sigma star. Now, let me state Mahily road theorem let I will be a language over sigma the following three statements regarding l are equivalent what are those statements statement 1 l is accepted by a DFA of course, once eventually if you prove that the finite automata is a character a character it captures the properties of regular languages that means it is it is characterizing regular languages. Then this theorem gives you a characterization for regular languages as I had mentioned what is the point here is there exist a right invariant equivalence relation tilde of finite index on sigma star such that l is the union of some of the equivalence classes of tilde. So, what we are trying to say here in the second point you can find a right invariant equivalence relation tilde such that l is and of course, it is a finite index such that l is union of some of its equivalence classes and number 3 is the equivalence relation tilde l as defined just now with respect to l because given a language l you can talk about tilde l as just we have defined the equivalence relation tilde l is a finite index. So, this theorem claims that these three are equivalent l is accepted by DFA and there is a right invariant equivalence relation tilde of finite index such that l is union of some of its equivalence classes and the equivalence relation tilde l is a finite index these three are equivalent. We prove 1 implies 2 and 2 implies 3 and 3 implies 1 to show that these three are equivalent to show 1 implies 2 assume l is accepted by DFA say a to be this q sigma delta q naught f. First what we will show that in point number 2 what we have to show there is a right invariant equivalence relation which is a finite index and l is union of some of its equivalence classes this is the point number 2 we have to prove. Here in this context what we will observe that that tilde into that is you know tilde a with respect of a will satisfy the tilde into that means what we have to observe that tilde is a finite index because we have already observed that tilde a is a right invariant equivalence relation on sigma star. So, what we have to observe the remaining two points one is it is a finite index and number 2 is l is union of some of its equivalence classes these two points we have to observe. So, tilde a you know already the claim is it is a finite index. So, to show it is a finite index first let me observe the following two points for x and sigma star delta cap of q naught x if it is equal to p then the equivalence class containing x is all those strings in sigma star if you put it if you put them in the initial state if you reach to p that is equivalent to x because if you remember the definition of tilde a two strings with respect to tilde a they are equivalent. If you put those two strings in the initial state of d f a here that is in q naught you should reach to the same state then we say those two are equivalent. That means if you take any string x what are the strings equivalent to x that is the equivalence class containing x is precisely all those strings in sigma star if you put them in the initial state q naught if you reach to the state p what is p here the state that you are reaching via x from the initial state q naught this is what is precisely the equivalence class containing x this is the point number 1 and point number 2 because here the state to which you are reaching that is characterizing the equivalence class. So, now corresponding to each state now let us look at all those strings let me call for a given q in q for a given state q consider c q the class of q I am calling is all those strings if you put them in the initial state q naught if you reach to the state q let me call it as c q this is an equivalence class of course this can be empty if no string can reach to this state q because if that is not reachable via using any string then of course this is this is empty let me consider empty also. So, in this context so what I am saying here this is an equivalence class of tilde a because what are all those strings what are all those strings if you put in the initial state to reach to q they are all equivalent to each other and this forms an equivalence class. Thus what you can understand equivalence classes of tilde a are completely determined by the states of a because from the point number 1 you understood that the state is coming into the picture. Now, take every state and set the class c q and collect all those strings this can be empty if that particular state is not reachable from the initial state, but wherever it is non empty all those strings which are reaching from the initial state to that particular state they are all equivalent and that is an equivalence class with respect to this tilde a from this 2 points what you can understand that this tilde a is completely determined by the states of a and moreover the number of equivalence class of tilde a is less than or equal to the number of states of a the reason why then every state gives an equivalence class to you of course, possibly empty in whenever the state is not reachable the equivalence class what I am calling here is empty set. So, thus the number corresponding to each state you get an equivalence class possibly an empty class some of the cases. So, thus the number of equivalence class of tilde a is less than or equal to the number of states of a and hence tilde a is a finite index because a has only finite number of states therefore, tilde a is a finite index of course, here index we mean index of an equivalence relation we mean the number