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DE MORGAN\'S law proof |
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Answer» Proof of De Morgan’s law: (A ∩ B)\' = A\' U B\'Let M = (A ∩ B)\' and N = A\' U B\'Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)\'⇒ x ∉ (A ∩ B)⇒ x ∉ A or x ∉ B⇒ x ∈ A\' or x ∈ B\'⇒ x ∈ A\' U B\'⇒ x ∈ NTherefore, M ⊂ N …………….. (i)Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A\' U B\'⇒ y ∈ A\' or y ∈ B\'⇒ y ∉ A or y ∉ B⇒ y ∉ (A ∩ B)⇒ y ∈ (A ∩ B)\'⇒ y ∈ MTherefore, N ⊂ M …………….. (ii)Now combine (i) and (ii) we get; M = N i.e. (A ∩ B)\' = A\' U B\' Proof of De Morgan’s law: (A U B)\' = A\' ∩ B\' Let P = (A U B)\' and Q = A\' ∩ B\'Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)\'⇒ x ∉ (A U B) ⇒ x ∉ A and x ∉ B⇒ x ∈ A\' and x ∈ B\'⇒ x ∈ A\' ∩ B\'⇒ x ∈ QTherefore, P ⊂ Q …………….. (i)Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A\' ∩ B\'⇒ y ∈ A\' and y ∈ B\'⇒ y ∉ A and y ∉ B⇒ y ∉ (A U B)⇒ y ∈ (A U B)\'⇒ y ∈ PTherefore, Q ⊂ P …………….. (ii)Now combine (i) and (ii) we get; P = Q i.e. (A U B)\' = A\' ∩ B\' |
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