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 ...