The distributive property of product operator holds for other set operators like union, intersection and difference operators. We write equations involving distribution of product operator for each of other operators as :
A
×
B
∪
C
=
A
×
B
∪
A
×
C
A
×
B
∪
C
=
A
×
B
∪
A
×
C
A
×
B
∩
C
=
A
×
B
∩
A
×
C
A
×
B
∩
C
=
A
×
B
∩
A
×
C
A
×
B
−
C
=
A
×
B
−
A
×
C
A
×
B
−
C
=
A
×
B
−
A
×
C
Here, sets “A”,”B” and “C” are non-empty sets. In order to ascertain distributive property product operator over other set operators we need to check validity of the equations given above.
We can check these relations proceeding from the defining statements. For the time being, we reason that sequence of operation on either side of the equation does not affect the “order” in the “ordered pair”. Hence, distributive property should hold for product operator over three named operators. Let us check this with an example :
A
=
{
a
,
b
}
,
B
=
{
1,2
}
a
n
d
C
=
{
2,3
}
A
=
{
a
,
b
}
,
B
=
{
1,2
}
a
n
d
C
=
{
2,3
}
1: For distribution over union operator
⇒
LHS
=
A
×
B
∪
C
=
{
a
,
b
}
×
{
1,2,3
}
⇒
LHS
=
A
×
B
∪
C
=
{
a
,
b
}
×
{
1,2,3
}
⇒
LHS
=
{
a
,
1
,
a
,
2
,
a
,
3
,
b
,
1
,
b
,
2
,
b
,
3
}
⇒
LHS
=
{
a
,
1
,
a
,
2
,
a
,
3
,
b
,
1
,
b
,
2
,
b
,
3
}
Similarly,
⇒
RHS
=
A
×
B
∪
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
∪
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
A
×
B
∪
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
∪
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
{
a
,
1
,
a
,
2
,
a
,
3
,
b
,
1
,
b
,
2
,
b
,
3
}
⇒
RHS
=
{
a
,
1
,
a
,
2
,
a
,
3
,
b
,
1
,
b
,
2
,
b
,
3
}
Hence,
⇒
A
×
B
∪
C
=
A
×
B
∪
A
×
C
⇒
A
×
B
∪
C
=
A
×
B
∪
A
×
C
2: For distribution over intersection operator
⇒
LHS
=
A
×
B
∩
C
=
{
a
,
b
}
×
{
2
}
⇒
LHS
=
A
×
B
∩
C
=
{
a
,
b
}
×
{
2
}
⇒
LHS
=
{
a
,
2
,
b
,
2
}
⇒
LHS
=
{
a
,
2
,
b
,
2
}
Similarly,
⇒
RHS
=
A
×
B
∩
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
∩
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
A
×
B
∩
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
∩
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
{
a
,
2
,
b
,
2
}
⇒
RHS
=
{
a
,
2
,
b
,
2
}
Hence,
⇒
A
×
B
∩
C
=
A
×
B
∩
A
×
C
⇒
A
×
B
∩
C
=
A
×
B
∩
A
×
C
3: For distribution over difference operator
⇒
LHS
=
A
×
B
−
C
=
{
a
,
b
}
×
{
1
}
⇒
LHS
=
A
×
B
−
C
=
{
a
,
b
}
×
{
1
}
⇒
LHS
=
{
a
,
1
,
b
,
1
}
⇒
LHS
=
{
a
,
1
,
b
,
1
}
Similarly,
⇒
RHS
=
A
×
B
−
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
−
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
A
×
B
−
A
×
C
=
{
a
,
1
,
a
,
2
,
b
,
1
,
b
,
2
}
−
{
a
,
2
,
a
,
3
,
b
,
2
,
b
,
3
}
⇒
RHS
=
{
a
,
1
,
b
,
1
}
⇒
RHS
=
{
a
,
1
,
b
,
1
}
Hence,
⇒
A
×
B
−
C
=
A
×
B
−
A
×
C
⇒
A
×
B
−
C
=
A
×
B
−
A
×
C
Let us consider an arbitrary ordered pair (x,y), which belongs to Cartesian product set “
A
×
B
∪
C
A
×
B
∪
C
”. Then,
⇒
x
,
y
∈
A
×
B
∪
C
⇒
x
,
y
∈
A
×
B
∪
C
By the definition of product of two sets,
⇒
x
∈
A
a
n
d
y
∈
B
∪
C
⇒
x
∈
A
a
n
d
y
∈
B
∪
C
By the definition of union of two sets,
⇒
x
∈
A
a
n
d
y
∈
B
o
r
y
∈
C
⇒
x
∈
A
a
n
d
y
∈
B
o
r
y
∈
C
⇒
x
∈
A
a
n
d
y
∈
B
o
r
x
∈
A
a
n
d
y
∈
C
⇒
x
∈
A
a
n
d
y
∈
B
o
r
x
∈
A
a
n
d
y
∈
C
⇒
x
,
y
∈
A
×
B
o
r
x
,
y
∈
A
×
C
⇒
x
,
y
∈
A
×
B
o
r
x
,
y
∈
A
×
C
By the definition of union of two sets,
⇒
x
,
y
∈
A
×
B
∪
A
×
C
⇒
x
,
y
∈
A
×
B
∪
A
×
C
But, we had started with "
A
×
B
∪
C
A
×
B
∪
C
" and used definitions to show that ordered pair “(x,y)” belongs to another set. It means that the other set consists of the elements of the first set – at the least. Thus,
⇒
A
×
B
∪
C
⊂
A
×
B
∪
A
×
C
⇒
A
×
B
∪
C
⊂
A
×
B
∪
A
×
C
Similarly, we can start with "
A
×
B
∪
A
×
C
A
×
B
∪
A
×
C
" and reach the conclusion that :
⇒
A
×
B
∪
A
×
C
⊂
A
×
B
∪
C
⇒
A
×
B
∪
A
×
C
⊂
A
×
B
∪
C
If sets are subsets of each other, then they are equal. Hence,
⇒
A
×
B
∪
C
=
A
×
B
∪
A
×
C
⇒
A
×
B
∪
C
=
A
×
B
∪
A
×
C
Proceeding in the same manner, we can also prove distribution of product operator over intersection and difference operators,
A
×
B
∩
C
=
A
×
B
∩
A
×
C
A
×
B
∩
C
=
A
×
B
∩
A
×
C
A
×
B
−
C
=
A
×
B
−
A
×
C
A
×
B
−
C
=
A
×
B
−
A
×
C