 Welcome back in continuation to the last lecture where we discussed reducing the given well formed formula in a prepositional logic to conjunctive and disjunctive normal forms where conjunctive normal form is considered to be conjunctions of disjunctions and disjunctive normal form is disjunctions of conjunctive normal forms. So given a well formed formula which is in the form of implication and negation etc and all so we will try to reduce it to a given normal form. So what is the use of reducing it to conjunctive and disjunctive normal forms this is also considered to be one of the important decision procedure method as usually in any decision procedure in decision procedure method in prepositional logic for example if you have taken into consideration truth table or semantic tabloids method or some other methods which we have considered so far in all these methods what we did is simply is that given a well formed formula we are able to check whether a given well formed formula is a tautology and as all of us know all tautologies are considered to be valid formulas and given for given a set of well formed formulas we also came to know with these decision procedure methods either truth table or semantic tabloids method or this method conjunctive normal forms reducing the given formula into conjunctive and disjunctive normal forms we can say that a given formula is satisfiable or not so under what conditions a given formula is going to be true and not only that thing we can also say that when a given formula is considered to be contingent etc and all. So what we essentially what we are essentially doing is that a given any complex formula we are trying to reduce it to its corresponding DNF there is disjunctions of conjunctions or conjunctions of disjunctions that is CNF so one of the important observations that we can make out by reducing the given well formed formula into a given CNF or DNF so you have to note that any given formula can be reduced to either CNF that is a conjunctive normal form or DNF that is disjunctive normal form. So these are some of the important observations which we can make out but before that so what I am essentially trying to do in this lecture is simply this that I will be talking about some examples of how to reduce the given formulas into conjunctive and disjunctive normal forms and then I will talk about some of the important properties of logic that is satisfiability consistency whether or not a given formula is valid etc and all and in the second part what I will be doing is as an application of reducing the given preposition logical formula into CNF and DNF so we will try to see its application in analyzing some simple digital switching circuits so what we will do in that thing is that a given a complex circuit so we will transmit it using the principles of preposition logic into essentially conjunctive and disjunctive either conjunctive or disjunctive normal form and then we reduce the given formula into a simple formula and that formula corresponds to a simple digital switching circuit. So again we reconstruct the circuit based on whatever simplified formula that we arrived at and from that you will reconstruct the digital circuit and then that will constitute to be a simple simplified form of a complex digital circuit and then we will also see with some examples so one of the important uses of logic is that it can also be applied in solving some kind of puzzles so we will also see with the help of CNF and DNF reducing the formula into CNF and DNF so we will see that some of the important problems such as nights and news etc which we have which we solved it already with the help of semantic tab blocks method those things can also be solved by using this reducing the given formula into CNF and DNF so essentially CNF reducing the given formula into CNF and DNF will serve as some kind of decision procedure method for knowing whether a given a formula is valid or invalid or when two groups of state and when group of statements are consistent to each other. So these are the following observations in continuation to the last lecture so these are some of the important observations that we can arrive at so a well form formula a after reducing it to conjunctive normal form or disjunctive normal form is valid if and only if it is in the case of conjunctive normal form if and only if each disjunctive clause in any conjunctive normal form representation of a contains a pair of complementary literals so what do you mean by a complementary literals suppose if you have a literal P it is considered to be positive literal and then the complementary of this one is not P usually we represent complementary as this particular kind of thing but instead of that we are using negation of P in Boolean logic we use complementary for this one and then for R we use in Boolean logic we use plus and for end it is a multiplication sign is used so these are the only differences between the Boolean logic and the preposition logic that we are trying to talk about so essentially they are more or less the same. So what is a CNF CNF is nothing but a conjunction of disjunctions now so first you write this thing conjunctions of what disjunctions so there are several disjunctions and all till the end so what do you mean by a disjunction it can be in this form P1 or P2 etc so each term in the conjunctive normal form is a disjunction so in this sense a well form formula is considered to be valid if and only if each disjunctive clause in the conjunctive normal form contains a literal and its negation and all so now observe this particular kind of formula so we have semantics of R so that is like