For any two sets A and B, prove that

A ∪ (B – A) = A ∪ B

Let x ϵ A ∪ (B –A) ⇒ x ϵ A or x ϵ (B – A)

⇒ x ϵ A or x ϵ B and x ∉ A

⇒ x ϵ B

⇒ x ϵ (A ∪ B) [∵ B ⊂ (A ∪ B)]

This is true for all x ϵ A ∪ (B–A)

∴ A∪(B–A)⊂(A∪B)……(1)

Conversely,

Let x ϵ (A ∪ B) ⇒ x ϵ A or x ϵ B

⇒ x ϵ A or x ϵ (B–A) [∵ B ⊂ (A ∪ B)]

⇒ x ϵ A ∪ (B–A)

∴ (A∪B)⊂ A∪(B–A)……(2)

From 1 and 2 we get…

A ∪ (B – A) = A ∪ B