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# If A, B are subsets of a set X then

(i) X- (A∪B) = (X – A) ∩ (X– B) or, (A∪B)′ = A′∩B′.

(ii) X- (A∩B) = (X – A) ∪ (X– B) or, (A∩B)′ = A′∪B′.

## Where we denote by A′, the complement of A.

Proof: We prove (i): Let x∈ X- (A∪B) be any element.

Then x ∈ X, x ∉ A∪B

⟹ x ∈ X, x ∉ A and x ∉ B

⟹ x ∈ X-A and x ∈ X – B

⟹ x ∈ (X – A) ∩ (X– B)

⟹ X- (A∩B) ⊆ (X – A) ∩ (X– B)

### Again, let y ∊ (X – A) ∩ (X– B) be any element.

Then y ∊ X – A and y ∊ X– B (or both)

⟹ y ∊ X, y ∉ A and y ∊ X, y ∉ B

⟹ y ∊ X, y ∉ A∪B

⟹ y ∊ – (A∪B)

⟹ (X – A) ∪ (X– B) ⊆ X- (A∪B)

And hence X- (A∪B) = (X – A) ∩ (X– B). (Proved)

Now we prove (ii): Let x∈ X- (A∩B) be any element.

Then x ∈ X, x ∉ A∩B

⟹ x ∈ X, x ∉ A or x ∉ B

⟹ x ∈ X-A or x ∈ X – B

⟹ x ∈ (X – A) ∪ (X– B)

⟹ X- (A∩B) ⊆ (X – A) ∪ (X– B)

Again, let y ∊ (X – A) ∪ (X– B) be any element.

Then y ∊ X – A or y ∊ X– B (or both)

⟹ y ∊ X, y ∉ A or y ∊ X, y ∉ B

⟹ y ∊ X, y ∉ A∩B

⟹ y ∊ – (A∩B)

⟹ (X – A) ∪ (X– B) ⊆ X- (A∩B)

And hence X- (A∩B) = (X – A) ∪ (X– B). (Proved)

# 2nd method: We prove (i):

Let, x ∊ (A ∪ B) ′

⟹ x ∉ A ∪ B

⟹ x ∉ A and x ∉ B

⟹ x A′ and x ∊ B′

⟹ x ∊ A′ and B′

⟹ x ∊ A′ ∩ B′

∴ (A ∪ B) ′ ⊂ A′ ∩ B′     —————————-(i)

# Again, let, x ∊ A′ ∩ B′

⟹ x ∊ A′ and x ∊ B′

⟹ x ∉ A and x ∉ B

⟹ x ∉ A ∪ B

⟹ x ∊ (A ∪ B) ′

∴ A′ ∩ B′ ⊂ (A ∪ B) ′         ————————-(ii)

From (i) and (ii) we get,

## (A ∪ B) ′ = A′ ∩ B′ (Proved)

Now we proved (ii): Let, x ∊ (A ∩ B) ′

⟹ x ∉ A ∩ B

⟹ x ∉ A or x ∉ B

⟹ x A′ or x ∊ B′

⟹ x ∊ A′ or B′

⟹ x ∊ A′ ∩ B′

∴ (A ∩ B) ′ ⊂ A′ ∪ B′     —————————-(i)

### Again, let, x ∊ A′ ∪ B′

⟹ x ∊ A′ or x ∊ B′

⟹ x ∉ A or x ∉ B

⟹ x ∉ A ∩ B

⟹ x ∊ (A ∩ B) ′

∴ A′ ∪ B′ ⊂ (A ∩B) ′         ————————-(ii)

From (i) and (ii) we get,

(A ∩ B) ′ = A′ ∪ B′ (Proved).

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