this you have A and you have B and these are the only values that it takes it takes this values and then alternative TF and all ARB the semantics of ARB that is truth meaning of a formula ARB is nothing but truth conditions of ARB and all so that is going to be false only when both disjuncts are false in all other cases it is going to be true so now this given formula D1 and D2 to DN so this is going to become true if and only if all the disjuncts are true this has to be true this has to be true and this has to be true then only your CNF is going to be true then you have established that a given conjunctive normal form is considered to be a tautology so and as usual you know all tautologies are also considered to be valid formulas so now one observation important observation is is that why we are reducing a given formula into this conjunctive normal form is that if it so happened that this disjunct has a literal and its negation something like P1 etc and all then this is always considered to be true only so this is nothing but a tautology and all so now suppose in your disjunction D you already have P1 or not P1 and there are some other letters that exist here it can be Q or yes whatever it is so now since this is already true so now it is like XR T XR some other formula like P something like that PR Q or something so now since in this disjunct in every disjunct this first formula is already true and all because of a literal and its negation is there that means always going to be tautology only so now irrespective of whether whatever follows after this thing is the true or false the whole thing is going to be true only because of this particular kind of thing so the first disjunct is already true observe these two cases and respect you of this whatever disjunct that follows after that one a literal and its negation suppose that happens to be true is also true and the next literal that occurs after the negation and its negation literal and its negation that is considered to be false then also it is going to be true only. So in that sense if each disjunct have to be true there should be at least a literal and its negation has to be there if a literal and its negation is there then that disjunction is automatically turning out to be true so in that sense if each disjunct it so happened that each disjunct has a literal and its negation then all the formulas all the disjunct that occurs in the congenitonomal form are going to be true in that sense this your given CNF is going to be true so in that sense a given formula is considered to be a tautology so let us consider some simple examples with which we will establish this particular kind of observation so we know that this particular kind of formula P implies Q implies P so that is a theorem or valid kind of thing is also called as paradox of material implication which we will not go into the details of this one but it is a it is a theorem in propositional logic suppose if I write like this theorem in general propositional logic so now as a first step in CNF in conjunctive normal form which is a conjunctions of disjunctions so what you will do is you will start eliminating this implication so you eliminate this implication by using this particular kind of rule x implies y is same as not x not x or y what essentially you are doing is you are reducing this implication to simple disjunction and its negation in the final formula what you will find is you will not find implication on this sign you will not get it but only signs that you will come across is negation R and conjunction so in that sense you are reducing the given formula into CNF so now as a first step what you will do is so now the whole thing is taken as x so now x implies y means not of P or Q implies P so this is the first step that we are trying to see so now this brackets should be there so now you reduce you eliminate this implication then you will get not Q or so this is not P or whatever is there here is the one which we have written and the second step this will become this so now what you can do is you can use distribution law and all or associative law this will become not P or not Q or P or not P not P or not Q and not P or P so now observe this particular kind of disjunction this is so this particular kind of disjunction observe this particular kind of thing it is in this form x or y so now since this is already true PR not P is also always considered to be true only so now irrespective of whether this formula is going to be TRF this is always going to be true only because of the semantics of disjunction so one particular disjunct is true then irrespective of whether this whole term is false or true is going to make the whole disjunct true it is in that sense so this formula is going to be a tautology so this is one way of representing one particular observation is that in a given well-formed formula suppose if it so happened that a literal and its negation occurs then that is considered to be a tautology and it is also considered to be a valid formula so similar kind of thing which one can do with the help of a truth table also for example suppose you know that this formula is always considered to be valid only that is the law of contra position P ? Q is nothing but not Q in place not so now from the truth table also one can draw the CNF and DNF so now there are two variables here P and Q that is why there are four entries that are possible so now then you need to take into consideration P ? Q then the next one is not Q and not P and then not Q ? not P and then the final one is this one the whole thing so that is P ? Q ? ? Q ? ? P so now just quickly we will construct the