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Given,

R = {(x, x+5) : x {0, 1, 2, 3, 4, 5} 

∴ R = {(0,0+5), (1,1+5), (2,2+5), (3,3+5), (4,4+5), (5,5+5)} 

⇒ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)} 

So, 

Domain of relation R = {0, 1, 2, 3, 4, 5} 

Range of relation R = {5, 6, 7, 8, 9, 10}

Here,

A = {1,2,4} ,B = {2,4,5} ,C = {2,5} 

So, 

A/C={1,2,4}/{2,5} 

= {1,4} 

Also,

B/C = {2,4,5}/{2,5} 

= {4} 

Now, 

(A /B)×(B/C) = {(x,y) : x ϵ (A/C) and y ϵ (B/C)} 

= {(1,4), (4,4)}

Given,

(a, b) R (c, d) a + d = b + c for all (a, b), (c, d)N x N 

i. (a, b) R (a, b) 

⇒ a + b = b + a for all (a, b) N x N 

∴ (a, b) R (a, b) for all (a, b) N x N 

ii. (a, b) R (c, d) 

⇒ a + d = b + c 

⇒ c + b = d + a 

⇒ (c, d) R (a, b) for all (c, d), (a, b) N x N 

iii. (a, b) R (c, d) and (c, d) R (e, f) 

a + d = b + c and c + f = d + e 

⇒ a + d + c + f = b + c + d + e 

⇒ a + f = b + c + d + e – c – d 

⇒ a + f = b + e

 ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) N × N

We know,

n(A×B) = n(A)×n(B) 

Similarly, 

n(A×B×C) = n(A) × n(B) × n(C) 

Here, 

n(A) = 3 and n(B) = 4 

n(A×A×C) = n(A)×n(A)×n(B) 

= 3×3×4 

= 36

Here, 

R = { (x, y) :x ,yϵ W, 2x+y = 8} 

= {(0,8),(1,6),(2,4),(3,2),(4,0)} 

Now we know, 

Domain is the set which consist all first elements of ordered pairs in relation R. 

So, 

Domain(R)= {0,1,2,3,4} 

Also we know, 

Range is the set which consist all second elements of ordered pairs in relation R. 

So,

Range(R) = {0,2,4,6,8}

Here, 

A×B ={(x,1) ,(y,2) ,(z,1)} and n(A) = 3, n(B) = 2 

We know,

A×B = {(x,y) : xϵA and yϵ B} 

∴ A = {x,y,z} and B = {1,2}

(i) If A and B are two nonempty sets, then any subset of the set (A × B) is said to a relation R from set A to set B. 

That means, if R be a relation from A to B then R ⊆ (A × B). 

Therefore, (x, y)∈R ⇒ (x, y)∈(A × B) 

That means x is in relation to y. Or we can write xRy. 

(ii) Let R be a relation from A to B. Then, the set containing all the first elements of the ordered pairs belonging to R is called Domain. 

For the relation R, Dom(R) = {x: (x, y)∈R} 

And the set containing all the second elements of the ordered pair belonging to R is called Range. 

For the relation R, Range(R) = {y: (x, y)∈R}

Given: A = {a, b, c, d,}, B = {c, d, e} and C = {d, e, f, g} 

(i) Need to prove: A × (B ∩ C) = (A × B) ∩ (A × C) 

Left hand side, 

(B ∩ C) = {d, e} 

⇒ A × (B ∩ C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} 

Right hand side,

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(A × C) = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)} 

Now, 

(A × B) ∩ (A × C) = {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} 

Here, right hand side and left hand side are equal. 

That means, A × (B ∩ C) = (A × B) ∩ (A × C) [Proved] 

(ii) Need to prove: A × (B – C) = (A × B) – (A × C) 

Left hand side, 

(B – C) = {c} 

⇒ A × (B – C) = {(a, c), (b, c), (c, c), (d, c)} 

Right hand side, 

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(A × C) = {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)} 

Therefore, (A × B) – (A × C) = {(a, c), (b, c), (c, c), (d, c)} 

Here, right hand side and left hand side are equal. 

