Answer:
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...
