# How to show $[(A'\cup B) \cap (A\cup B')] \cap C = (A'\cap B' \cap C) \cup (A \cap B \cap C)$?

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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By $A'$ do you mean the complement of $A$? It is usually denoted as $A^c$. – Asaf Karagila Oct 23 '11 at 9:49
I'm thinking that my book left out a set of parenthesis... – Robert S. Barnes Oct 23 '11 at 9:52
I know. The book I'm learning from (Israeli OpenU) denotes complement using A'. It's an old book I guess. – Robert S. Barnes Oct 23 '11 at 9:54
I added in what seem to be the missing parens... – Robert S. Barnes Oct 23 '11 at 9:58
Intersection is associative, so actually the added parenthesis are of no consequence. – Asaf Karagila Oct 23 '11 at 10:05

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 ;-) – Robert S. Barnes Oct 23 '11 at 10:11

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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+1 That's the step that Phira was missing, took me a minute to figure out what he was doing... – Robert S. Barnes Oct 23 '11 at 10:30
@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. – Phira Oct 23 '11 at 10:40

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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