That means, A × (B – C) = (A × B) – (A × C) [Proved]

(iii) Need to prove: 

(A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) 

Left hand side, 

(A × B) = {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)} 

(B × A) = {(c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, a), (e, b), (e, c), (e, d)} 

Now, (A × B) ∩ (B × A) = {(c, c), (c, d), (d, c), (d, d)} 

Right hand side, 

(A ∩ B) = {c, d} 

So, (A ∩ B) × (A ∩ B) = {(c, c), (c, d), (d, c), (d, d)} 

Here, right hand side and left hand side are equal. 

That means, (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) [Proved]

Given: 

A = {x ϵ N: x ≤ 3} 

Here, N denotes the set of natural numbers. 

∴ A = {1, 2, 3} 

[∵ It is given that the value of x is less than 3 and natural numbers which are less than 3 are 1 and 2] 

and B = {x ϵ W: x < 2} 

Here, W denotes the set of whole numbers (non – negative integers). 

∴ B = {0, 1} 

[∵ It is given that x < 2 and the whole numbers which are less than 2 are 0 and 1] 

So, A × B = {1, 2, 3} × {0, 1} - 

[By the definition of equality of ordered pairs .i.e. the corresponding first elements are equal and the second elements are also equal, but here the pair (1, 0) is not equal to the pair (0, 1)]

(i) Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3}

L. H. S = A × (B ⋃ C) 

By the definition of the union of two sets, (B ⋃ C) = {2, 3, 4} 

= {1, 3, 5} × {2, 3, 4} 

Now, by the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

= {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)} 

R. H. S = (A × B) ⋃ (A × C) 

Now, A × B = {1, 3, 5} × {3, 4} 

= {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} 

and A × C = {1, 3, 5} × {2, 3} 

= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)} 

Now, we have to find (A × B) ⋃ (A × C) 

So, by the definition of the union of two sets, 

(A × B) ⋃ (A × C) = {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)} 

= L. H. S 

∴ L. H. S = R. H. S is verified 

(ii) Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3} 

L. H. S = A × (B ⋂ C) 

By the definition of the intersection of two sets, (B ⋂ C) = {3} 

= {1, 3, 5} × {3} 

Now, by the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q}

= {(1, 3), (3, 3), (5, 3)}

R. H. S = (A × B) ⋂ (A × C) 

Now, A × B = {1, 3, 5} × {3, 4}

= {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} 

and A × C = {1, 3, 5} × {2, 3} 

= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)} 

Now, we have to find (A × B) ⋂ (A × C) 

So, by the definition of the intersection of two sets, 

(A × B) ⋂ (A × C) = {(1, 3), (3, 3), (5, 3)} 

= L. H. S 

∴ L. H. S = R. H. S is verified

Given: A, B and C three sets are given. 

Need to prove: A × (B – C) = (A × B) – (A × C) 

Let us consider, (x, y)∈A × (B – C) 

⇒ x∈A and y∈(B – C ) 

⇒ x∈A and (y∈B and y ∉ C) 

⇒ (x∈A and y∈B) and (x∈A and y ∉ C)

⇒ (x, y)∈(A × B) and (x, y) ∉ (A × C) 

⇒ (x, y)∈(A × B) – (A × C) 

From this we can conclude that,

⇒ A × (B – C) ⊆ (A × B) – (A × C) ...... (1)

Let us consider again, (a, b)∈(A × B) – (A × C) 

⇒ (a, b)∈(A × B) and (a, b) ∉ (A × C) 

⇒ (a∈A and b ∈B) and (a∈A and b ∉ C)

⇒ a∈A and (b∈B and b ∉ C) 

⇒ a∈A and b∈(B – C) 

⇒ (a, b)∈A × (B ∪ C)

From this, we can conclude that, 

⇒ (A × B) – (A × C) ⊆ A × (B – C) ......... (2)

Now by the definition of set we can say that, from (1) and (2), 

A × (B – C) = (A × B) – (A × C) [Proved]

B. reflexive and symmetric

∵ According to the condition |x – y| ≤ 1

⇒ R={(1,1),(2,1), (1,2), (2,2), … (n,n), (n+1,n), (n,n+1) …∞ }

⇒ (a,a) ∈ R → Reflexive

⇒ (a,b) ∈ R and (b,a) ∈ R →Symmetric

(c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)} 

(d) R = {(2, 3), (3, 5), (2, 5)}.

