Answer:
A ∪ (B \ C) = (A ∪ B) \ (C \ A).
Explanation:
Set B \ C represents the set of all elements of B except the element of C. It means B \ C = B - C.
Let A, B and C be sets.
To prove: A∪(B\C) = (A∪B) \ (C\A).
Proof :
First we need to prove that A ∪ (B\C) ⊆ (A∪B) \ (C\A).
Let x∈A∪(B\C) , then x∈A or x∈(B\C) .
Case 1: If x∈A, then
x∈A then x ∉ C\A and x∈A then x∈A∪B
Since x∈A∪B but x∉C\A, therefore x∈(A∪B) \ (C\A).
Case 2: If x∈B\C , then
If x∈B\C then x∈B but x ∉ C
x∈B then x∈A∪B and x ∉ C then x ∉ C\A
Since x∈A ∪ B but x ∉ C\A , therefore x∈(A∪B) \ (C\A).
From case 1 and case 2 we can say that x∈ (A ∪ B) \ (C \ A). So,
A ∪ (B \ C) ⊆ (A ∪ B) \ (C \ A). ..... (1)
Now we need to prove that (A ∪ B) \ (C \ A). ⊆ A ∪ (B \ C)
Let x∈(A ∪ B) \ (C \ A) , then x∈A ∪ B but x∉C\A
x∈ A∪B then x∈ A or x∈ B
Case 1: If x∈ A
If x∈A then x∈A∪(B \ C)
Case 2: If x∈ B
Suppose that x ∉ A and x∈ C, then since x ∉ A we have that x ∈ C\A, a contradiction. Therefore x ∉ C.
Since x∈ B and x ∉ C. then x ∈ B\C, hence x∈ A ∪ (B \ C)
From case 1 and case 2 we can say that x∈ A ∪ (B \ C) . So,
(A ∪ B) \ (C \ A). ⊆ A ∪ (B \ C) .... (2)
Using (1) and (2) we get
A ∪ (B \ C) = (A ∪ B) \ (C \ A).
Hence proved.