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)
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)
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
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.
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}