R and S are two symmetric relations on set A

(i) To prove: R ⋂ S is symmetric

Symmetric: For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R ⋂ S

⇒ (a, b) ∈ R and (a, b) ∈ S

⇒ (b, a) ∈ R and (b, a) ∈ S

[∴ R and S are symmetric]

⇒ (b, a) ∈ R ⋂ S

⇒ R ⋂ S is symmetric

To prove: R ⋃ S is symmetric

Symmetric: For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let (a, b) ∈ R ⋃ S

⇒ (a, b) ∈ R or (a, b) ∈ S

⇒ (b, a) ∈ R or (b, a) ∈ S

[∴ R and S are symmetric]

⇒ (b, a) ∈ R ⋃ S

⇒ R ⋃ S is symmetric

(ii) R and S are two relations on a such that R is reflexive.

To prove : R ⋃ S is reflexive

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Suppose R ⋃ S is not reflexive.

This means that there is a ∈ R ⋃ S such that (a, a) ∉ R ⋃ S

Since a ∈ R ⋃ S,

∴ a ∈ R or a ∈ S

If a ∈ R, then (a, a) ∈ R

[∵ R is reflexive]

⇒ (a, a) ∈ R ⋃ S

Hence, R ⋃ S is reflexive

Given x and y are integers, i.e x,y ∈ Z.

∴ Domain of R = Set of all first elements in the relation.

= Values of ‘x’ which are in the relation.

= Z (Integers)

Range of R=Set of all second elements in the relation.

=Values of ‘y’ which are in the relation.

= Z(Integers)

Since, x²+ y² ≤ 4 and x,y are integers;

⇒ R= {(0,0), (1,0), (0,1), (-1,0), (0,-1), (1,1), (-1,-1), (-1,1),

(1,-1), (0,2), (0,-2), (2,0), (-2,0)}

⇒ Domain of R= {-2,-1,0,1,2}

Given as

A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

Let us find the (A × B) ∩ (B × C)

(A × B) = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(B × C) = {3, 4} × {4, 5, 6}

= {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

Hence (A × B) ∩ (B × C) = {(3, 4)}

Given as

A = {1, 2, 3} and B = {2, 4}

Then, let us find the A × B, B × A, A × A, (A × B) ∩ (B × A)

A × B = {1, 2, 3} × {2, 4}

= {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}

B × A = {2, 4} × {1, 2, 3}

= {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

A × A = {1, 2, 3} × {1, 2, 3}

= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

B × B = {2, 4} × {2, 4}

= {(2, 2), (2, 4), (4, 2), (4, 4)}

The intersection of two sets represents common elements of both the sets

Therefore,

(A × B) ∩ (B × A) = {(2, 2)}

Given, 

A = {1, 2} 

To find : A × A × A 

Firstly, 

We will find Cartesian product of A with A. 

A × A = {(1, 1), (1, 2), (2, 1), (2, 2)} 

Now, 

Cartesian product of A x A with A.

∴ A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

Given,

A = {1, 2} and B = {1, 3} 

To find: A × B, B × A 

A × B = {(1, 1), (1, 3), (2, 1), (2, 3)} 

B × A = {(1, 1), (1, 2), (3, 1), (3, 2)}

Given A = {1, 2, 3, 4, 5, 6, 7} and R = {(a, b) : both a and b are either odd or even number}

Therefore,

R = {(1, 1), (1, 3), (1, 5), (1, 7), (3, 3), (3, 5), (3, 7), (5, 5), (5, 7), (7, 7), (7, 5), (7, 3), (5, 3), (6, 1), (5, 1), (3, 1), (2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)}

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Here (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7) all ∈ R

From the relation R it is seen that R is reflexive.

Symmetric: For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

From the relation R, it is seen that R is symmetric.