truth table and all so since there are only two way variables in the truth table there are only 2 to the power of n entries will be there 2 to the power of n rows is going to be there so there are four rows which are possible so now in this case T T F and F T F alternative T and alternative F is the one which you write so now P ? Q is going to be true is going to be false only in this particular kind of case in all other situations so that is going to be true so in all other cases it is going to be T so this is all we know about material implication that is P ? Q is nothing but not P or Q so now not Q is exactly the opposite of these things so now whenever it becomes T it becomes false whenever it is F it is T and F and T and not P this one this is F F exactly the opposite of these values and all when P takes value T not P takes value F so now not Q implies not P so now this formula is going to be false only in this case so that is when the antecedent is true the consequence is false this is going to be false in all other cases it becomes T so now you need to observe the implication of these two things so that is going to be your final formula so why we have what is that we are essentially trying to do is that from the truth table also you can make out you can write these sanctions as DNF and CNF so now this one now we need to check whether there is any formula in which P ? Q is T and not Q ? not P is false so now this is the antecedent in which your three T's are there so now we need to check whether there are there is any formula in which your antecedent is true and the consequence is false so now you will observe that there is no way no row in which you have P ? Q T and not Q ? not P false it is in that sense all are going to be T only so all these things are turning out to be true so now what we can do is this thing so from the truth table one can construct one can construct a DNF let us see what how we do it so now for the DNF you take into consideration all the things which are true all the rows are true only so now we start constructing this conjunctive normal form for this one so now what are the values for this one what are the rows in which it is go it is true in all the rows it is true only so now it is like the P and Q P and Q are so now the next formula is our P and not Q are so now this one so that is not P whenever it is P is false it is represented as not P and then not P and Q this one both are false not P and so what essentially we are we did is this thing that so it is like D1 C1 or C2 or C3 etc so now what is this C1 it is in the distinctive normal form so where each distinct is a conjunction so this is like to be written in this way D1 or D2 or D3 so where each distinct is a conjunctive so how did we essentially write this one so under what condition this whole formula is going to be true so now when P takes value T and Q takes value T that is one particular kind of condition under which this formula is going to be true and then you have P and not Q in that case it is going to be T and you have to list out all the rows in which the this formula is going to be T so this is what we have at the end so that is in the form of distinctive normal form D1 or D2 or D3 where each D1 is a conjunct so this is the distinctive normal form so now from the truth table one can make out we can write down the corresponding DNF or CNF so that is one way of doing it that is what is the important observation that we can make out here so from the truth table also one can construct a corresponding DNF or CNF so now the second observation is that a given well form formula is considered to be unsatisfiable if and only if each conjunctive clause in any DNF representation of X contain a pair of complementary literals so it so happen that in a DNF distinctive normal form it so happen that you have both P and not P is there in any one of this formulas in each and every descent then all decisions will turn out to be false in that case a given DNF is going to be false then it is said to be unsatisfiable. So for example in this case P ? Q and Q ? R and P and ? R so that is considered to be unsatisfiable because in the last term in particular P and ? R so that is having a problem so that is why that form the term is going to be false that makes the hall is makes the conjunct false in all so when a formula X contains a pair of literal and it is complementary that is not P then in the given form that the whole formula is going to be false in all so that makes this form unsatisfiable now so let us see why it is a case that P ? Q and P ? Q and Q ? R and P ? P and ? R so you add to this one why it is considered to be unsatisfiable so now it is like this thing first one can be written as this one now we are trying to say why it is considered to be unsatisfiable and all let us see whether it is unsatisfiable or not so now the first statement can be written as this one P R not P R Q and the second one is not Q R R with the second one and then the third one is P and ? R so this needs to be transformed into this one not P R Q now the second you write it in this way not Q R R and so this can be written as this one use De Morgan's law and then convert it into that is its corresponding disjunctive form so it is P and ? R will become not P R R so why it has become like this P and ? R is nothing but if you take the negation into consideration you take the negation of P and R so now you see this one not of not P is P only not of disjunction is conjunction and this is not R so this is same as this one so now why this formula is considered to be unsatisfiable so now we need to further reduce this particular kind of