Prove that X*(Y-Z) = (X*Y)-(X*Z)

how to proof this X*(Y-Z) = (X*Y)-(X*Z) is true or not without taking examples,can we draw venn diagram for this ?

Mahesh Kumar @maheshkumars 2 Oct 2015 04:55 pm


PS:INTERSECTION(^) in set,which is AND(*) in Algebra.And an number represent a region.             

Kuldeep Narwariya @kuldeepnarwa 2 Oct 2015 04:43 pm

here you are considering * as an intersection but in my question i am declare * as a Cartesian during the designing of venn diagram can we consider cartesian product as an it right way to solve this.....?

Mahesh Kumar @maheshkumars 2 Oct 2015 04:54 pm

yes.In set A.B represents both A and B should present,which is nothing but intersection.

Pritam Prasun @pritam 2 Oct 2015 05:13 pm


Cartesian Product and Intersection are two entirely different thing and Cartesian Product can not be represented by Venn Diagram. Please see the following link mentioned by Ranita.

Mahesh Kumar @maheshkumars 2 Oct 2015 05:47 pm

@pritam:Ya.But Cartesian product is commonly applied for sets(or an element can contain more than one values) .And here i assumed it as simple Algebra(a variable can contain a unique value at a time),so i think Cartesian product will be equivalent  to the product(multiplication).So product(multiplication) can be represented using venn diagrams as intersection. 

Virtual GATE @virtualgate 2 Oct 2015 03:52 pm

X.(Y−Z) can be drawn as follows:
 <dot>  =
Similarly, (X.Y)−(X.Z) can be drawn like this:
<minus> =

Hence, we can say that X.(Y−Z) = (X.Y)−(X.Z)

Kuldeep Narwariya @kuldeepnarwa 2 Oct 2015 04:17 pm

you are taking * as an intersection but it is the Cartesian product of  X and i thing this is not the right way to proof it.

Ranita Biswas @ranita 2 Oct 2015 04:45 pm

I don't think Cartesian product can be represented using Venn diagrams. You can try the methods proposed in this link to prove the given expression without using examples.

Piyush Balwani @piyushbalwani 3 Oct 2015 01:13 am

I can tell you the formal proof,but without venn diagram.

(x,y)is all elements that belongs to A×(B∪C)
Obviously, x belongs to A and y belongs to (B∪C)
x∈A and y∈(B∪C)
x∈A and (y∈B or y∈C)
(x∈A and y∈B) or ( x∈A and y∈C)
(x,y)∈A×B or(x,y)∈A×C    (Since, A×B contains all element of type (A,B))
Since,(x,y) belongs to (A×B)∪(A×C),means atleast (A×B)∪(A×C) contains (x,y) which is all elements that belongs to A×(B∪C)
therefore, A×(B∪C)⊂(A×B)∪(A×C)

Similarily take (x,y): all elements that belongs to (A×B)∪(A×C) and derive in same way (x,y)∈A×(B∪C)
and conclude (A×B)∪(A×C) ⊂ A×(B∪C).

Hence,A×(B∪C)=(A×B)∪(A×C).(because sets are subsets of each other, then they are equal.)