 Hello and welcome to the session. In this session we discussed the following question that says, let e be the set containing the elements 1 and 2 show that the system containing the set pe union intersection and complement is a Boolean algebra where this pe is the power set of the set e. A system consisting of the set b with operations union intersection complement is a Boolean algebra the following laws satisfied. The first law is the closure law then we have the commutative law then next is the associative law then we have the distributive law then identity law and next we have the inverse law. So if all these laws are satisfied then the given system is a Boolean algebra. This is the key idea that we use in this question. Let us now proceed with the solution we are given a set e with elements 1 and 2 then this pe is the power set of the set e so it would contain the elements 5, single term 1, single term 2 and the set with elements 1 and 2 this is the power set of e. We have to show that this is some consisting of the set pe that is the power set of e then union intersection complement is a Boolean algebra so this means we will consider the operations of union intersection and complement. First we will make the tables corresponding to the operations of union intersection and complement. This first table shows the operation of union these are the elements of the power set of e. Now 5 union 5 would be 5, single term 1 union 5 would be single term 1, single term 2 union 5 would be single term 2 set with elements 1 and 2 union 5 would be set with elements 1 and 2 in the same way we will fill the rest of this table. So this is the table showing the operation of union now consider this table showing the operation of intersection now 5 intersection 5 is 5, single term 1 intersection 5 is 5, single term 2 intersection 5 is 5, set with elements 1 and 2 intersection 5 is 5. In the same way we will fill the rest of this table so this is the table showing the operation of intersection this third table shows the operation of complement. Now complement of 5 would be the set with elements 1, 2 then complement of single term 1 is single term 2, complement of single term 2 is single term 1 and complement of the set with elements 1 and 2 is 5. Now we will consider the laws one by one first we have the closure law. Now consider the element single term 1 which belongs to the power set of e also consider single term 2 which belongs to the power set of e now single term 1 union single term 2 is equal to we observe this from table 1 single term 1 union single term 2 is the set with elements 1, 2 this also belongs to the power set of e now consider single term 1 intersection single term 2 would be equal to we observe this from the table 2 and this also belongs to the power set of e so as single term 1 and single term 2 belongs to the power set of e so the union also belongs to the power set of e and the intersection also belongs to the power set e. Thus we can say for any two elements belonging to the power set of e the union and intersection also belongs to the power set of e the closure law is this file. Next we consider the commutative law in this we will check if the operations union and intersection are commutative or not or consider single term 1 single term 2 belonging to the power set of e now single term 1 union single term 2 would be equal to from the table 1 it would be equal to set with elements 1, 2 now single term 2 union single term 1 would be equal to the set with elements 1, 2 this is again from the table 1 these two are equal that is single term 1 union single term 2 is equal to single term 2 union single term 1 now consider single term 1 intersection single term 2 is equal to 5 this is from the table 2 now single term 2 intersection single term 1 is equal to again 5 so these two are also equal that is we have single term 1 intersection single term 2 is equal to single term 2 intersection single term 1. Therefore we can say for any elements a and b belonging to the power set of e a union b is equal to b union a a intersection b is equal to b intersection a and thus we can say that the commutative law next we have the associated law consider three elements single term 1 single term 2 which belong to the power set of e first we consider 5 union single term 1 union single term 2 the whole now this is equal to 5 union now from the table 1 single term 1 union single term 2 is the set with elements 1, 2 now again from the table 1 5 union set with elements 1, 2 is equal to the set with elements 1, 2 now consider 5 union single term 1 the whole union single term 2 now from table 1 we have 5 union single term 1 is single term 1 now single term 1 union single term 2 would be equal to the set with elements 1, 2 so these two are equal therefore we say that 5 union single term 1 union single term 2 is equal to 5 union single term 1 the whole union single term 2 now let's see for intersection 5 intersection single term 1 intersection single term 2 now this would be equal to 5 intersection now from table 2 single term 1 intersection single term 2 is equal to 5 so now 5 intersection 5 from table 2 we have this is equal to 5 in the same way consider 5 