of equivalence classes of that relation. Now, we want to prove that the second point l is union of some of its equivalence classes here this tilde a is equivalence classes. Now, consider this l by definition this is all those x and sigma star if you put them in the initial state you will be reaching to the to one of the final states that means, delta cap of q naught x is in f that is union of these sets because if you reach to a particular final state because for all final states I am considering these sets some of the non final some of the final states may not be reachable in which class this set is empty. So, now just we have proved that this set is an equivalence class we are calling it as C p corresponding to a state p and thus you understand that l is union of C p's for p belongs to f as desired. And thus we have observed that 1 implies 2 that means we could identify a right invariant equivalence relation here that is tilde a which is a finite index number 1 and number 2 l is the language l is union of some of its equivalence classes we have proved 1 implies 2. Now, let us consider 2 implies 3 to prove 2 implies 3 what we have to observe that tilde l is a finite index the number of equivalence classes of tilde l is finite. Suppose, tilde is an equivalence as in 2 that means tilde is a right invariant equivalence relation of finite index number 1 and number 2 is and number 2 is l is union of some of this its equivalence classes. So, that is what I am assuming assuming tilde is an equivalence as in 2 point number 2 of the theorem we show that tilde is a refinement of tilde l what is refinement. If you suppose this is the set under consideration and if you consider an equivalence relation equivalence relation you know equivalence relation partitions the set. Now, if you further refine this relation that means if you break this equivalence classes further for example, maybe you are breaking this direction or whatever. Whatever the classes here I am getting you can clearly see that what are the partition that I am getting here each equivalence class that means equivalence class here I mean the portions that are shown with the boundaries of this cross lines with the vertical lines. So, this portion is this is a new equivalence class for example here this is a new equivalence class these equivalence classes are contained in the original equivalence classes because they are subsets of the because we are partitioning further partitioning this. So, now we say this is a refinement we are further refining this equivalence classes. So, that means if you take any 2 elements in this set if they are related with respect to new equivalence relation they are related to the original with respect to original relation because new equivalence class is a subset of original equivalence is a subset of one of original equivalence class. So, what we are trying to show that this tilde is a refinement of tilde l that means once we show that this is a refinement as I have explained that the number of equivalence classes of tilde is more than the number of I mean more or equal to the number of equivalence classes of tilde l. That means a index of tilde is greater than or equal to index of tilde l as it is given that tilde is a finite index tilde l is also is a finite index. So, once you we show that tilde is a refinement of tilde l then we are through now suppose we have as I have mentioned that if x related to y with respect to tilde we have to show that x related to with respect to tilde l also to observe that this is tilde is a refinement of tilde l. Now, for that purpose for x y in sigma star suppose x related to y with respect to tilde to show x related to with y with respect to tilde l what we have to show for all that if you concatenate that from its right side this is in l if and only if y that is in l. Now, since tilde is right invariant that is what is hypothesis given into the point number 2 of the theorem since tilde is right invariant what we have x z related to y z for all that and in point number 2 we have this information also l is union of some of its equivalence classes some of the equivalence classes of tilde. That means whenever x z related to y z that means x z is in l then y z should also be in l. So, that is we have for all that x z is in l if and only if y z is in l that is what is the desired property we look for to show that x related to y with respect to tilde l and hence what we have observed x related to y with respect to tilde implies x related to y with respect to tilde l also and hence tilde is a refinement of tilde l and thus tilde l is a finite index. So, this is this gives you 2 implies 3 now we prove 3 implies 1 in 3 what we have tilde l is a finite index let us assume that and what we have to show we have to show that there is a d f a to accept l l is accepted by some d f a. Now, we construct the d f a in this portion 3 implies 1 now construct a l by taking q sigma delta q naught f of course, here I have to give you that q the state set you consider it as sigma star by tilde l that means the partition of sigma star with respect to the equivalence relation tilde l hence all those equivalence classes bracket x x and sigma star that is what we consider it as state set in a l and what is the initial state you consider the equivalence class containing epsilon the empty string and then you consider the final states are as here all those equivalence classes with which are in you know if you take any string in l the equivalence class containing that x you consider as a final state and now you