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For all sets A, B and C, show that (A – B) ∩ (A – C) = A – (B ∪ C)

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According to the question,

There are three sets A, B and C

To show:

(A – B) ∩ (A – C) = A – (B ∪ C)

Let x ∈ (A – B) ∩ (A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

⇒ (x ∈ A and x ∉ B) and (x ∈ A and x ∉ C)

⇒ x ∈ A and (x ∉ B and x ∉ C)

⇒ x ∈ A and x ∉ (B ∪ C)

⇒ x ∈ A – (B ∪ C)

⇒ (A – B) ∩ (A – C) ⊂ A – (B ∪ C) …(i)

Let y ∈ A – (B ∪ C)

⇒ y ∈ A and y ∉ (B ∪ C)

⇒ y ∈ A and (y ∉ B and y ∉ C)

⇒ (y ∈ A and y ∉ B) and (y ∈ A and y ∉ C)

⇒ y ∈ (A – B) and y ∈ (A – C)

⇒ y ∈ (A – B) ∩ (A – C)

⇒ A – (B ∪ C) ⊂ (A – B) ∩ (A – C) …(ii)

We know that,

If P ⊂ Q and Q ⊂ P

Then, P = Q

Therefore, from equations (i) and (ii),

A – (B ∪ C) = (A – B) ∩ (A – C)

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