welcome to lecture eight in this lecture we will continue our discussion of Min terms and then we will begin to talk about Max terms we don't really have too much more to say about men terms but I would like to point out one thing uh let's suppose that we have the following expression for the function f of the Boolean variables X Y and Z let's suppose maybe that u f is given by a formula such as this now I hope that you all agree that this is a minterm expansion for f in other words each one of these expressions is a minterm because each one is a a simple product of all three Boolean variables in either complemented or uncomplemented form and uh of course just as a quick review uh if we wanted to write this in the Little M notation remember the way that we do this is to look at each individual minterm and find out when it will be equal to one so the first one would be one when X is z y is 1 and Z is zero the second one would be one when all three variables are one and the last one would be one when X is one y is zero and Z is one and uh if we now interpret those uh numbers as binary numbers then this is two and 111 is 7 and one1 is 5 so if we wanted to write this function in Little M notation we would have M2 or M5 or M7 which could also be written even more compactly like that now that's just a very quick review of Min terms but the uh new point that I wanted to make here was simply to emphasize that the minterm expansion is a sum of products expression we can see a sum from the fact that we have the plus signs and E what what is being added are products but not just any products these products have to be men terms or else this would not be a minterm expansion in other words for instance if this first term were X Prime y instead of xpre y z Prime if it were just X Prime Y and everything else was the same then this would still be an sop expression it would be still the sum of products but it would not be a minterm expansion in that case because X Prime Y is not a Min term since there are three Boolean variables available but uh so any any minterm expansion maybe I should put the word any here just to emphasize any minterm expansion is an sop the sum of products expression so having said that I think we're now ready to start talking about Max terms and essentially everything that we said about Min terms if we take the complement of that we will get a true expression about Max terms and uh I'll now illustrate what I mean by that first of all recall that a minterm is equal to one only for one combination of the Boolean variables um I'll just say variables and zero otherwise well a Max term is equal to zero only for one combination of the variables and one of the wise now what does a Max term look like well let's um let's give an example this is an example of a Max term expansion for a function each one of these terms is a Max term Just Like A Min term is a simple product of all all of the available Boolean variables in complemented or uncomplemented form so is a Max term a simple sum of all of the available Boolean variables in either complemented or uncomplemented form so since each one of these parenthetical Expressions has is a encloses A or or is a simple sum and in each case all of the variables are there in either complimented or in or uncomplemented form then each one of these is a Max term now another Point uh recall that we name men terms according to when they're equal to one in fact that's that's really just about the only way we could name them if you think think about it because a men term uh is equal to zero for many different combinations of the variables so we couldn't name A Min term according to when it's equal to zero because we'd have many different choices for the name so we name it uh Min terms according to when they're equal to one because there's only one choice then so each minterm will have a unique label well likewise for Max terms since they are usually equal to one we can't name them according to when they're equal to one we have to name them according to when they're equal to zero which again is just for a unique combination of the variables so let's uh think about that and look at this expression we have here for this function what about this first max term here well um whenever uh Y is equal to 1 this Max turn term will have a value of one whenever Z is equal to 1 it will have a value of one whenever X is equal to Z it will have a value of one the only situation in which this first term will have a value of zero is when X is 1 Y is z and Z is zero similarly for the second Max term that one will have a value of zero only when X is z Y is one and Z is one and finally the last Max term here will have a value of zero only when X is 1 Y is zero and Z is 1 so if we interpret these numbers as unary numbers then this is the number four this is the number three and this is the number five so this function could also be written as the product M3 and notice for a Max term we use a capital M instead of a lowercase M so capital M ended with capital uh excuse me Capital M3 ended with capital M4 ended with capital M five and if we wanted to put that in still more compact notation uh now instead of using the summation symbol we use the symbol for a product which is uh this symbol here and we would say capital m 3 4 5 now one interesting thing to note about U Min terms and Max terms is the following uh suppose um we have the three Boolean variables X Y and Z okay if that is the case then I think that you will all agree that um for instance x y z Prime is a minterm and X prime or Y prime or Z is a Max term I think you all agree with that because uh notice that uh XY Z Prime is a simple product of all of the available Boolean variables and either I'm sorry about