# Prove that for all sets $A$, $B$, and $C$, if $A\cap{B}=\emptyset$ and $A\cap{C}=\emptyset$, then $A\cap({B}\cup {C})=\emptyset$.

Prove that for all sets $$A$$, $$B$$, and $$C$$, if $$A\cap{B}=\emptyset$$ and $$A\cap{C}=\emptyset$$, then $$A\cap({B}\cup {C})=\emptyset$$.

I know that this is obviously true, however I'm not sure how to prove it.

Any help is appreciated, thanks!

$$A\cap({B}\cup {C})=(A\cap B)\cup (A\cap C)=\emptyset\cup\emptyset=\emptyset.$$

Suppose not, suppose if $$A \cap B = \emptyset$$ and $$B \cap C = \emptyset$$ then $$A \cap (B \cup C) \ne \emptyset$$.

Then there must be some $$x \in (A \cap B) \cup (B \cap C)$$ by the distributive property of intersection over unions.

So then there must be some $$x \in (A \cap B)$$ or $$x \in (B \cap C)$$ by the definition of the union.

However, the premise states that both $$A \cap B = \emptyset$$ and $$B \cap C = \emptyset$$, so it is impossible for an $$x$$ to be in either of these sets.

Therefore, we reach a contradiction - it must be the case if $$A \cap B = \emptyset$$ and $$B \cap C = \emptyset$$ then $$A \cap (B \cup C) = \emptyset$$.

EDIT:

For clarification, the trick here is to notice that you can:

1. Distribute $$A$$ over $$B \cup C$$ in order to get the two terms $$(A \cap B)$$ and $$(B \cap C$$).

2. Use contradiction with the properties of the union to show that an element in the consequent would imply an element in the two sets in the antecedent.

I think in terms of thinking about it, the "enlightening" moment comes from realizing you can distribute the intersection in the way above. The rest follows in a very straight forward, logical fashion.