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Prove that if $A \cup B$ $\space\subseteq $ $\space C \cup D$, $A \cap B$=$\emptyset$ and $C\subseteq A$, then $B \subseteq D.$

I tried working around with this for a while and reached this conclusion, however I do not know if I am making any sense at all:

Let $y \in$ $A \cup B$ $\subseteq $ $ C \cup D$

Since $(y \in A$ or $y \in B\space)$ $\subseteq \space$ $( y \in C$ or $y \in D)$ and because $A \cap B$=$\emptyset$ which means that A and B are disjoint, $C\subseteq A$ implies that $B \nsubseteq C$ because A and B are disjoint. However since $A \cup B\space$ $\subseteq \space$ $ C \cup D$, $B$ MUST be a subset of $D$ as $B \nsubseteq C$.

Sorry if I restated the same thing a few times.

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  • $\begingroup$ Use \cap for intersection and \cup for union. $\endgroup$ – Lost Dec 13 '13 at 1:09
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    $\begingroup$ Your argument looks correct, but probably could use some trimming. $\endgroup$ – Lost Dec 13 '13 at 1:16
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$\forall x \in B$.

$x \in B, A\cap B = \emptyset \Rightarrow x \not\in A$.

$x \in B,A \cup B \subseteq C\cup D \Rightarrow x \in C \cup D$.

$x \not\in A, C\subseteq A \Rightarrow x \not\in C$.

$ x \in C \cup D, x \not\in C \Rightarrow x \in D$.

Therefore, $B \subseteq D$.

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