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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!

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

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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.

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