 Welcome to the session on the topic set theory under the course discrete mathematical structures at second year of information technology engineering semester one. At the end of this session students will be able to apply principles of set theory to solve problems involving sets. So earlier we have learnt the concepts of set theory in terms of the different types of sets along with its notations. Then we have learnt about set operations namely union, intersection, compliment so on and so forth. So in this session we are going to solve problems based on the operations on sets. So here comes the first example if A1 is equal to 1, 2, 3, A2 is equal to set of 1, 2, 3 and A3 is equal to 1, 2, 3 then show that A1, A2 and A3 are mutually disjoint. Now in this type of question first of all we know that what are mutually disjoint sets. We have seen a property by which you can say that the given sets are mutually disjoint and the property says if two sets let us say A and B are mutually disjoint then they satisfy the condition A intersection B that is A intersection B must be equal to Psi the empty set. So we will apply the same property here and try to judge whether the given sets are mutually disjoint. So if you observe if I try to take A1 intersection A2 okay and we observe that it results into there are no common elements between A1 and A2 so it is equal to Psi. Similarly if I take the intersection between other two pairs like A2 intersection A3 you will observe that it also results into Psi similarly the third pair A1 intersection A3 also results into intersection giving Psi as the answer. So we have shown that all three intersections results into an empty set Psi hence proved that is the given sets are mutually disjoint. So that is how we show the given any number of sets and you have been asked to prove that they are mutually disjoint then you have to simply take intersection between every possible pair of the sets and show that it is equal to Psi the empty set. We will just move on to the second example show that A subset of B is equivalent to A intersection B is equal to A. Now what does this mean we have been given the condition that A is a subset of B and then we have to prove that it is equivalent to saying that if you take intersection between A and B it must be same as A itself okay so to prove such identities or the given properties we will write the solution in this way okay we know that or we may note that for any x where x is an element from set A we write it as x belongs to A conditional x belongs to B is equivalent to saying x belongs to A and that is conjunction with x belongs to B by conditional x belongs to A okay and it simply follows from so we say which follows from the identity that we use is P conditional Q is equivalent to P conjunction Q by conditional okay. Now A subset of B is equivalent to all those elements x such that x belongs to A conditional x belongs to B while A intersection B is equal to A is equivalent to saying that set of elements x such that x belongs to A and x belongs to B by conditional x belongs to A okay so if you observe the solution once again we say for any x you may note down the given property as if x belongs to A it simply implies x belongs to B and it is equivalent saying that x belongs to A and x belongs to B is by conditional x belongs to A and we have assumed this from the property that P conditional Q is equivalent to P and Q by conditional P and hence we have shown that if A is a subset of B it is equivalently if you take intersection of A and B it simply results into the same set A itself okay. So with this we move on to the next example. Example 3 is S equal to A, B, P, Q and Q is equal to A, P, T then find S union Q and S intersection Q so we can directly apply these operations and find out that the solution is S union Q results into the common elements as well as remaining elements from the set so we get A, B, P, Q and T so it is simply the common as well as uncommon elements coming from S and Q similarly S intersection Q results into the common elements and that is only A, P so this is the solution. So this way we have solved certain examples and you have got an idea how to go about solving problems based on sets so here is an assignment for all of you given A equal to 2, 3, 4, B equal to 1, 2 and C equal to 4, 5, 6 find A plus B, B plus C, A plus B plus C and A plus B plus B plus C where plus denotes the Boolean sum operator. So you have to apply the Boolean sum and find out the answers for the given sets. Thank you.