FalseLet A = {0, 1} and B = {1, 2}∴ A ∪ B = {0, 1, 2}P(A) = {Φ, {0}, {1}, {0, 1}}P(B) = {Φ, {1}, {2}, {1, 2}}P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}∴ P(A) ∪ P(B) ≠ P(A ∪ B)
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