have to give a transition function in case of d f a. Let me define delta from q cross sigma to q q cross sigma to q defined by delta of bracket x at a we assign it to bracket x a the equivalence class containing the string x a of course, for all states bracket x and for all a in sigma we have to understand that this is a d f a to understand this is d f a what do you require this of course, sigma is given to be alphabet that is a finite set alphabet means I have what I have to understand that q is a finite set number 1 and number 2 this delta is a function from q cross sigma to q that means we have to look at we have to prove the well defined s of delta. So, that this is a d f a so to observe these two points of course, since tilde l is a finite index since tilde l is a finite index we can quickly see that the number of equivalence classes of tilde l is finite and thus the partition that means the number of equivalence classes here is finite. So, q is a finite set very quickly that q is a finite set to understand that this delta is well defined how is delta defined this is defined as to show this delta is well defined consider bracket x equal to bracket y consider two states for bracket y bracket x in q these are the states consider and consider arbitrary a in sigma the claim is bracket x a is equal to bracket y a because what we have to show delta of bracket x at a is equal to delta of bracket y at a that is what we have to show that bracket x is equal to bracket y a. So, that means what we have to show with respect to to show these two equivalence classes are same we have to show that this x a with respect to tilde l it is related to y a that is what we have to show. And what is tilde a tilde a you know already it is a right invariant equivalence relation that means, if you concatenate any string on it is on right hand side the resultant strings are also equivalent. Here a is in sigma that is a string you can of length one if you if you concatenate a to x as well as y they should also be related with respect to tilde a because tilde l because tilde l is a right invariant equivalence relation. And hence you have this property once you have this property these two strings are in the same equivalence class that means, the equivalence classes containing x a and y a are same. And thus what do you have delta of a bracket x at a is equal to delta of bracket y at a. And thus you can understand this delta is well defined this is well defined. So, once you understand that this delta is a map from q cross sigma to q this with this assignment we have this a l is a d f a deterministic finite automaton. So, we have constructed d f a we prove that the language accepted with this d f a is l the claim is l of a l is equal to l. For that purpose we show that delta cap of q naught w is equal to bracket w. So, the state the equivalence class the state the resultant state will be the equivalence class containing w this serves the purpose. If you show this this point this is sufficient because if you take any string w is in l if and only if bracket w is in f the reason why w is in l if and only if bracket w is in f. How this is once we have proved this thing delta cap of q naught w equal to bracket w once we prove this the initial state here of course is delta cap of we are proving this that is what we are proving. If this state is in final state then w is in f that is straight forward if this is w is in l and whenever w is in l what is the requirement whenever you put this w in the initial state you should reach to a state that is in f. So, with this criteria with this criteria delta cap of q naught w is equal to bracket w if you can prove then w is in l if and only if bracket w is in f is clear. And thus what will happen the language accepted by a l is equal to p l we prove this assertion by induction on the length of w. So, induction basis is clear for strings of length 0 because if you take string of length 0 that is epsilon if you put in this that is what is q naught by definition of delta cap of course in any d f a here q naught is bracket epsilon thus we have the basis of the induction that is for those strings of length 0 of course empty string. Moreover you can in fact observe that by definition of this delta this assertion is true for those strings of length 1 also because delta we have defined with this condition. Now, if you consider delta cap of q naught at a for the strings of length a that means that is delta cap is nothing else but delta and delta is defined as so and this string is nothing else but a epsilon a is nothing else but a that is you are getting a equivalence class containing a. And hence what are the assertion we have that is true for all those strings of length 0 and all those strings of length 1. So, induction basis from this point you can observe for inductive step you consider a string in sigma star some x and consider a in sigma what I have to observe the delta cap of q naught x a is equal to bracket x a that is what we have to prove. So, consider delta cap of q naught x a this equal to this that is the definition of delta cap by inductive hypothesis we have this delta cap of q naught at x is bracket x that is delta of bracket x at a by definition of delta you know this is bracket x a and thus what we have delta cap of q naught x a is equal bracket x a. And hence by induction what we have delta cap of q naught at w is equal to bracket w for all w in sigma star. And hence any string is in l if you put it in the initial state of this particular automaton you will be reaching to a final