formula and then you will see why it is the case that it is going to be unsatisfiable or not so now there are several ways of showing that whether or not this particular kind of formula is unsatisfiable or not so first there is a method which we have we came to know that is a semantic tableaux method using that you can see whether it is unsatisfiable or not so now you can write it in this way P and ? R is represented in this way so now this this formula is checked in all so now we are using semantic tableaux method just to see whether it is unsatisfiable or not so when these three formulas are going to be unsatisfiable when you construct a tree for this formulas in all if all the branches closes then that is considered to be unsatisfiable if at least one branch is open so that is considered to be that is considered to be satisfiable so now we will come back to this particular kind of format little bit later so now we are checking it with semantic tableaux method whether it is considered to be satisfiable or not so now this is not Q and R so this in this way you can write it so now you have R here and not R here so this is contradictory to each other so this closes and then whatever is left is this one not P and Q so now P and not P closes and you have Q and not Q this also closes so now from this method we can make make out that these three statements are inconsistent to each other so now according to our this thing so this is in CNF it is a conjunction of disjunctions and all so this formula if it has to be false and all if this is going to be false if all this at least one of this disjunction is false and all then it is going to make the whole conjunct false when you say that P and Q is false at least one conjunct is false then you can say that the whole thing is false so there are ways to say that so now what we are observation tells us is that a well-formed formula is said to be unsatisfiable if each conjunctive clause in any DNF representation of X contains a pair of complementary literals and all but here the formula is in CNF so now we need to convert this formula into corresponding DNF and all so if you take the negation of this whole formula then it will be converted into the corresponding DNF and all so that is not PRQ and not QRR and not PRR so now the negation of this one is this so now this will become not of not PRQ now negation of conjunction will become disjunction and then each term will be like this and negation of conjunction will become disjunction here and again negation of not P and R so now so what essentially we are trying to do is the formula is in CNF but we are trying to convert it into disjunctive normal form so once you convert into disjunctive normal form then we are we use we make use of the observation that is a term which consists of both a literal and it is negation then the conjunct has that term will be also now so now this translates into you have to use DeMorgan's laws then it will become P and not Q not of not PSP not of Q is not so now this is R now this will become Q and not R now R this is P and not so now this is in the disjunctive normal form that is D1 or D2 or D3 where each disjunct is a conjunct like in this first case D1 is nothing but P and not Q and D2 is Q and not R and P D3 is P and not so now one of the important observation is this that whenever your disjunct that is which is nothing but a conjunct here P and not Q a literal and negation and all P and not Q in each one of these terms a literal is there and the unnegated form and the negated form is there and all if you have Q and it is not R and then P not R etc. So that makes all these things false in any given DNF is DNF of your formula in this case so in your DNF each conjunct is having a literal and its negation or sometimes it need not have to be its own negation but negated formula and unnegated formula and unnegated literal then obviously that disjunct D1 has to be false and hence and since each disjunct is false that makes the whole formula false enough so that is what is the important observation that we can make out a well form formula is considered to be unsatisfiable that means the formula is going to be false if and only if each conjunctive clause in any DNF so that is here in this case P and not Q Q and not R P and not R you will see negation and unnegated term and all unnegated term is followed by a negated term so in that case all the D1 D2 D3 are going to be false and hence the formula is going to be false that is why it is considered to be unsatisfiable so we can also show that a given formula is valid or not by again reducing it into conjunctive or disjunctive normal form so this is some of the important observations which one can make out so these are the two important things which are directly related to satisfiability of a given formula a formula in conjunctive normal form is considered to be valid if and only if each of its disjunctions because in conjunctive normal forms is conjunctions of disjunctions so in each of such kind of disjunctions it contains a pair of complementary literals like P not P and on so then each conjunct is going to be true that means all the I mean the whole formula is going to be true that means tautology and hence is a valid formula and conversely a formula in DNF like this the one which we have expanded on the board is considered to be unsatisfiable that means is going to become the formula is going to become false that means takes the value 0 if and only if each of its conjuncts contain a pair of complementary literals like P or not P or P and not Q etc. So the idea here is that if one formula is one literal is true and the other literal