Transitive: For Transitivity, we need to prove that-

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

[I (a, b) are odd and (b, c) are odd then (a, c) are also odd numbers]

From the relation R, it is seen that R is transitive too.

Also, from the relation R, it is seen that {1, 3, 5, 7} are related with each other only and {2, 4, 6} are related with each other .

Given: A = {2, 3} and B = {4, 5} 

To find: A × B

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {2, 3} and B = {4, 5}. So,

 A × B = (2, 3) × (4, 5) 

= {(2, 4), (2, 5), (3, 4), (3, 5)} 

∴Number of elements of A × B = n = 4 

Number of subsets of A × B = 2ⁿ 

= 2⁴ 

= 2 × 2 × 2 × 2 

= 16 

∴, the set A × B has 16 subsets.

Given as

(a/3 + 1, b – 2/3) = (5/3, 1/3)

From the definition of equality of ordered pairs,

Let us solve for a and b

a/3 + 1 = 5/3 and b – 2/3 = 1/3

a/3 = 5/3 – 1 and b = 1/3 + 2/3

a/3 = (5-3)/3 and b = (1+2)/3

a/3 = 2/3 and b = 3/3

a = 2(3)/3 and b = 1

a = 2 and b = 1

∴ Values of a and b are, a = 2 and b = 1

(ii) Given as

(x + 1, 1) = (3y, y – 1)

From the definition of equality of ordered pairs,

Let us solve for x and y

x + 1 = 3y and 1 = y – 1

x = 3y – 1 and y = 1 + 1

x = 3y – 1 and y = 2

Here, y = 2 we can substitute in

x = 3y – 1

= 3(2) – 1

= 6 – 1

= 5
Hence, the values of x and y are, x = 5 and y = 2

Given as a ∈ {- 1, 2, 3, 4, 5} and b ∈ {0, 3, 6},

To find the ordered pair (a, b) such that a + b = 5

Now the ordered pair (a, b) such that a + b = 5 are as follows

(a, b) ∈ {(- 1, 6), (2, 3), (5, 0)}

Given,

(x+1,1) = (3y, y - 1)

To find : values of a and b 

By the definition of equality of ordered pairs, we have 

x + 1 = 3y and 1 = y - 1 

⇒ x = 3y - 1 and y = 2

So,

x = 3(2) - 1 

= 6 - 1 = 5 

⇒ x = 5 and y = 2

Reflexivity-

Consider (a, b) as an arbitrary element of N × N

Here, (a, b) ∈ N × N where (a, b) ∈ N

It can be written as

a + b = b + a

We get (a, b) R (a, b) for all (a, b) ∈ N × N

Hence, R is reflexive on N × N

Symmetry-

Consider (a, b), (c, d) ∈ N × N such that (a, b) R (c, d)

It can be written as

a + d = b + c and c + b = d + a

We get

(c, d) R (a, b)

(a, b) R (c, d) => (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N

Hence, R is symmetric on N × N.

Transitive-

Consider (a, b), (c, d), (r, f) ∈ N × N such that (a, b) R (c, d) and (c, d) R (e, f)

It can be written as

a + d = b + c and c + f = d + e

By adding both

(a + d) + (c + f) = (b + c) + (d + e)

On further calculation

a + f = b + e where (a, b) R (e, f)

So (a, b) R (c, d) and (c, d) R (e, f) we get (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N

Hence, R is transitive on N × N.

Therefore, R is reflexive, symmetric and transitive is an equivalence relation on N × N.

We know that a = ± b and a² = b² are equal

It is given that {(a, b): a, b ∈ S and a² = b²}

Here, (a, a) ∈ R and a² = a² is true.

Hence, R is reflexive where (a, b) ∈ R

We know that a² = b² and b² = a² where (b, a) ∈ R

Hence, R is symmetric.

If (a, b) ∈ R and (b, c) ∈ R

We know that a² = b² and b² = c² we get a² = c² where (a, c) ∈ R

Hence, R is transitive.

A = {1, 2} 

A × A = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} 

A × A × A = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)} 

Therefore 

A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

It is given that

R = {(a, b): a > b}

Reflexive-

If a ∈ R we get (a > b)

So (a, b) ∉ R for a ∉ R

Hence, R is not reflexive.