state precisely the equivalence class containing the string. And thus the language accepted by this automaton a l is l and hence l is accepted by some d f a of course, here a l this completes the proof of Mahill Neyrod theorem. So, we have proved that 1 implies 2 2 implies 3 and 3 implies 1 thus the condition the statements 1 2 and 3 are equivalent. Now, let me give you a remark the proof of 2 implies 3 shows that the number of states of any d f a accepting l is greater than or equal to the index of tilde l given a language l consider any d f a accepting l in 2 implies 3 what we have proved the number of states of that d f a a is greater than or equal to the index of tilde l that is what we have proved. Because to observe that tilde l is a finite index we have proved this point in 2 implies 3. And in the proof 3 implies 1 we have provided a d f a namely a l with the number of states equal to the index of tilde l. Because in that a l the states are essentially the equivalence classes of tilde l. So, that means the states what we have considered they are equivalence classes that is the number of equivalence classes is equal to the number of states of that a l. So, from these 2 points you can understand that a l is a minimum state d f a accepting l. So, given l you know what is tilde l from tilde l you know the equivalence classes from those equivalence making those equivalence classes as states we have constructed a l. And since any d f a accepting l should have the number of states of any d f a accepting l should have more or equal states than the index of tilde l. And a l is in fact having the number of states equal to the index of tilde l we can conclude that a l is minimum state d f a accepting l. So, we have from main linear node theorem we have a minimum state d f a accepting l that is what is the conclusion at this point of time. Now, let me give 1 or 2 examples as applications of main linear node theorem, because main linear node theorem is characterizing once again I am emphasizing that main linear node theorem is characterizing the languages that are accepted by d f a eventually what we point out that main linear node theorem characterizing regular languages. So, let me consider this example you know this is a regular language you can give a regular expression for this consider the language x in consider the language with all those strings having a b as substringer what is the regular expression for this this is any string is here is of the form x a b y for any x and y in over a b. And thus the regular expression for this is a plus b star a b a plus b star this you know and hence this is a regular language. And you know d f a accepting this language also that we have already discussed have constructed and from main linear node theorem we should understand that the tilde l should be of finite index. So, let us calculate the number of equivalence class of tilde l in this example to understand using main linear node theorem this is a regular I mean this is a language actually by d f a. First observe that the strings epsilon a and a b are not equivalent to each other let us observe this point. So, to show epsilon a are not equivalent to each other we have to find a string we have to find a string such that if you concatenate that string on the right to its right hand side of epsilon and to a one string should be in l one string should not be in other string should not be in l that is how we have to find. Now, in particular suppose if you consider b the string b epsilon b is b you know b is not in l, but if you concatenate b to a on its right hand side that is a b that is in l. So, you can quickly observe from this that a and epsilon are not equivalent because we could identify a string b here that concatenating b to the right hand side of both the strings one string we are getting in l other string we are not getting in l and thus a and epsilon are not equivalent with respect to tilde l. Similarly, we observe that this epsilon and a b are not equivalent if you consider any string which is not containing a b if you concatenate that to epsilon right hand side you will get of course, it is a same string and thus that is not in l and if you concatenate that string any string to a b on the right hand side anyway that is having a b and thus that is a string in l. So, you get strings one is in l other is not in l, so that you can understand that epsilon and a b are not equivalent. So, what I am trying to say here is to show a and a b are not equivalent with respect to tilde l you choose any string x in which a b is not a substring for which a b is not a substring. Suppose, if you consider this concatenate x to epsilon right hand side this is resultant string is x only and if you concatenate x to a b on its right hand side resultant string anyway is having a b a substring. You see this string x which is epsilon x is not having a b a substring, so that is not in l whereas, this string is always in l as a b is substring of this resultant string and hence we observe that epsilon and a b are not equivalent any string x a general string I have mentioned in particular here we are considering b if you choose b b is not having a b a substring and you can understand that epsilon and a b are not equivalent. Now, we also observe that a and a b are not equivalent here I have taken the string a if you concatenate a to its right side then of course, the resultant string is a a that is not in l if you concatenate a to a b at the right hand side that is a anyway string of l and thus a a when this little a distinguishes