in this video we are going to discuss loss offsets theory so first let us talk about idempotent laws there are two laws one is a union a is equal to a set a union said a is equal to a that is if we take union of a set with itself the result is the set itself and the second idempotent law is a intersection a is equal to a that is if we take intersection of any set with itself the result is the set itself now we are going to prove the first law that is a union a is equal to a so by definition a union a is a set of X such that X belongs to a or X belongs to a basically wherever you see Union offsets we use the word or and for intersection we use the word and these two points must be kept in mind well so there is a union between a and a so we used the word or so a union a is a set of X such that X belongs to a or X belongs to a clearly this there is a repetition of the same statement X belongs to a or X belongs to a that these two states statements are identical so we may write these two repeated statement only once as X cell that X belongs to a and by definition this is the set itself a set of X at that X belongs to me that is the set a so we have proved that a union a is equal to a now let's prove the second idempotent law that has a intersection a is equal to a so by definition a intersection a is the set of X such that X belongs to a intersection then we have to use the word and X belongs to a because in intersection we take the common element so if an element belongs to intersection of the two sets that must belong both of the sets so we use the word end X belongs to a the Fassett end X must also belongs to second set which is also a so we write X belongs to a and X belongs to a again this is the repetition of the same statement so we may write it as X set of X as that X belongs to a again by definition this is the set a so we have proved the second idea important law now the second laws our identity laws identity laws are also two laws first identity law is a union Phi is equal to a and the second identity law is a intersection U is equal to a fie is a null set and capital u is a universal set now we are going to prove the first identity law that is a union Phi is equal to a so by definition a union Phi is a set of X such that X belongs to either of these two sets so we may write X belongs to a for union or X belongs to Phi and if we observe this statement attentively what we may infer from this X belongs to a or X belongs to Phi and as you know Phi is a null set which has no element so X may not belong to Phi so this statement is equivalent to the statement a set of X such that X belongs to a again this is the definition of set a a set of X where X belongs to a that is the set a itself so we have proved that a union Phi is equal to set a itself so we have proved first identity law now we are going to prove the second identity law which is a intersection U is equal to a a intersection universal set u is equal to a by definition a intersection Universal set is a set of X such that X belongs to a end because there is intersection and for intersection we use the word end because if an element belongs to a intersection you that must belong both of the sets so we should say X belongs to a and X must belongs to Universal set you and since Universal set must have all the elements of a and there may be some other elements which are not in a and if an element belongs both of these two says that is belongs to a and belongs to you Universal said that must belongs to a because a as a whole is contained by Universal said so for this statement we may write only a set of X that X belongs to a because if an element belongs to a that must belongs to universal said because universal set contains all the elements of a so this is sufficient to write the set of X such that X belongs to a and again by definition this is the set a itself so we have proved that a intersection U is equal to a for going ahead it is very important to recall one important property which we have discussed in the case of subsets that if two sets are equal then those says must be subsets of each other because a contains all the elements of B and B contains all the elements of a and by the definition of a subset if a contains all the elements of B then be said to be subset of a and if B contains all the elements of a then a is said to be a subset of a so if we need to prove two annua sets equal to each other we have to prove that that these sets are subsets of each other here is a subset of B and B is a subset of a then it is sufficient to prove that both the and we are equal to each other now we move on to the third laws of Destiny which are commutative laws the first commutative law is a union B is equal to B Union a that is in the case of union of two sets the sequence of the stats is not taken into account and the second one is intersection B is equal to B intersection a so similarly as the sequence of sets in the case of a union is not important in the case of intersection to the sequence of the sets doesn't matter so we are going to prove the first commutative law that is a union B is equal to B Union a to prove this law we take an arbitrary element X which belongs to a union B we let X belongs to a union B to the left hand side and by definition it implies that X belongs to a for a union B using our word X belongs to B so X belongs to a union B then X belongs to a or X belongs to B it implies that you know where you know well this statement may also be written as X belongs to B or X belongs to a it implies that and if we write this statement back in the form of Union we may also write it write it as X to be union a because of the word or X belongs to B or that is Union a X belongs to a so X belongs to B Union a so we have taken arbitrary element X belongs to a union B and we have proved that X belongs to B Union a that is if X belongs to a union B that must also belongs to B Union a so from this we infer that a union B is a subset of B Union a let it be equation number one now let Y belongs to B Union a now we take an arbitrary element from the right hand side that is B Union a and again as we have done for a union B this statement may be written as X belongs to B or sorry Y belongs to B or Y belongs to a it implies that the same a statement may also be written as Y belongs to a or Y belongs to B it implies that if we write this statement collectively we may write it as Y belongs to a union B because of the word or we use Union operation so we have taken an arbitrary element Y from B Union a and we have proved that Y must also belongs to a union B so we infer from this that B Union a is a subset of a union B I repeat if Y belongs to B Union a it must also belongs to a union B that means a union B must contain all the elements of B Union a so we may write it as B Union a is a subset of a union B let it be equation number one so if we compare the question number one and to a union B is a subset of B Union a and B Union a is a subset of a union B so it is only possible when a union B and B Union a are equal sets so we may write it as from 1 & 2 a union B is equal to B Union a that's it we have proved first commutative law now the second commutative law you better try yourself now the next laws are associative laws the first associative law is a union B grouped in a parenthesis Union C is equal to a union parenthesis B Union C and the second associative law is parenthesis a intersection B intersection C is equal to a dissection parenthesis B intersection C so now we