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If $A\cap B = \emptyset$, prove that: $A\cup B'=B'$

My Attempt: Let $x$ be an element of $A\cup B'$. Then $$x\in A\cup B' \implies x\in A \ \textrm {or} \ x \in B'$$ $$\implies x\in B'$$ So, $A\cup B' \subset B'$.

Again, Let $x$ be an element of $B'$ $$x\in B' \implies x\in U \ \textrm{and} \ x\notin B$$ where $U$ is the universal set.

How do I proceed?

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  • $\begingroup$ The converse is easy, see the that $ A \subset A \cup B$ $\endgroup$ – Thiago Nascimento Jul 28 '17 at 3:40
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Alternative approach:

From your Previous question, you have proven that if $A \cap B = \emptyset$, then we have $B \cap A' = B$

From De Morgan's law, $$B'=(B \cap A')'=B' \cup A$$

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If you have $A\cap B=\emptyset$ then this implies that $A\subseteq B'$, so the conclution follow.

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The first half needs work. You say: $x \in A$ or $x \in B'$, therefore $x\in B'$. That goes a bit quick. Sure, if $x \in B'$ then $x \in B'$, but what if $x \in A$? You still need to show that that also means that $x \in B'$ by using $A \cap B = \emptyset$. Also, once you have shown this, you have that $A \cup B' \color{red} \subseteq B'$

The second half is actually the easy half. If $x \in B'$, then you certainly have that $x \in A$ or $x \in B'$, and hence $B' \subseteq A \cup B'$

Another approach is to do this purely algebraically:

$$B'= B' \cup \emptyset = B' \cup (A \cap B)= (B' \cup A) \cap (B \cup B') = (A \cup B') \cap U = A \cup B'$$

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1) Prove $A\cap B=\emptyset \implies A\subset B'$

2) prove $A\subset C \implies A\cup C=C $.

1) is easy. If $x\in A $ then $x\in B $ would mean $x\in A\cap B $ which is impossible so $x\in A\implies x\in B'$.

2) is trivial. $x\in A \cup C$ means $x\in A\subset C$ or $x\in C $. Either way, $x\in C $ so $C\subset A\cup C\subset C $.

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