For all sets A, B and C, show that (A – B) ∩ (A – C) = A – (B ∪ C)
Given: There are three sets A, B and C
To prove: (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:
P ⊂ Q and Q ⊂ P ⇒ P = Q
From (i) and (ii):
A – (B ∪ C) = (A – B) ∩ (A – C)
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