that okay as I was saying XYZ Prime is a simple product of all the available Boolean variables in complemented or uncomplemented form and X prime or Y prime or Z is a simple sum of all of those Boolean variables in either complemented or uncomplemented form uh so x y z Prime is a Min term X prime or Y prime or Z is a Max term now uh I suggest you stop for a moment and what i' like you to do during this break is to find out the name for that men term and the name or label if you wish to call it that find the label for the minterm and the label for the max term and then uh turn the then resume the video okay well I hope thatth you had luck with that and uh let's see how we would proceed to do what you were asked to do remember that A Min term is named according to when is equal to one and this m term will be equal to one when X is one and Y is 1 and Z is zero and if we interpret 1 one Z is a binary number it would be the binary number six and therefore we can say that x y z Prime is equal to Little M because it's a m term six now let's look over here at the max term recall that Max terms are named according to when they are equal to zero and X prime or Y prime or Z will only be equal to zero when X is 1 Y is 1 and Z is zero well that's the same combination we had before and of course then this is the binary number six and so we have X prime or Y prime or Z is equal to capital M 6 and now an interesting feature that I'd like to to show you is this I claim that the complement of little M6 is equal to Capital m6 and consequently that the complement of capital M 6 is little M6 I really don't have to prove that second statement if I can prove the first statement because it would follow immediately because we know that if we take the complement twice in other words if if uh M6 comp little M6 compliment is indeed Capital m6 and then we took the compliment one more time on each side then uh we would on the left hand side we would get back little m6 and on the right hand side we' get Capital M6 Prime so that would say little M6 is equal to Capital M6 Prime which is the same as this statement just reversed so really we only have to prove this first statement and so let's see how that would go well so uh we want to uh prove that x y z Prime that's l M6 Prime is equal to X prime or Y prime or Z and I hope that you immediately recognize that this proof is just a one-step proof this is simply one of demorgan's theorems this is the one that says the complement of a product product is equal to the sum of the individual complement and uh the the other statement up here would be the other demorgan theem which would say the complement of a product is the sum of individual compliments so in general so I'll just say um use uh de Morgan de Morgan's uh theorems in general um little M sub I prime is equal to capital M sub I and thus capital M sub I prime is Little M sub now this uh leads to some interesting ideas let's suppose that um we had as we uh well well let's just say this let's suppose we have G of XYZ is equal to uh X Prime y Prime Z or x y z prime or X Y Prime Z Prime and suppose we were asked to find the max term expansion for G Prime of XYZ well this is very very easy to do uh in the following way we We Begin by noting that g is equal to now let's I want to find out the labels for these men these are this is a minterm expansion for G and so um let's find out what Ms these are we have 0 0 1 1 1 0 1 0 0 and if we interpret those as binary numbers this would be the binary number one this would be six and this this would be four so G is equal to little M1 or with little M4 or with little M6 thus G complement is the complement of M1 or with M4 or with M6 but if we use the Morgan's theorem it tells us that the complement of this Psalm is equal to the product of the individual complement so it would be M1 complement product with M4 complement product with M6 complement but remember that M1 complement is capital M1 little M4 complement is capital M4 and little M6 compliment is capital M6 X and so we are already done G Prime of XYZ is equal to M1 Capital M1 ended with capital M4 ended with capital M6 which could also be written in this form and that is a Max term expansion for G Prime just that quickly now let's look at another question let's let's continue with this same function uh and let's pose another question now uh rather than finding a Max term expression for G Prime let's find a Max term expression expansion sorry find a Max term expansion before G now previously when we've had um a situation like this where we start with a sum of products expression and we want to get a product of sum Expressions after all remember um or actually I notice that the max term expansion I should have pointed this out earlier actually um right here I'll say um uh we'll we'll point out right here this is a product of s's expression and that's true of all Max term expansions all Max term expansions are p s Expressions just as all minterm expansions are sop Expressions so returning then to uh what we're talking about down here at the bottom um you're now we're given this function G which is a sum of products expression and and in fact it's more than that it's a uh midterm expansion and now I'm asking you to find the max term expansion for this same function and we know that the max term expansion is a product of sums expression well in the past when I've asked you to find a product of sums expression from a sum of products expression the way that we've done this is to take the complement twice and then use the Morgan theorems to simplify and we proceeded in that fashion but there's something there's a much easier way to proceed with this in this case and I'd like to show it to you now so recall again that