a and a b and the therefore, a and a b are not equivalent because to show two strings are equivalent with respect to tilde l if you pick up any string from sigma star concatenating that string right side of both the strings the resultant strings both should be in l or should not be in l. So, with that criteria we have to cross check and from these three points I hope you can you understood that these three are distinguishable each other and thus what is the conclusion here these three strings should be in different equivalence classes. So, let me say epsilon is in the equivalence class bracket epsilon the equivalence class containing epsilon let me call bracket a for the equivalence class containing a bracket a b for this. So, let me assume these three classes are there from this. Now, what we are going to show that if you take any other string in sigma star that string should be in one of these equivalence classes thus precisely we will have these three equivalence classes and hence we can understand that tilde l is a finite index let me under this point also pick up a string x and now I see the property of x in two cases whether a b is a substring of x or a b is not a substring of x. If a b is a substring of x then x clearly will be in the clearly suppose a b is a substring of x let me use this notation a b less than equal to x I mean a b is a substring of x. So, what we are trying to say that x is related to a b so that this x will be in bracket a b. This is very quick because to x if you concatenate any string z whether it has a b or not having a b since already there is a b in x in the resultant string x that you will have a b a substring and since with the same property if you concatenate the z with to a b this resultant string is also having a b a substring thus this is in l and this is also in l irrespective of whatever that z you are concatenating and hence these two are equivalent x and a b are equivalent. So, if you take any string x in a b star if it is having a b a substring that clearly will be in bracket a b now if you consider the other case that if a b is not a substring of x then what are the possibilities if a b is not a substring of x then let me see x may be of the form a power n or may be b power n in which a b is not there or when you have some a's and b's since a b should not be there you can have some number of b's followed by some number of a's but of course after that you should not have any b's after a's because if you have any b after following to this a's then you will always get a b a substring thus I can categorize those strings x for which a b is not a substring means it may be a power n may be b power n result of the form b power n or b power n a power n these three cases. Now, in each case we can discuss that these strings will fall in some of the equivalence classes among bracket epsilon bracket a of course whether it will be in bracket a b or not you will observe here for n greater than equal to 1 because if it is empty string that is one situation that I will consider in the case of b power n for n greater than equal to 0 here at least 1 a for n greater than equal to 1 and the other case I will consider. So, for n greater than equal to 1 if you consider x 2 b of the form a power n in this case x will be in bracket a as I have discussed so far you can take it as an exercise and observe that x will be in bracket a in this case and if you take x which is of the form b power n of course n greater than equal to 0 also you can put where in which case that is epsilon that will be in bracket epsilon. And if x has some a's and b's as I had mentioned x must be of the form x must be of the form this b power n a power n for m n greater than equal to 1 in this case you can understand that this will be in bracket a thus this tilde l has exactly 3 equivalence classes and hence it is a finite index and the conclusion from Mahler's theorem that this language can be accepted by a DFA of course you know already that this language a DFA accepting this language. So, let me consider one more example we know this language a power n b power n for m greater than equal to 1 so far we have not constructed any of course we have a grammar for this but not a regular grammar we have constructed a contrasted grammar for this in this example we observe that tilde l for this language l a power n b power n n greater than equal to 1 we observe that tilde l is of the index of tilde l is infinite it is not finite. And thus using Mahler's theorem you can say this is not this cannot be accepted by any DFA and as we are saying that DFA characterizes the properties of regular languages of course from this we can conclude that a power n the language a power n b power n greater than equal to 1 is not a regular language. So, let me observe this point we show that the index of tilde l is not finite for instance you consider these two strings a power n b power m these two strings from sigma star here sigma is a b with m different from n consider these two strings they are not equivalent to they are not equivalent to equivalent with respect to tilde l because if you consider b power n if you concatenate b power n to a power n on its right hand side that is an element of l whereas, since m different from n a power m b power n is not an element of l. And thus for each n there has to be one equivalence class to accommodate a power n the string a power n. And thus this tilde l is not a finite index and therefore, the conclusion from Mahler's theorem is that there cannot be any DFA accepting this language.