184k views
4 votes
Prove or disprove. for all sets A,B, and c, (A⋃B)⋂C ⊆A ⋃(B⋂C)

User Trinition
by
5.4k points

1 Answer

4 votes

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...

User Neelima
by
5.2k points