is considered to be false it makes the whole formula whole D1 false only since all disjunctions disjuncts are false and the given disjunctive normal form whatever formula that exists in the disjunctive normal form D1 or D2 D3 all are false so hence D1 D2 D3 the DNF is going to be false that means it is going to be unsatisfiable. So this we make use of it in solving puzzles also while solving the Knights and Naves puzzles we translate the statements into the appropriate language of prepositional logic and then we say that either we write it in the CNF or DNF usually we write it in the CNF conjunctions of disjunctions then if whether or not we will check whether or not a given formula is satisfiable enough so usually Knights and Naves puzzles are translated into those problems in which is usually satisfiable and all that makes some of these sentences true and all so then we will look for the solution whether or not given person who is talking about something is a Knight or even we will come to it little while from now so now this is an important theorem which is worthwhile to mention here a clause P1 or P2 or P3 etc is considered to be valid if there exists some I and J such that the PI is equivalent to not PJ I mean the literal literal and its negation exist in a given formula then obviously it is going to be valid a conjunctive normal form C1 C2 C1 and C2 C3 where each C1 is a disjunct is going to be valid if each of this clauses CI that is D1 or D2 D3 are true if all these things are true C1 is true and C2 C3 all are true then obviously CNF is going to be true that is a tautology and hence it is a valid formula so now in the examples that are there here not PRQ PRR so that is considered to be valid because if you rearrange it in a certain way then it will become PR not PQRR so at least if PR not P is already true we know that it is a tautology so then the whole formula is obviously going to be true irrespective of whether P the other preposition variables whether it takes true true or false it is going to be true so that makes the whole formula true that means all true prepositions are considered to be tautologies in the second case also we have a literal negation in the in C1 and C2 so a literal negation is there that is why that makes C1 true for example C1 is nothing but not PRQ or P so in that PR not P is already there irrespective of whether Q is true or false the first C1 the first term C1 is going to be obviously true in the same way in the second term R and not a literal negation is there is always going to be true hence both the terms are true it is going to be a tautology and obviously all tautologies are considered to be valid formulas so now let us try to solve some simple examples with while making use of this particular kind of idea that is a literal negation exist in a given CNF then the formula is going to be a tautology and hence it is a valid formula so now in these kinds of puzzles we are trying to look for the satisfiability or unsatisfiability of a given formulas and all so what we essentially we do is like this let us consider simple example which we already discussed it in the context of semantic tableaux method but again we try to do the same thing with the help of converting this particular kind of formulas into conjunctive and disjunctive normal forms so here is a story which goes like this there was a robbery in which lots of goods were stolen and it so happened that the robbers left in a truck and it is also known to us that these are the some of the things which are known to us nobody else could have been involved other than only these three persons that is a B C so no D is involved in this particular kind of that information we are sure of and the second case is this that C never commits a claim without a spot is patient so wherever C is there you can assure you can surely say that he is already there so now the third thing which we already know is this particular kind of B does not know how to drive that means if B comes alone out of all these things stolen goods and all you cannot flee so you cannot run away because you do not know how to drive the truck and all that means he needs help of either A and C so now with this information we are to whether or not we need to show whether A is innocent or a is guilty so now this problem requires some kind of representation so that is when I say that a B C etc now is guilty B is guilty C is guilty if I say not a not B not C then a is innocent not B means B is innocent not C means C is innocent so that particular kind of representation is what is needed in the beginning of solving this particular kind of problem so what essentially we are trying to do is we translate the given English language sentence into appropriately into the language of propositional logic and then we try to convert it into CNF and DNF and from that you can make out when it is going to be satisfiable and all is the one which we are trying to look for so now the first statement can be represented in this way nobody else could have been involved means either A or A is involved or B is guilty or C is involved as one either one of them is guilty and all C never commits a crime without A's participation that means C implies A so now third statement B does not know how to drive so B if B has to be accompanied with A there is a first sentence or B has to be accompanied by C that's why B and C so now we have set of formulas and all the first one is obviously in particular kind of format A or B or C let us consider it as a term C 1 and the next one you need to change it into appropriately into a corresponding form so now given these formulas we are trying