1. Prove that
a) $\displaystyle (A-B) \cap C = (A \cap C) - (B \cap C)$;
b) (A symmetric difference B) = $\displaystyle (A U B) - (A \cap B)$
1. Prove that
a) $\displaystyle (A-B) \cap C = (A \cap C) - (B \cap C)$;
b) (A symmetric difference B) = $\displaystyle (A U B) - (A \cap B)$
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 $\displaystyle x \in (A - B) \cap C \implies x \in (A \cap C) - (B \cap C)$ AND $\displaystyle x \in (A \cap C) - (B \cap C) \implies x \in (A - B) \cap C$
can you continue?
Hello,
$\displaystyle (A \cap C)-(B \cap C)=(A \cap C) \cap (B \cap C)'$ (definition of the difference)
$\displaystyle (A \cap C)-(B \cap C)=(A \cap C) \cap (B' \cup C')$ (de Morgan's law)
$\displaystyle (A \cap C)-(B \cap C)=[(A \cap C) \cap C'] \cup [(A \cap C) \cap B']$ (distributivity)
$\displaystyle (A \cap C)-(B \cap C)=[A \cap (C \cap C')] \cup [(A \cap B') \cap C]$ (associativity)
But $\displaystyle C \cap C'=\emptyset$
Hence $\displaystyle A \cap (C \cap C')=\emptyset$
And $\displaystyle A \cap B'=A-B$ (definition of the difference)
So now you have :
$\displaystyle (A \cap C)-(B \cap C)=\emptyset \cup [(A-B) \cap C]=\boxed{(A-B) \cap C}$