1. Prove that

a) $(A-B) \cap C = (A \cap C) - (B \cap C)$;

b) (A symmetric difference B) = $(A U B) - (A \cap B)$

2. Originally Posted by ninano1205
1. Prove that

a) $(A-B) \cap C = (A \cap C) - (B \cap C)$;

b) (A symmetric difference B) = $(A U B) - (A \cap B)$
for both, show that each side is a subset of the other. do you know how to do this?

for instance, in the first, show that $x \in (A - B) \cap C \implies x \in (A \cap C) - (B \cap C)$ AND $x \in (A \cap C) - (B \cap C) \implies x \in (A - B) \cap C$

can you continue?

3. Thank you for the response. But I know till u proceeded, my problem is after that. For the first part, I don't know how to prove LHS is a subset of RHS.

4. Hello,

$(A \cap C)-(B \cap C)=(A \cap C) \cap (B \cap C)'$ (definition of the difference)

$(A \cap C)-(B \cap C)=(A \cap C) \cap (B' \cup C')$ (de Morgan's law)

$(A \cap C)-(B \cap C)=[(A \cap C) \cap C'] \cup [(A \cap C) \cap B']$ (distributivity)

$(A \cap C)-(B \cap C)=[A \cap (C \cap C')] \cup [(A \cap B') \cap C]$ (associativity)

But $C \cap C'=\emptyset$
Hence $A \cap (C \cap C')=\emptyset$

And $A \cap B'=A-B$ (definition of the difference)

So now you have :

$(A \cap C)-(B \cap C)=\emptyset \cup [(A-B) \cap C]=\boxed{(A-B) \cap C}$