the function under consideration if we bring this down is equal to little M1 plus little M4 plus little M6 now what that means is that if we want to find the truth table for this function it is extremely easy to find we just um put the columns for X Y and Z and then as usual we list all the possible combinations 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 one 1 1 1 0 and 1 one one and now we know that the function since it's the sum of the in terms M1 and M4 and m6 and we know that M1 is only equal to 1 for the combination 0 01 M4 is only equal to one for the combination 1 0 0 and M6 is only equal to 1 for the combination 1 1 and 0 then the function G itself which is the sum of those three terms will only be equal to one when you have one of those three cases at all other times it will be equal to zero so in other words the the truth table for G is very simply just Z 1 0 0 1 0 1 0 you see we have a one in the uh Row for the binary number one a one in the row for the binary number four and a one for the binary number six or the as you say binary number that represent six so this is the true table for this function now uh our task again is to find the max term expression or excuse Max term expansion for G and I'm just going to go ahead and write down the answer and then we'll explain why it's true so it turns out that g of x y and z is equal to Capital m0 anded with capital M2 ended with capital M3 anded with capital M5 and capital M7 in other words we just look for when the function is equal to zero and that tells us which Max terms will be present in its Max term expans iion now the reasoning behind this is as follows remember that these Max terms each one of them will be equal to one most of the time in fact each one of them will be equal to one all of the time except in one possible combination of the variables well the function G in order to be in order for it to be equal to zero one at least one of these terms has to be zero and so the five possible combinations we see that g is equal to 5 excuse me G is equal to zero in five different possible combinations and so we have these five Max terms the m0 max term is zero for the first combination 0 0 0 M2 is z for the combination 0 1 0 M3 for 011 M5 for 101 and M7 for 111 and so these are precisely the situations when G will be equal to zero and this will be our answer notice that we didn't have to do any algebra at all we just uh looked at the truth table and in fact after you get accustomed to this I don't think you'd even need to do that you could simply say if if the men terms are 1 4 and six then the max terms will be 0 2 3 5 and [Music] 7 Let's uh actually verify that this is true now uh generally I won't I wouldn't ask you to do that because this is going to be a little bit of algebra but I think it's worth the effort in this case so let's uh see how that would go so we want to verify that little M1 or with little M4 or with little M6 is the same as capital m0 Ed with capital M2 Ed with capital M3 ended with capital M5 ended with capital M7 now since I'm uh requesting you to do this uh algebraically then we'll go back to the method we've looked at before which is to use the Morgan theorems and uh we'll see how this goes so we have um so I'll say uh proof so little M1 or with little M4 ored with little m6 and remember we take the compliment twice that's the uh technique that we're going to use here well uh if we take the the first complement then this is a situation where we say that the complement of a sum is the product of the individual complements so we have M1 complement anded with M4 complement ended with M6 compliment and then I'll keep the other compliment outside and once again now in this first step I've used the fact that the complement of a sum is the product of the individual compliments now let's go ahead and use the uh information that we've learned here uh the complement of little M1 is capital M1 the complement of little M4 is capital M4 and the complement of little MC 6 is capital M6 so this is what we have at this point now let's go ahead and expand each one of those terms now I'll show you the way that I like to do this we know that each one of [Music] those terms since each one of them is a Max term each one of them must be a sum of all three Boolean variables in complimented or uncomplemented form we know that because that's the definition of a Max term so I'm going to go ahead and write X or Y or Z for each one of them I have X or Y or Z for M1 X or Y or Z for M4 X or Y or Z for M6 now uh the Second Step this is just the way I like to proceed you might like to proceed in another way but I'm just showing you one way to do it now I want to look at the uh binary number for one which will be 0 0 1 the binary number for four will be 1 0 0 and the binary number for 6 will be 1 1 0 and now finally we knowe that in order for each sum to be equal to zero which after all is the condition we want because remember Max terms are named according to when they are equal to zero so in over for each term each one of these sums to be equal to zero what do we do we need to complement any variable that's equal to one so in the first one we need to complement the Z in the second one we need to complement X and in the last one we need to complement both X and Y so I'm saying here that Capital M1 is X or Y or Z Prime that Capital M4 is X prime or Y or Z and the capital M6 is X prime or Y prime or Z so let me write that down again now without the uh numbers so we have X or Y or Z Prime X prime or Y or z x prime or Y prime or Z and then this is all going to get complemented now we're ready to go ahead and um multiply this out however uh there is something uh interesting we can do here uh and let's uh consider