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A intersection (B union C) |
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Answer» We need to PROVE that, A ∩ (B ∪ C) = (A ∩ B) U (A ∩ C) LET, X ∈ {A ∩ (B ∪ C)} ⇒ (x ∈ A) ∧ {x ∈ (B ∪ C)} ⇒ (x ∈ A) ∧ { (x ∈ B) ∨ (x ∈ C) } ⇒ {(x ∈ A) ∧ (x ∈ B)} ∨ {(x ∈ A) ∧ (x ∈ C)} ⇒ {x ∈ (A ∩ B)} ∨ {x ∈ (A ∩ C)} ⇒ x ∈ {(A ∩ B) U (A ∩ C)} ∴ A ∩ (B ∪ C) ⊆ (A ∩ B) U (A ∩ C) ...(i) Again, let y ∈ {(A ∩ B) U (A ∩ C)} ⇒ {y ∈ (A ∩ B)} ∨ {y ∈ (A ∩ C)} ⇒ {(y ∈ A) ∧ (y ∈ B)} ∨ {(y ∈ A) ∧ (y ∈ C)} ⇒ (y ∈ A) ∧ { (y ∈ B) ∨ (y ∈ C) } ⇒ (y ∈ A) ∧ {y ∈ (B ∪ C)} ⇒ y ∈ {A ∩ (B ∪ C)} ∴ (A ∩ B) U (A ∩ C) ⊆ A ∩ (B ∪ C) ...(ii) ∴ From (i) and (ii), we get A ∩ (B ∪ C) = (A ∩ B) U (A ∩ C) Hence, PROVED. # |
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