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Question Number 26889 by shiva123 last updated on 31/Dec/17
Answered by prakash jain last updated on 31/Dec/17
3.A∩(B−C)=(A∩B)−(A∩C)x∈A∩(B−C)⇒x∈A∧x∈(B−C)⇒x∈A∧x∈B∧x∉C⇒x∈(A∩B)∧x∉(A∩C)⇒x∈(A∩B)−(A∩C)⇒(A∩B)−(A∩C)⊇A∩(B−C)(I)x∈(A∩B)−(A∩C)⇒x∈(A∩B)∧x∉(A∩C)⇒x∈A∧x∈B∧x∉(A∩C)⇒x∈A∧x∈B∧x∉C⇒x∈A∧x∈(B−C)⇒x∈A∩(B−C)⇒(A∩B)−(A∩C)⊆A∩(B−C)(II)From(I)and(II)(A∩B)−(A∩C)=A∩(B−C)◼
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