to check whether by redoing these formulas into conjunctive or designative normal forms we are trying to check whether we are trying to find out whether A is guilty or not so now let us try to convert this formula into corresponding normal formula so now this is the thing which we have so these are the three formulas that we have so that is first one is A or B or C convention the next statement is what is that C implies A and the third one is B implies B and A this bracket should be there here particular or B and C so now it looks like that it is like in this particular kind of format C 1 C 2 C C 1 and C 2 and C 3 so that is why it is called as CNF I mean you cannot call it a CNF at this stage but we need to convert these things each and every formula into conjunctions of disjunctions so this already all disjuncts only so now this is as it is A or B or C so this is a disjunct only and you need to write this thing as not C or A and you need to convert this thing into corresponding normal form so that is so now this whole thing is taken as X sorry Y and this as X so then B implies X implies Y is not X or Y so that is not B or the whole thing B and A or B and C so what we are essentially doing is we are reducing this formula into its corresponding normal form so these things are already transformed into corresponding disjunctions so D1 and D2 etc and all so now this is as it is A or B or C and not C or A and now this is the one you have to use the associative law and all so then this will become not B or B and A or is the first one or not B or B and so this is the second kind of thing so now we have to further reduce it into this particular kind of form so now you have to use distribution laws here so this unit I have to do anything here it will remain as it is so now we need to convert this thing to particular kind of thing not B or B the first one and not B or A so this is what it reduces to or now this one is not B or B and not B so this formula reduces into this one so now so now we know that B or not B is obviously it is going to be T only so you do not have to worry much about it so this is always going to be T usually you represent it as this letter T so in the same way here not B or B is also going to be T so these terms will vanish and all here so now what will remain here is not B or A or the other one here is this one now this is this particular kind of connection and not B or C so now this formula is reduces to obvious tautology and all it does not make any sense now this is not B or A or whatever is there here whatever left is not B or C so now so what are the formulas that we have now here so now you need to see this box and all you have A or B or C and not C or A and so not C A and these are the formulas not B or A or B or so now this further reduces to this thing so why what is what essentially we are trying to do is that a given formulas C1 C2 water C3 and all we are trying to reduce it into its corresponding normal forms and all and once you reduce it into some kind of format either CNF or DNF then we can talk about whether this formula is going to be satisfiable or not so not B or A or not B or C so now this reduces to A or B or C and not C or A is as it is now here we can use some kind of associative law and all then we can say that A or this not B goes inside not B or not B C so now not B or not B same as one not B only so this is A or B or C and not C or A and then A or C so now we have everything is in the form of conjunctive normal form each C1 C2 and C3 so now we need to inspect that a literal net's negation is there here it doesn't matter even here also a negation and a negated form is there that makes these two formulas true and then so now it is further reduces to this one so there is a way to talk about this thing once you convert this formulas into a given conjunctive normal forms and all here is a role which we use B or C not C or A so these two resolves into simple formula that is A or B and C or not C is going to be true only so then this reduces to this one so this is nothing but A or B so now in the same way A or B or C the first term and the last term that is A or not B or C so now this resolves into a and B or not B is B and not B cancels and all and then whatever whatever remains is C A or A or C so this is another term so now A or C and this term not C or A we will end it in a while from now and not C or A so this translates to A why because A or C and not C or A because this translates into I mean this reduces to simply it resolves into A so now we got this final thing that we got A is the case and all that means A has to be guilty so what essentially we did is clearly is that first you translated the English language sentence into appropriately into the propositional logic and then we reduce the non normal forms into normal forms and then we came across this particular kind of formula A or B or C not C or A if you further simplify it and all so then ultimately we have these two literals A or B or C and not C or A these resolves into A or B or A and A or A is nothing but A only it becomes B so this particular kind of method is what is called as resolution refutation method which we will talk about it in the next class so how whenever you one of the advantages of reducing the given formula into conjunctive normal formula is this particular kind of thing so once you come across any conjunctive normal form then you can use this resolution and refutation method to further reduce it into its corresponding formulas so these two resolves into A or C and A or C and not C or A so this becomes true only in the case that A has to be true and all these conjuncts