intersection single term 1 the whole intersection single term 2 now this would be equal to from table 2 we have 5 intersection single term 1 is 5 so now 5 intersection single term 2 from table 2 is equal to 5 again so these two are equal that is we have 5 intersection single term 1 intersection single term 2 the whole is equal to 5 intersection single term 1 the whole intersection single term 2 therefore we can say for any elements say A B C which belongs to the power set of E we have A union B union C the whole is equal to A union B the whole union C also A intersection B intersection C the whole is equal to A intersection B the whole intersection C and from this we can say that the associative law is satisfied next we consider the distributive law and let's see if this is satisfied or not again we consider the elements say 5 single term 1 single term 2 now consider 5 union single term 1 intersection single term 2 the whole now this is equal to 5 union now from table 2 single term 1 intersection single term 2 is 5 from table 1 5 union 5 is 5 so this is equal to 5 next we consider 5 union single term 1 the whole intersection 5 union single term 2 the whole now from table 1 5 union single term 1 is single term 1 now this intersection 5 union single term 2 is single term 2 now from table 2 single term 1 intersection single term 2 is 5 so this is equal to 5 now as these two are equal therefore we say that 5 union single term 1 intersection single term 2 the whole is equal to 5 union single term 1 the whole intersection 5 union single term 2 the whole next consider next consider 5 intersection single term 1 union single term 2 the whole now this is equal to 5 intersection from table 1 we have single term 1 union single term 2 is the set with elements 1 2 and from table 2 5 intersection the set with elements 1 and 2 is equal to 5 now consider 5 intersection single term 1 the whole union 5 intersection single term 2 the whole now from table 2 5 intersection single term 1 is 5 now 5 union 5 intersection single term 2 is again 5 now 5 union 5 is 5 so these two are equal therefore we observe that 5 intersection single term 1 union single term 2 the whole is equal to 5 intersection single term 1 the whole union 5 intersection single term 2 the whole thus we can say that if A B and C are three elements which belong to the power set of E then A intersection B union C the whole is equal to A intersection B the whole union A intersection C the whole and A union B intersection C the whole is equal to A union B the whole intersection A union C the whole and thus we now say that the distributed law the identity law let's see if this is satisfied or not consider element A belonging to the power set of E now from the table 1 we observe that any element A which belongs to the power set of E union 5 is equal to A like 5 union 5 is 5 single term 1 union 5 is single term 1 single term 2 union 5 is single term 2 set with elements 1 and 2 union 5 is set with elements 1 and 2 and this would be also equal to 5 union A and therefore we can say that this 5 is the identity element for the operation of union now from the table 2 we observe that the element A intersection the set with elements 1 2 is equal to the set A which is same as the set with elements 1 2 intersection the element A so if we take A as 5 then 5 intersection the set with elements 1 2 is 5 itself single term 1 intersection the set with elements 1 2 is single term 1 itself in the same way we can consider the other elements also so here the set with elements 1 2 is the identity element for the operation of intersection now finally we consider the inverse law now consider 5 belonging to the power set of E now complement to 5 is he set with elements 1 2 this is from the table 3 now consider 5 union 5 complement that is 5 union set with elements 1 2 for the table 1 we observe that this is equal to the set with elements 1 2 that is 5 union 5 complement is equal to the identity element for intersection and this would also be equal to 5 complement union 5 so we can say any element A of the power set E there exists its inverse A complement which belongs to the power set E such that A union A complement is equal to the set with elements 1 2 in this case which is the identity element for intersection and this is same as A complement union A now consider 5 in the section 5 complement that is 5 in the section the set with elements 1 2 which is the complement of 5 the from table 2 we observe that this would be equal to 5 and this would also be equal to 5 complement intersection 5 now where this 5 is the identity element for the operation of union so we can say for any element A of the power set E there exists A complement which belongs to the power set E such that A intersection A complement is equal to the identity element for the operation union which is same as A complement intersection A so you can now say that the inverse law this is also called the complement law and this is satisfied so hence all the laws are satisfied so we can say that the system consisting of the power set E union intersection and complement is a Boolean algebra so this completes the question hope we have understood the solution of this question