Prove or disprove. for all sets a, b, and c, (a⋃b)⋂c ⊆a ⋃(b⋂c)

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...

29 mil

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