it we can we can make our work a little bit easier in the following way now I don't expect you to see this right away uh so don't worry if it didn't stick out but it is something that's interesting to note for these last two terms I'm going to write the uh the second one is X prime or Z or Y and the last one is X prime or Z or Y Prime now uh over here on the right hand side I'll explain why I'm doing that notice that if we have a or Y A where we're assuming that a is Boolean and Y is Boolean as well if we have a or Y ended with a or Y Prime and we multiply this out we would get a a which is just a or a y Prime which would be ABS uh absorbed into a so we don't have to write it down or with a Y which also would be absorbed into a and finally or y y Prime but y y Prime is a number ended with its compant so that's zero so a or Y ended with a or Y Prime is simply a now that's going to be of significance to us in a few minutes but it's also of significance to us now because if we if uh we look at these last two terms and we think of X prime or Y as playing the role of a then the second term is a or Y and the the third term is a or Y Prime so we can immediately say that the product of those two terms is simply X prime or Z and once again as as I mentioned I wouldn't expect you uh to notice that right away um you know is probably faster just just to rather than looking for something like that at every step it might be faster just to go ahead and multiply out uh when you're doing an algebraic problem like this but since this uh since this is an identity uh right here this is an identity that we're going to need in just a few moments I thought I'd go ahead and introduce it at that point so now in our algebra we've got down to this point and uh it's it's relatively easy now to multiply out what remains uh X or X Prime of course is zero uh excuse me I think I said X or X Prime I meant x and x Prime so x and x Prime is zero we have x z or X Prime y or y z or X Prime Z Prime and uh then finally uh the last term would be Z Prime Z but Z Prime Z of course is zero because that's a a variable complemented a variable ended with its complement so when we multiply all of that out we're left with this and now we uh to get the product of sums form we will use the Morgan's theorem that says the complement of a sum is the product of the individual complements so we have x z complement X Prime y complement y z comp compliment X Prime Z Prime complement and now for each one of those terms we can use the demorgan theorem for the complement of a product being the sum of individual complement and so we have X prime or Z Prime X or Y Prime y prime or Z Prime X or Z so now we have a product of sums expression but if you'll go back to the beginning remember we're trying to find not simply a product of of sums expression for um the function G we're trying to find the max term expansion and if we look at this not a single one of these terms is a Max term uh we have variables missing in every case and so the question arises is well then how can we or is there any way to easily turn this product of sums expression into a Max term expansion and the answer is yes and the key to doing that is once again using this identity that we derived up here so let me repeat it down here we have a oh don't get it in a moment we have a a or Y ended with a or Y Prime is equal to a now we're going to use this um in the um previous case up above when we were doing this problem we use this problem we use this principle you might say going from left to right but now we're going to use it going from right to left left I'll show you what I mean in just a moment so we have X prime or Z Prime is our first term well I'm going to change that to X prime or Z prime or Y end it with X prime or Z prime or Y Prime here I'm thinking of X prime or Z Prime as being a and so this term right here is equal to the product of these two terms using this identity right here and you can see why I'm doing that because each one of these factors here is a Max term X prime or Z Prime is not a Max term but each one of these is a Max term so let me do one more and then I'll ask you to complete this process uh X or y Prime is the same thing as X or Y prime or Z ended with X or Y prime or Z Prime and now you go ahead and try to do the remaining two and then we'll finish this up okay I hope that the uh next two terms you wrote down were X or Y prime or Z Prime and X prime or Y prime or Z Prime and then finally for the last term I hope you wrote down X or Z or Y ended with X or Z or Y Prime so now we have a uh an expression that has six Max terms in it and you might think then okay then this is a Max term expansion but let's be careful about this uh let's um make sure that no Max term is being repeated here and the easiest way to do that is to go ahead and put each one of these Max terms in its order and the correct order for its variables and then to assign the uh labels that we know about so let's see how this will go uh the first one is X prime or Y or Z Prime the next one is X prime or Y prime or Z Prime the next one is X or Y prime or Z the next is X or Y Prime or C Prime the next one is X well we notice actually immediately notice that this term X or Y prime or Z Prime is the same as this X or Y prime or Z Prime and any Boolean expression anded with itself is itself so there's no need for us to write it down again so I'll just skip this term here the X the second X or Y prime or Z Prime term I'll just skip that and go down to this so we have X Prime or Y prime or Z Prime X or Y or Z and the last term will be X or Y prime or Z and now we want to uh for each one of those terms we want to find each one of these is a Max term and let's find the correct