Question

prove that (A-B) U (B-A) =(A U B) - (A intersection B)

Answer 1

Let A={a,b,c} & B={a,b,d}

LHS=> A-B={c}  and B-A={d} 

Thus, (A-B) U (B-A)={c,d}

RHS=>  A U B= {a,b,c,d}

A ∩ B = {c,d}

Thus, (A U B) - (A ∩ B) = {c,d}

LHS=RHS

Hope it helps.

Cheers!!!

Answer 2

(A U B) - (A ∩ B) = (A U B) ∩ (A ∩ B)' = Definition of set minus
(A U B) ∩ (A' U B') = De Morgan's Law
[(A U B) ∩ A'] U [(A U B) ∩ B'] = Intersection Distributes over Union
[(A ∩ A') U (B ∩ A')] U [(A ∩ B') U (B ∩ B')] = Intersection Distributes over Union
[Ø U (B ∩ A')] U [(A ∩ B') U Ø] = Intersection of a set with its complement is empty
(B ∩ A') U (A ∩ B') = Union of a set with empty is that set
(B - A) U (A - B) = Definition of Set Minus
(A - B) U (B - A) Union is Commutative

HOPE IT HELPS ✌