 this session I am Priyanka Bidla assistant professor in ENTC engineering today we will see the numerical on Boolean algebra this is the learning outcome of this video lecture at the end of this session students will be able to solve the Boolean expression using Boolean algebra and De Morgan's theorem these are the contents now before moving towards pause this video for few seconds and you have to recall what are the laws and rules for solving the Boolean algebra expression so actually these laws are used for solving the expression so first law is commutative law so there are the two laws of addition for two variables here there are the two or gates and here I am giving the two inputs A and B so the resultant output is A or B and if we are giving B and A inputs to the OR gate here the resultant output of the OR gate is B or A so here we write A plus B is equal to B plus A or we can say A or B is equal to B or A so now we see the law of multiplication for the two variables so for multiplication we have used the AND gates so here A and B are the inputs of AND gate and the resultant output is A ANDing B and suppose B and A are the inputs of the AND gate then resultant output is B and A so the meaning is what A ANDing with B is equal to B ANDing with A so this is commutative law then associative law so here addition of three variables you can see in this figure B and C are the inputs of the OR gates it generates B plus C then A and the B plus C are given to the OR gate and it generated A plus B plus C similarly here A and B are the inputs of OR gate it generates A plus B or we can say A or B it is applied to this OR gate ORing with C and here it generates A plus B plus C so for addition of three variables is written as A plus in bracket B plus C is equal to A plus B plus C then similarly you have to write down for the multiplication instead of OR gates here I have used the AND gates so B and C are the inputs of the AND gate it generate BC then A and BC are given to the AND gate and it generates ABC this is equivalent to A and B are given to this AND gate it generates AB then AB and C are given to this AND gate it generates ABC so in both way you can write down so this is associative law now distributive law here there is a combination of AND and OR gate so B and C are the inputs of OR gate it generates B OR with C and then applied to AND gate with A input this is equivalent to this output how it will be A and B are given to this AND gate it generates AB then A and C are given to this AND gate it generates AC then AB and AC are applied to this OR gate and it generates the Y that is what we can write down in this way A ANDing with B or C is equivalent to what A into B or A ANDing with B or A ANDing with C so this is distributive law so with the help of these laws we have to solve the examples now before moving towards first we will see the De Morgan's theorem so De Morgan's theorem is nothing but here this theorem explains that AB complement is equal to A complement ORing with B complement or we can say A ORing with B complement of that is equal to A complement and B complement means what the De Morgan's theorem here this is known as the NAND means here this is NAND gate okay and NAND is converted into the negative OR gate here this is NOR gate it's the output of the NOR gate okay and it converts to the negative AND gate so this theorem explains that the complement of the product of all the term is equal to sum of the complement of the individual term or we can say complement of the sum of the term you can see here complement of the sum of the term is equal to product of complement of individual terms right so this is the De Morgan's theorem now we will see the rules of the Boolean algebra so complement rule is nothing but A complement of complement is equal to 1 then these are the rules for AND gate so A AND 0 will get the 0 A AND 1 will get 1 A ANDing with A here we are getting A then A AND A complement is equal to 0 so these are the rules for OR gate and this is De Morgan's rule already we have seen it's A plus B complement is equal to A complement and B complement now here A B complement is equal to A complement or B complement so these rules are useful for solving the example now we will see the example based on Boolean algebra so first example A ORing B and A ORing C so you have to multiply with this A into A plus A into C plus A into B plus B into C now A into A is what? A remaining equation same then take the common term that is A in bracket 1 plus C plus A B plus B C so here 1 plus C in place of that here we can write down 1 remaining equation same then again take common term in between them that is A in bracket 1 plus B plus B into C so again for 1 plus B we are getting 1 and the output is A plus B C second example A B C plus A complement plus A B complement C so A into C take the common term which one is remaining in bracket B plus B complement plus A complement so B plus B complement is equal to 1 then here we can write down A into C plus A complement so here instead of A I am writing double complement this is the complement rule double complement right so here the result is A complement plus C so A double double complement of A is always equal to 1 so here I am writing A complement plus C now third example A plus C into A into D plus A rebar plus A C plus C take the common term here then which is remaining D plus D complement plus A into C plus C now in place of D plus D complement I am writing here 1 yes then take the common term that is A and then you have to write down in bracket A plus C plus C plus C so again you have to simplify A into A plus C plus C use the rule then again you have to multiply A into A plus A into C plus C take the common term C is the common term which is remaining A plus 1 so here we will get A plus C then fourth example A into B into C complement plus A B C complement let us assume that A B C complement is equal to P so this equation becomes P plus P complement so we know that P plus P complement is equal to 1 so output of this equation is 1 now next example Y is equal to A plus B A plus B complement into A complement plus C so you have to first multiply first two brackets then use the rules yes B into B complement is 0 so in place of that I have written here 0 then A 1 plus B complement plus 1 plus A B in second bracket A complement plus C then you have to multiply A into 1 plus B again in second bracket A complement plus C so here we are getting the result A into A complement plus A into C then this is the last example A complement B complement C complement plus A complement B complement C plus A complement C complement so take the common term and C plus C complement is equal to 1 we know that right now again take the common term from this A complement is the common term so B complement plus C complement is remaining so this is your here we can use the De Morgan's theorem yes B complement plus C complement is equal to B C complement so Y is equal to A complement B C complement so in this way with the help of laws of Boolean expression and De Morgan's theorem we can solve the equation we can solve the complicated equations in easy form these are the references of this video lecture thank you