labels and remember again a Max term is labeled according to when it's equal to zero so we want 1 0 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 0 1 0 so if we look at these uh numbers uh as if we look at these as uh binary numbers this is 5 7 2 3 7 0 2 and you see at least I find it particularly easy once we've identified these by number you know found the labels for them then it's obvious to see that we have a couple of other repetitions we have the two uh repeated so we can get rid of one of those and we have seven repeated so we can get one of rid of one of those and so now you can write down Capital M5 capital M 7 capital m 2 capital m 3 capital m0 and as we've said before we write these uh we we traditionally write these in ascending order so we have capital m0 capital M2 Capital M3 Capital M5 Capital M7 and if we compare that to what we wanted at the top they indeed 02357 02 3 5 and 7 so we have proven what we wanted to prove and uh of course it was a lot of algebra we wouldn't ordinarily want to do that it's so much easier just to look at the truth table as we did right here and uh identify if we want to find again if we want to find a minterm expansion for a function look for when it's equal to one if we want to find Max term expansion for a function look for when it's equal to zero and if you'll do that you'll be in great shape but uh the reason that I think it was useful to look at this algebraic uh work is to we see the importance of this identity right here that a or Y ended with a or Y Prime is the same as a and that's particularly important down here when we want to uh turn this product of sums expression into a Max term expansion so uh that concludes that notion I'd like to give you uh one more quick problem uh before looking at the uh test problems for this lecture so let's uh suppose that the truth table for f of XYZ is as follows okay suppose then that that's the truth table for app and let's suppose that we were uh charged in the following way find the um minterm and Max term expansions for both F and F Prime now I hope that you see that this is very easy and so I very strongly encourage you to stop the video for a moment try to do this on your own and then come back okay so here we are ready to do this uh final problem in the lecture before learning about the uh test for this lecture and uh the way that I would proceed is follow is as follows since um we're going to need F Prime in addition to F I think I would go ahead and start by adding one more column to the trueu table I'll say F Prime and of course nothing could be easier than finding F Prime from F uh is simply you know zero complement is one and one complement is zero so we get we get that for f Prime now now we want to find the minterm and Max term expansions for both of these functions well uh minterm expansion for f remember when we're looking at Min terms or when we're looking for Min term expansion we look for when the function is equal to one and so we come over here and we have uh Little M one or with little M2 or with little M5 or with little M 7 that's the minterm expansion for f uh since I'm running out of space I'll come up here ma next term expansion 4 F Well recall that for ma term expansion we look when the uh function is equal to zero and uh f is equal to 0 or zero so we have capital m0 capital M3 Capital M4 and capital m6 and notice actually to be real honest with you once you get this first one the minterm expansion then you notice that the maxterm expansion the the um subscripts that appear are just the ones that don't appear in the men term expansion so really once we have this first expression we don't even need to glance at the truth table anymore once you get experienced you'll find out how to do that very quickly and then for f Prime well F Prime is just the complement of F and so therefore the minterm expansion for f Prime will simply be little m0 or with little M3 or with little M4 board with little m6 and likewise the max term expansion for f Prime is capital M1 anded with capital M2 anded with capital M5 anded with capital M7 and and now of course I was to get these to do what I just did I use the fact that the midterm and Max term uh well the men terms are are complements of the corresponding Max term and vice versa but you can also again for the minterm expansion for fime if you want to you can just look at when the function is equal to one fime is equal to one at zero 3 4 and six so that's why we have 0 3 4 6 and down here for the max term expansion look when it's equal uh to zero and frime is equal to 0 at 1 2 five and 7 so we have 1 2 five and 7even here so that's how easy it is to find a minterm or Max term expansion for a function if you have this truth table nothing could be easier so now we'll look at uh the test problems for this lecture so the uh two test problems for this lecture are as follows in 8.1 it says find the minterm expansion for the function f of XY Z is equal to XY or Z and your choices are a capital m0 ended with capital M2 ended with capital M4 B little M1 or with little M3 or with little M5 or with little M6 or with little M7 C little m0 or little M2 or little M4 or D Capital M1 order with capital M3 or with capital M5 order with capital M6 or with capital M 7 that's 8.1 and then 8.2 is to find the max term expansion for the very same function and the choices are a capital m0 anded capital M2 anded Capital M4 B little M1 or with little M3 or with little M5 or with little M6 or little M7 see the Lim zero ended with the Lim two ended little M4 and finally uh D Capital M1 ended with capital M3 ended with capital M5 ended with capital M6 ended with capital M7 and that concludes uh well good luck of your work and that concludes lecture 8