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Statement: Let $A$ and $B$ be subsets of $R$. Show that $(A \cup B)'=A'\cap B'$.

My attempt: To show that two sets are equal to one another we have to show that $(A \cup B)' \subset A'\cap B'$ and $ A'\cap B'\subset (A \cup B)'$. Beginning with the former, suppose that $x \in (A \cup B)'$ then $x\notin A \cup B$. This implies that $x\notin A$ or $x\notin B$, which in turn should imply that $x\in A'$ or $x\in B'$. However the author of my textbook states that: $$x\notin A \text{ or } x\notin B \Rightarrow x\in A' \text{ and } x\in B'$$ Whereas, I think that the following should be true: $$x\notin A \text{ or } x\notin B \Rightarrow x\in A' \text{ or } x\in B'$$

Why did the or change into an and?

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  • $\begingroup$ Alright, I got it. I guess I was just cconfused about the words...Please right this as an answer so that I can accept it. $\endgroup$ – model_checker Sep 29 '16 at 7:48
  • $\begingroup$ If the refrigerator is not in the apartment composed of kitchen and bathroom it means that you cannot find it neither in the kitchen nor in the bathroom. $\endgroup$ – Mauro ALLEGRANZA Sep 29 '16 at 7:49
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NO : $x∉A∪B$ means that $x∉A$ and $x∉B$.

To be element of the union of two sets means to belong at least to one of them.

Thus, if $x$ does not belong to the union of $A$ and $B$, it means that $x$ cannot be element neither of $A$ nor of $B$.

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$(A \cup B)'$ can be written as $\neg [(x \in A) \vee (x \in B)] \implies \neg (x \in A) \wedge \neg(x \in B) = A' \cap B'$

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  • 1
    $\begingroup$ This is wrong... you are using the DeMorgan laws to prove them... Its circular. $\endgroup$ – Masacroso Sep 29 '16 at 8:39
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    $\begingroup$ @Masacroso It is not intended to be a proof. He asked "Why did the or change into an and?". I thought it would be easier to see why that might be the case if it was expressed in this way. $\endgroup$ – Frejo Sep 29 '16 at 8:48

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