Examples on De Morgan’s law: 1. If U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}. Proof of De Morgan's law: (X ∩ Y)' = X' U Y'. Solution: We know, U = {j, k, l, m, n} X = {j, k, m} Y = {k, m, n} (X ∩ Y) = {j, k, m} ∩ {k, m, n} = {k, m} Therefore, (X ∩ Y)' = {j, l, n} ……………….. (i) Again, X = {j, k, m} so, X' = {l, n} and Y = {k, m, n} so, Y' = {j, l} X' ∪ Y' = {l, n} ∪ {j, l} Therefore, X' ∪ Y' = {j, l, n} ……………….. (ii) Combining (i)and (ii) we get; (X ∩ Y)' = X' U Y'. Proved 2. Let U = {1, 2, 3, 4, 5, 6, 7, 8}, P = {4, 5, 6} and Q = {5, 6, 8}. Show that (P ∪ Q)' = P' ∩ Q'. Solution: We know, U = {1, 2, 3, 4, 5, 6, 7, 8} P = {4, 5, 6} Q = {5, 6, 8} P ∪ Q = {4, 5, 6} ∪ {5, 6, 8} = {4, 5, 6, 8} Therefore, (P ∪ Q)' = {1, 2, 3, 7} ……………….. (i) Now P = {4, 5, 6} so, P' = {1, 2, 3, 7, 8} and Q = {5, 6, 8} so, Q' = {1, 2, 3, 4, 7} P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7} Therefore, P' ∩ Q' = {1, 2, 3, 7} ……………….. (ii) Combining (i)and (ii) we get; (P ∪ Q)' = P' ∩ Q'. Proved