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14 March, 02:50

Prove or disprove. for all sets A, B, and c, (A⋃B) ⋂C ⊆A ⋃ (B⋂C)

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  1. 14 March, 05:45
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    Let x ∈ (A∪B) ∩ C

    ⇒ x ∈ (A ∪ B) and x ∈ C

    ⇒ x ∈ A or x ∈ B and x ∈ C

    ⇒ x ∈ A or x ∈ B∩C

    ⇒ x ∈ A ∪ (B∩C)

    Now, x ∈ A ∪ (B∩C)

    ⇒ x ∈ A or x ∈ B ∩ C

    ⇒ x ∈ A or x ∈ B and x ∈ C

    ⇒ x ∈ (A∪B) and x ∈ C

    ⇒ x ∈ (A∪B) ∩ C

    Since, x shows the an arbitrary element,

    ⇒ A ∪ (B∩C) = (A∪B) ∩ C

    ∵ A set always contains itself,

    ⇒ A ∪ (B∩C) ⊆ A ∪ (B∩C)

    ⇒ (A⋃B) ⋂C ⊆A ⋃ (B⋂C)

    Hence, proved ...
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