Question

If A, B and C are any three sets then A – (B ∪ C) is equal to

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Answer 1

Answer:

Let x be any element of A - (B ∩ C). Then, x ∈ A - (B ∩ C)

⇒ x ∈ A and x ∉ (B ∩ C)

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

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

⇒ x ∈ A and x ∉ (B ∩ C)⇒ x ∈ A and (x ∉ B or x ∉ C)⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)⇒ x ∈ (A - B) or x ∈ (A - C)

⇒ x ∈ A and x ∉ (B ∩ C)⇒ x ∈ A and (x ∉ B or x ∉ C)⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)⇒ x ∈ (A - B) or x ∈ (A - C)⇒ x ∈ (A - B) ∪ x ∈ (A - C)

⇒ x ∈ A and x ∉ (B ∩ C)⇒ x ∈ A and (x ∉ B or x ∉ C)⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)⇒ x ∈ (A - B) or x ∈ (A - C)⇒ x ∈ (A - B) ∪ x ∈ (A - C)∴ A - (B ∩ C) ⊆ (A - B) ∪ (A - C)

⇒ x ∈ A and x ∉ (B ∩ C)⇒ x ∈ A and (x ∉ B or x ∉ C)⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)⇒ x ∈ (A - B) or x ∈ (A - C)⇒ x ∈ (A - B) ∪ x ∈ (A - C)∴ A - (B ∩ C) ⊆ (A - B) ∪ (A - C)Similarly, (A - B) ∪ (A - C) ⊆ A - (B ∩ C)

⇒ x ∈ A and x ∉ (B ∩ C)⇒ x ∈ A and (x ∉ B or x ∉ C)⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)⇒ x ∈ (A - B) or x ∈ (A - C)⇒ x ∈ (A - B) ∪ x ∈ (A - C)∴ A - (B ∩ C) ⊆ (A - B) ∪ (A - C)Similarly, (A - B) ∪ (A - C) ⊆ A - (B ∩ C)Hence, A - (B ∩ C) = (A - B) ∪ (A - C)

Step-by-step explanation:

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Answer 2

If A, B and C are any three sets then A – (B ∪ C) is equal to (A – B) ∩ (A – C).

Step 1: Solve the equation.

Given-  find A -  (B ∪ C).

We know that , X - Y = X ∩ Y'                     (Y' is the complement of Y)

So,   A – (B U C) = A ∩ (B U C)'

= A ∩ B' ∩ C'                                                      (∴ De Morgan’s Law)

And

(A – B)∩(A - C) = (A ∩ B')∩(A ∩ C')

= A ∩ B' ∩ C'

So, we got A–(B U C) = (A–B) ∩ (A – C)