Symmetric-

We know that

(3, 4) ∈ R but (4, 3) ∉ R

Hence, R is not symmetric.

Transitive-

Consider (a, b) ∈ R and (b, c) ∈ R

It can be written as

a > b and b > c

So we get

a > c where (a, c) ∈ R

Hence, R is transitive.

Therefore, R is transitive but neither reflexive nor symmetric.

To prove : (A ∪ B) × C = (A × C) ∪ (B × C) 

Proof : Let (x, y) be an arbitrary element of (A ∪ B) × C. 

⇒ (x, y) ∈ (A ∪ B) C 

Since,(x, y) are elements of Cartesian product of (A ∪ B) × C 

⇒ x ∈ (A ∪ B) and y ∈ C 

⇒ (x ∈ A or x∈B) and y ∈ C 

⇒ (x ∈ A and y ∈ C) or (x ∈ Band y ∈ C) 

⇒ (x, y) ∈ A × C or (x, y) ∈ B × C 

⇒ (x, y) ∈ (A × C) ∪ (B × C) …1 

Let (x, y) be an arbitrary element of (A × C) ∪ (B × C). 

⇒ (x, y) ∈ (A × C) ∪ (B × C) 

⇒ (x, y) ∈ (A × C) or (x, y) ∈ (B × C) 

⇒ (x ∈ A and y ∈ C) or (x ∈ Band y ϵ C) 

⇒ (x ∈ A or x ∈ B) and y ∈ C 

⇒ x ∈ (A ∪ B) and y ∈ C 

⇒ (x, y) ∈ (A ∪ B) × C …2 

From 1 and 2, we get : 

(A ∪ B) × C = (A × C) ∪ (B × C)

(i) Given: A ⊆ B 

Need to prove: A × C ⊆ B × C 

Let us consider, (x, y)∈(A × C) 

That means, x∈A and y∈C 

Here given, A ⊆ B 

That means, x will surely be in the set B as A is the subset of B and x∈A. 

So, we can write x∈B 

Therefore, x∈B and y∈C 

⇒ (x, y)∈(B × C) 

Hence, we can surely conclude that, 

A × C ⊆ B × C [Proved] 

(ii) Given: A ⊆ B and C ⊆ D

Need to prove: A × C ⊆ B × D 

Let us consider, (x, y)∈(A × C) 

That means, x∈A and y∈C 

Here given, A ⊆ B and C ⊆ D 

So, we can say, x∈B and y∈D (x, y)∈(B × D) 

Therefore, we can say that, A × C ⊆ B × D [Proved]

A = {3, 4}, B = {4, 5} and C = {5, 6} 

B × C = {(4, 5), (4, 6), (5, 5), (5, 6)} 

A × (B × C) = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}

Let A and B be any two sets such that 

A × B = {(a, b): a ϵ A, b ϵ B} 

Now, 

B × A = {(b, a): a ϵ A, b ϵ B}

A × B = B × A 

(a, b) = (b, a) 

We can see that this is possible only when the ordered pairs are equal. 

Therefore, 

a = b and b = a 

Hence, Proved

Given: A ⊆ B 

Then, A = B at some value 

Multiplying by C both sides, we get, 

A × C = B × C 

Hence, Proved.

Option : (C)

Since, 

R is a relation from set A to set B, 

Therefore, 

It will always be a subset of A x B.

Given: A × B ⊆ C × D and A × B ≠ ϕ 

Need to prove: A ⊆ C and B ⊆ D 

Let us consider, (x, y) (A × B)......... (1) 

⇒ (x, y)∈(C × D) [as A × B ⊆ C × D] .......... (2)

From (1) we can say that,

x∈ A and y∈B .......... (a) 

From (2) we can say that, 

x∈C and y∈D ....... (b) 

Comparing (a) and (b) we can say that, 

⇒ x∈A and x∈C 

⇒ A ⊆ C Again, 

⇒ y∈B and y∈D 

⇒ B ⊆ D [Proved]