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