have to be true only when A has to be otherwise it is going to be false so ultimately what we got is this particular kind of solution that observe this one A is considered to be the case that means A is considered to be guilty so now this is the way in which one can solve some of the important puzzles and all so now we will move on to one simple example where this particular kind of thing is used so this can also be used in analyzing some simple digital circuits and all so I will talk of essentially about the basic idea of that month and I will try to solve one example based on how a given complex digital switching circuit can be translated into a simple kind of formula and all so now usually in the digital switching circuits whenever you have this particular kind of thing a this and then be suppose if these two switches are in a series then usually you write it as a and b usually in the case of Boolean interpretation it is multiplication of B so whenever two switches are in there are arranged in a parallel kind of thing let us say there are two more switches like this D and all then you write it like this one as the first one is in the series that's why you write it as a and B and these two are in parallel so that's why you use this particular kind of symbol or in the simple digital switching circuits you can call it as plus operation so now for example if not D is there so here the notation is this that you usually take care take it in this way the moment I write B that means B is closed you switched it on or if I write not B that means B is open the switch is off so that is the only difference here so this one can be written as B and not because B and not D are in a series so you can write it in this way so what you will essentially do is that given a simple digital switching circuit we translated it appropriately into a given propositional logic and then we try to simplify this formula and then the simplified formula is corresponds to simplified kind of circuit let us consider one simple example with this we will end this lecture so here is this example so now you have a switching circuit a like this and you have B and then is connected in this sense see so these two are connected parallely a and this one and not be is the one which is there here now you generate some kind of output in so this is the input and you will get some kind of output in so this diagram essentially says that the C and A are connected parallely and of course a and B are in series C and A are connected parallely and then not B is also connected in a series so now this is written in terms of propositional logic like this so the first formula is written as this thing since a and B are in series that means both switches have to be on so that current passes through it now second one or it should be like this C so this is going to be the right kind of diagram so now this is in parallel so that is why you write it in this way C or a and now this is in series that is why it is not so now here are the two formulas and now these formulas are further reduces to this particular kind of thing this is as it is a and B are C are a and a and B are not so now we use distributive law and all suppose if you take this as X and this as Y so now the first term a and B are C are not a that is the first term and now a and B are not be so now this is going to be your second term so now we need to rearrange it a little bit and then this will become like this a R B R C R A as it is and this will become a R B a R not be so the first one and and a R sorry B R not be so now again used distributive law and all so this will become a R not be and a B R not be so now so this changes into this particular kind of thing a and B are C are a are not be so because B are B is obviously true and all you can ignore this particular kind of thing so now you have a and B C are a and a are not so with this try to end it in all so now observe these two terms a and B a and B are C are a and a are not be so now again we need to use some kind of associative law and all so now this will become a R B C are a and here be this thing so now use distributive law on this one it will become a and B are a are C a are not be a are not be then a C so ultimately this reduces to this particular kind of thing since a are B and you are the same so now so this will become a are not be will become a only because of law of absorption so this is a are C and what else is the case what is left and a are not be so now this can be written as a are C and not be so this is what we have reduced it into so now a complex final remark is that a complex digital circuit when it is transformed in by using the principles of preposition law this transformed into this particular kind of thing so now you start worrying about your circuit and all so now this says that you have a switching circuit A and then it moves to goes to like this and then not be so now this is going to be your simplified circuit corresponding to this one so what is given here is this that a given complex circuit can be reduced into a simple switching circuit in all so in this lecture what we did is simply this that we given a well-formed formula we try to reduce it into conjunctive and disjunctive normal forms and then we talked about we solved some kind of puzzles by using this car is reducing it into conjunctive and disjunctive normal forms and then we try to see whether we can simplify the complex digital circuit into a simple switching digital circuit so in the next class we will be talking about a resolution refutation method which is considered to be an outcome of reducing the given formula into a conjunctive normal form.