in the last presentation we saw the introduction to the Boolean Algebra I explained you this example and also we did three rules in the Boolean algebra the first one was the complement rule then we saw four rules in the end operation a and a is equal to a a and 0 is equal to0 a and 1 is equal to a a and a complement is equal to zero in the same way we did for the or operation a or a is equal to a A or 0 is equal to a A or 1 a very important one is equal to 1 and a or a complement is equal to 1 so this was the rules that we completed in the last presentation now we will move to the fourth rule distributive law this is one of the most important laws in the Boolean algebra and we will solve a lot of problem depending upon this Rule and according to me none of the Boolean algebra problem asked in the exam will not have this rule involved you have to use this rule to solve the problem so this one is very important and we will see two rules in this the first one is a and b or c this is equal to a and b or a and and C this is known to everyone no need to explain and the second one is little bit weird one a r b and c now this one is equal to a r b and a r c this is very important very important and it is nothing different from this because here I was having and operator and at this point I have R operator here I was having R operator but here I'm having and operator so the operator is interchanged so I will interchange the operator in the result also the end is changed to the r the r is changed to the end in the same way the end is changed to the R so in this way you can remember it it is very simple and we use this distributive lot when we have a or a complement B most of the student will see as the minimized form because they cannot take or operation they cannot take and operation so what will they do they will conclude that this one is the minimum form and this is our answer but definitely it is not the case you can further break it into this form a or a complement from here you can see we have a and b is replaced by a complement and C is replaced by B so I'm just using this distributive law A or B in the same way I have a or a complement and A or B A or B and you very well know from the r operation a or a complement is equal to 1 so we have we have a or a complement as one and a or b and one and a or b one and A or B from here you can see 1 and a is equal to a so if I say this A or B is X let's say this A or B is X so I have 1 and X so this is equal to x and x is a or b so I have a or b so directly when I have a or a complement B I will just eliminate this a complement and I will have a or b this is the most important part of the bu algebra very very very important and we will use this a lot along with the rules that I have already explained so this one is important there is one more form of this particular type if I have a complement or A and B then I will eliminate this a and simply I will have a complement or a B so this two things you have to keep in your mind while solving the Boolean algebra and I hope you will do now we will move to the fifth rule that is our commutative law the commutative law buan algebra does follows the commutative law if I have a or b then it can be written as b or a there is no difference in these two things and the same way if I have a and b then it is equal to b and a this is a simple commutative law and the sixth one the sixth one is associative law associative law and in this if I Have A and B and C then it is definitely equal to a b c what we did we took the end of operation of A and B first then with C but in this case we took B and C first then with a and there is no difference in both the things whatever be the output in this case is same as this one so this is associative and commutative law I told you the distributive law and there is one more important thing which is the priority the priority this is no law but the priority I'm going to tell you the priority of the not is the greatest then we have and then we have and and then we have R so first we have to solve the not operation the complement one then we have to solve the end then we have to solve the r this is a important rule to remember when you solve the Boolean algebra and the last rule the seventh one is the dorgan law the D Morgans law it's very simple if you have a or b and then you want to take its complement it's just a complement law then it is equal to a complement and B compliment we just complemented the variables then we have a complement and B complement and we changed the operator to end we were having or and we change it to and in the same way if I have a and b and then I want to take the complement it is equal to a complement I took the complement of the variable and I will replace this and by or operator and I will the compliment of the variable so this is the deogan law it is just a simple way to take the complement and uh we have completed all the rules for the boan algebra now we will see one example the example that I took initially the umbrella one in which I had to take the umbrella and I told you that this Y is same as a or b and c complement and I'm going to reduce this one to this by using the Boolean J so let's do it Y is equal to b a c complement or B complement a c complement or BC complement and the first thing that we have to do is to solve the complement I have already explained you about the priority and in the priority we have to first solve the not that is the complement then and then or and we will solve the complement by using the deogan law so let's try to solve the complement here but uh you can see there is nothing to solve in the complement we don't have anything like a or C complement we have simply the complement of the variable and you cannot solve it further then we will try to solve for the end operation we will see if we have a and a a and zero a and one a and a complement we will try to have this combination here but unfortunately we don't have then finally we will move for the r and I can see this AC complement this AC complement is common in the first two literals so I will take common AC complement and in bracket I will have b or B complement and this one is written as it is b c complement so what if I have b or B complement what is the result b or B complement what is the result let me show you the result from here A or a complement is equal to 1 so if I have b or B compliment definitely it will be equal to 1 so Y is equal to AC compliment and one this b or B compliment is one or BC compliment if anything is ended with one like X is ended with one the result is X from here you can see a is ended with one we have one so if I say this a c complement is X I'm solving it here then Y is X X and 1 or BC complement X and 1 is equal to X or BC complement X is AC complement so I will write a c complement or BC complement this is y now these two things are definitely equal if you want to use the distributive law here then it is equal to a c complement I'm just using the distributive law or BC complement so AC complement or BC comp complement is the same as what we have got here so in this way you have to solve for the Boolean algebra you have to use the rules that I have explained you it is better if you write these rules at some point and try to solve few more problems in the next presentation I will explain two or three problems and that will help you to solve the Boolean algebra and you can see we have reduced the number of gates here we were having more gates as compared to this one so this is all for this presentation see you in the next one