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Prove: If A and B are disjoint and B and C are disjoint so $A\cup C$ and B are disjoint

We know that $A\cap B=\emptyset \wedge B\cap C=\emptyset \rightarrow (A\cap B)\cap (B\cap C)= \emptyset \rightarrow (A\cap B)=\emptyset \wedge (B\cap C)= \emptyset$

On the other hand $(A\cup C)\cap B= \emptyset \rightarrow (A\cap B)\cup (C \cap B)=\emptyset \rightarrow (A\cap B)=\emptyset \vee (C\cap B)=\emptyset$

How can I prove it?

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In this answer we look at the elements of the sets.

Let $b\in B$.

Then $b\notin A$ (since $A$ and $B$ are disjoint) and $b\notin C$ (since $B$ and $C$ are disjoint).

So it is legal to conclude that $b\notin A\cup C$.

Proved is now that the sets $B$ and $A\cup C$ are disjoint.

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  • $\begingroup$ So now if I take an element $b\in B$ it will not be in $A\cup C$ and we are done? $\endgroup$ – gbox Jul 2 '15 at 9:21
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    $\begingroup$ Yes. If $B$ and $A\cup C$ are not disjoint then they must have an element in common. This proves that that is not the case. $\endgroup$ – drhab Jul 2 '15 at 10:34
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Notice that $B \cap (A \cup C) = (B \cap A) \cup (B \cap C) = \emptyset \cup \emptyset = \emptyset$.

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  • $\begingroup$ And $(B \cap A) \cup (B \cap C) = \emptyset$ is only true iff both are $\emptyset$ ? $\endgroup$ – gbox Jul 1 '15 at 18:08
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    $\begingroup$ Affirmative. One direction of the implication is obvious, namely the direction we need for this problem. $\endgroup$ – Sloan Jul 1 '15 at 18:17
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You're very close. Note that your second line of reasoning still works if you turn the arrows around: $$(A \cap B) = \emptyset \lor (C \cap B) = \emptyset \to (A \cap B) \cup (C \cap B) = \emptyset \to (A \cup C) \cap B = \emptyset.$$

Now you just need to show that $(A \cap B) = \emptyset) \land (B \cap C) = \emptyset \to (A \cap B) = \emptyset \lor (C \cap B)=\emptyset$.

Can you see why this is true?

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