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Using set algebra how do I get from:

[(A'\cup B) \cap (A\cup B')] \cap C

to

(A'\cap B' \cap C) \cup (A \cap B \cap C)

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    Intersection is associative, so actually the added parenthesis are of no consequence.2011-10-23

3 Answers 3

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I like to use '$+$' for $\cup$ and '$.$' for $\cap$

Then your expression becomes (a'+b).(a+b').c=(a'a+a'b'+ab+bb').c=(0+ab+a'b'+0)=(ab+a'b').c=abc+a'b'c

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The trick is to get from (A'\cup B)\cap(A\cup B') to (A'\cap B')\cup(A\cap B); once you’ve done this, the $C$ will almost take care of itself. It’s mostly just distributivity. In gruesome detail: \begin{align*} (A'\cup B)\cap(A\cup B') &= (A'\cap(A\cup B'))\cup (B\cap(A\cup B'))\\ &= ((A'\cap A)\cup(A'\cap B'))\cup ((B\cap A)\cup(B\cap B'))\\ &= (\varnothing \cup (A'\cap B'))\cup ((B\cap A)\cup \varnothing))\\ &= (A'\cap B')\cup(B\cap A) \end{align*}

Now just use commutativity of $\cap$ and take care of the $C$; you’ll use distributivity again.

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    @Robert Please don't speak of me as "he" and please don't say that I was "missing" a step. Applying the distributive law to two parentheses is not further from the axioms than your use of intersections of three sets in your problem statement. What is acceptable depends on context which you have not given.2011-10-23
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You apply the distributive law twice:

\begin{align} (A'\cup B) \cap (A\cup B') \cap C &=((A'\cap A) \cup (A' \cap B') \cup (B\cap A)\cup(B\cap B') )\cap C\\ &=((A' \cap B') \cup (B\cap A))\cap C\\ &=(A' \cap B'\cap C) \cup (B\cap A\cap C) \end{align}

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    Wow, that's quite a few steps to leave out :-) You actually left out a step also ;-)2011-10-23