are going to prove the first associative law so let X belongs to a union B Union C now we have taken an arbitrary element X belonging to the left hand side a union B Union C it implies that by definition we may write it as X belongs to a union B or X belongs to C we have used we have used or for this Union operation so next X belongs to a union B may also be written as as X belongs to a or X belongs to B and the next this or as it is this X belongs to C as it is so in the next step we may write it as X belongs to a as it is or as it is and we combine these two statements as X belongs to B Union C so it implies that now we may combine these two statements for the word or so it implies X belongs to a union for or B Union C so taking an arbitrary element from a union B paranthesis Union C we have proved that X belongs to a Union parenthesis B Union C that means if X belongs to a parenthesis a union B Union C it must belong it must belong to a union parenthesis B Union C so it implies that this set is a subset of this earth so a union B Union C is a subset of a union B Union C let it be question number one now we take another arbitrary element Y belonging to the right hand side this time a a union B Union C so again we may write it as Y belongs to a or Y belongs to parenthesis B Union C it implies that Y belongs to a or as it is and now expand this statement as Y belongs to B or Y belongs to C now in the next step we may combine these two statements as Y belongs to a union B and or as it is and Y belongs to C as it is in the next step we may combine these two statements for the word or Y belongs to a union B for all humanity Union Y so we have taken by is an arbitrary element of a union B Union C and it implies that Y belongs to a union B Union C it means if I belongs to this estate this said it must also belongs to this set so we may infer from this as a union B Union C that is this set is a subset of this set a union B Union C so let it be question number two now if we observe equation number one and equation number two a union B Union C is a subset of a Union parenthesis B Union C and from to a Union parenthesis B Union C is a subset of parenthesis a union B Union C so this is only possible when these two sets are equal to each other from 1 & 2 we get a parenthesis a union B Union C is equal to a union parenthesis B Union C so in this way we prove the first associative law now the second associative law you should try yourself now the next laws to be discussed are distributive laws the first distributive law is a Union parenthesis B intersection C is equal to parenthesis a union B intersection parenthesis a Union C and the second distributive law is a intersection parenthesis B Union C is equal to parenthesis a intersection B Union parenthesis a intersection C now we are going to prove the first distributive law that is a Union parenthesis intersection c is equal to parentheses a union B intersection parentheses a union C so let X belongs to a Union parenthesis B intersection C you have taken an arbitrary element X from a union B intersection C it implies that X belongs to a and for this Union we may take the word or X belongs to parenthesis B intersection C it implies that X belongs to a as it is or as it is and we may write this statement as X belongs to B and for this intersection we may use an word X belongs to C X belongs to B intersection C implies that as belongs to B and X belongs to C it implies that now we may write this statement as in two parts X belongs to a or X belongs to B and X belongs to a or X belongs to C now it implies that this statement may be written as X belongs to a union B the parenthesis because of our word we use the operation Union and this end as it is and this these two statements may also be written X belongs to a union C because of this word or it implies that now these two statements are connected with and so these two statements may be written as X belongs to parentheses a union B intersection four end parenthesis a union C so we have taken an arbitrary element X from a Union parenthesis B intersection C and we have proved that X belongs to a union B intersection a union C it means a union B intersection C is a subset of parenthesis a union B intersection a parenthesis a union C let it be a question of a 1 similarly we may prove that a union B intersection a union C is a subset of a union B intersection C taking Y as an arbitrary element from right hand side as we have done in the case of associative laws and commutative laws so from these two results if this set is a subset of this set and this set is a subset of this set it implies that both the sets must be equal so from 1 & 2 let me write a union B intersection C is equal to a union B intersection a union C that's fruit so we have proved the first distributive law in this way and now you should try the second distributive law yourself now the next laws to be discussed are B Morgan's laws the first D Morgan law is a union B complement is equal to a complement intersection B complement that is if there is Union inside the parenthesis whose complement we are taking then on the right hand side there should be intersection between their individual complements and the second one is parenthesis a intersection B complement is equal to a complement Union B complement again if inside the bracket there is intersection operation whose complement we have take we have to take that is equal to the union of the individual complements of the sets so these are called the Morgan's law and we are going to prove the first de Morgan's law Let X belongs to a union B complement it implies that X does not belong to a union B as you know the complement of a set is a set which contains all the elements of universal set which are not in the set itself whose complement we are taking so if X belongs to a Khan a union B complement that may not be in the set a union B so if X belongs to a union B complement then X must not belong to a union B it applies that X does not belong to a and for the symbol is does not belong to for union operation we use the word end and for the operation intersection we use the word or this process is reversed as for the symbol belongs to Union means or an intersection means end for does not belong to Union means end and intersection means or so X does not belong to a union B we may write it as X belongs to a and X does not belongs to B it implies that now as you see if X does not belongs to does not belong to a X must belongs to a compliment and if X is not an element of B X must be an element of B complement so X belongs to B complement now because these symbols are belongs to now n stands for u intersection so implies that X belongs to a complement intersection B complement so if X belongs to parenthesis a union B complement then it must belong to a complement intersection B complement it implies that a union the compliment is a subset of a complement intersection B complement let it be question number one and now in the same way we may prove that a complement intersection B complement is a subset of a union B complement and we'll make it as equation number two and then comparing the equation 1 and 2 from equation from one and two a union B complement is equal to a complement intersection B complement because it is only possible when these two sets are equal to each other that this is a subset of this and this is a subset of this so in this way we have proved the first de Morgan's law and the second de Morgan's law you should try yourself