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Is the following simplification correct?

\begin{align*} [(A \cup B \cup C)\cap(A\cup B)] - [(A \cup(B-C))\cap A] &= (A \cup B) - [(A \cap A) \cup (A \cap(B-C))]\\ &= (A \cup B) - [A \cup (A \cap (B-C))] \\ &= [(A \cup B) - A] \cup [(A \cup B) - (B-C)]\\ &= B \cup [(A \cup B) - (B-C)]\\ &= B \cup [(A \cup B)\cap (U-(B-C))]\\ &= [B \cup (A \cup B)] \cap [B \cup (U-(B-C))]\\ &= B \cap[B \cup (B-C)^{'}]\\ &= B \cap [B \cup (B \cup C')]\\ &= B \cap (B \cup C')\\ &= B. \end{align*}

The textbook I'm using doesn't have any solutions. I went over my work a couple of times, but I can't tell if I'm correct or not. Any help would be appreciated.

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It is not necessarily equal to $B$. Your error was to assume $(A \cup B) - A = B$, going from line 3 to line 4.

The following fact simplifies the problem significantly: if $X \subseteq Y$ then $X \cap Y = X$. Since $A \cup B \subseteq A \cup B \cup C$, and $A \subseteq A \cup (B - C)$, we have $$[(A \cup B \cup C) \cap (A \cup B)] - [(A \cup (B - C)) \cap A] = (A \cup B) - A$$ This is only equal to $B$ when $A$ and $B$ are disjoint.

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  • $\begingroup$ Thank you. Your answer was very useful, I didn't stop to think about the inclusion of the sets before going into the algebra. I'll keep it in mind. $\endgroup$
    – Eduardo M.
    Jun 16 '16 at 18:22

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