Prove A∩(B∩C)=(A∩B)∩C
Letx∈A∩(B∩C)x \in A \cap (B \cap C)x∈A∩(B∩C)
⟹ x∈A\implies x \in A⟹x∈A and x∈B∩Cx \in B \cap Cx∈B∩C
⟹ x∈A\implies x \in A⟹x∈A and x∈Bx \in Bx∈B and x∈Cx \in Cx∈C
⟹ x∈A∩B\implies x\in A \cap B⟹x∈A∩B and x∈Cx \in Cx∈C
⟹ x∈(A∩B)∩C\implies x \in (A\cap B) \cap C⟹x∈(A∩B)∩C
⟹ A∩(B∩C)⊂(A∩B)∩C\implies A\cap (B \cap C) \sub (A\cap B) \cap C⟹A∩(B∩C)⊂(A∩B)∩C
Conversely, let y∈(A∩B)∩Cy \in (A\cap B) \cap Cy∈(A∩B)∩C
⟹ y∈A∩B\implies y \in A \cap B⟹y∈A∩B and y∈Cy \in Cy∈C
⟹ y∈A\implies y \in A⟹y∈A and y∈By \in By∈B and y∈Cy \in Cy∈C
⟹ y∈A\implies y\in A⟹y∈A and y∈B∩Cy\in B\cap Cy∈B∩C
⟹ y∈A∩(B∩C)\implies y\in A \cap (B\cap C)⟹y∈A∩(B∩C)
⟹ (A∩B)∩C⊂A∩(B∩C)\implies (A\cap B) \cap C \sub A \cap (B \cap C)⟹(A∩B)∩C⊂A∩(B∩C)
Hence, A∩(B∩C)=(A∩B)∩CA \cap (B \cap C) = (A\cap B) \cap CA∩(B∩C)=(A∩B)∩C
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