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 February 25th, 2014, 05:31 PM #1 Newbie   Joined: Feb 2014 Posts: 3 Thanks: 0 Simplify set expression Good evening everyone. I'm new to discrete math, and I've been struggling with this problem for two days now. The problem is to simplify the following: ((B U C) ? (A - B) ? (B ? C' )' I used ` to denote complement, I couldn't find a symbol for it in character map. So I have been looking at this problem, and I made some progress with the set difference laws, DeMorgan's laws, and the double complement law, but I really can't seem to figure it out past that. I really appreciate any help that anyone can give me.
 February 25th, 2014, 05:32 PM #2 Newbie   Joined: Feb 2014 Posts: 3 Thanks: 0 Re: Simplify set expression Missed a ) in that, sorry: ((B U C) ? (A - B)) ? (B ? C' )'
February 25th, 2014, 07:20 PM   #3
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Re: Simplify set expression

Quote:
 Originally Posted by Zyous Missed a ) in that, sorry: ((B U C) ? (A - B)) ? (B ? C' )'
OK, let's see what we can do.

A - B contains no elements from B, so neither can (B U C) ? (A - B) and hence this equals C ? (A - B).

(B ? C' )' is B' U C by De Morgan's law, so the whole expression simplifies to (C ? (A - B)) ? (B' U C). But all elements on the left are in C, so this is just (C ? (A - B)) ? C = C ? (A - B). That's pretty much as simple as you can get, unless you want to replace - with its definition.

February 26th, 2014, 01:04 PM   #4
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Re: Simplify set expression

Hello, Zyous!

Quote:
 $\text{Simplify: }\:\big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B\,\cap\,C')'$

$\begin{array}{ccccc}& \big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B\,\cap\,C')' && \text{Given} \\ \\ \\
= & \big[(B\,\cup\,C) \,\cap\,(A\,\cap\,B')\big]\,\cap\,(B\,\cap\,C')' && \text{Def. subtr'n} \\ \\ \\
= & (B\,\cup\,C)\,\cap\,A\,\cap\,B'\,\cap\,(B' \cup\,C) && \text{DeMorgan} \\ \\ \\
= & A\,\cap\,B'\,\cap\,(B\,\cup\,C)\,\cap\,(B' \,\cap\,C) && \text{Comm., Assoc.} \\ \\ \\
= & A\,\cap\,B'\,\cap\,\big[(B\,\cap\,B')\cup\,C\big] && \text{Distr.}\\ \\ \\
= & A\,\cap\,B'\,\cap\,(\emptyset \,\cup\,C) && P\,\cap\,P' \,=\,\emptyset \\ \\ \\
= & A\,\cap\,B'\,\cap\,C && \emptyset\,\cup\,P \,=\,P \end{array}$

 February 26th, 2014, 01:31 PM #5 Newbie   Joined: Feb 2014 Posts: 3 Thanks: 0 Re: Simplify set expression Wow thank you very much for the help! If I can bother you by asking the second part of the question, it would be very helpful. The second part is to see how the answer would change if the first question were instead ((B U C) ? (A - B)) ? (B' ? C)', with the compliment moved from the C to the B in at the end of the problem. Again, thanks a lot for the detailed answer, it's really helping me see how this process works!
February 26th, 2014, 09:34 PM   #6
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Re: Simplify set expression

Hello, Zyous!

Quote:
 $\text{Simplify: }\:\big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B'\,\cap\,C)'$

$\begin{array}{ccccc}& \big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B'\,\cap\,C)' && \text{Given} \\ \\ \\
= & \big[(B\,\cup\,C) \,\cap\,(A\,\cap\,B')\big]\,\cap\,(B'\,\cap\,C)' && \text{Def. subtr'n} \\ \\ \\
= & (B\,\cup\,C)\,\cap\,(A\,\cap\,B')\,\cap\,(B\,\ cup\,C') && \text{DeMorgan} \\ \\ \\
= & (A\,\cap\,B')\,\cap\,(B\,\cup\,C)\,\cap\,(B\,\ cap\,C') && \text{Comm., Assoc.} \\ \\ \\
= & (A\,\cap\,B')\,\cap\,\big[(B\,\cup\,(C\,\cap\,C')\big] && \text{Distr.}\\ \\ \\
= & (A\,\cap\,B')\,\cap\,(B\,\cup\,\emptyset) && P\,\cap\,P' \,=\,\emptyset \\ \\ \\
= & (A\,\cap\,B')\,\cap\,B && P\,\cup\,\emptyset \,=\,P \\ \\ \\
= & A\,\cap\,(B'\,\cap\,B) && \text{Assoc.} \\ \\ \\
= & A\,\cap\,\emptyset && P'\,\cap\,P \,=\,\emptyset \\ \\ \\
= & \emptyset && P\,\cap\,\emptyset\,=\,\emptyset \end{array}$

 February 27th, 2014, 02:02 AM #7 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Simplify set expression Good stuff, although: $B' \cap B = \emptyset = f?$
 February 27th, 2014, 02:52 AM #8 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407 Re: Simplify set expression Absolutely correct, Pero! [color=beige]. . [/color](I must have had Logic on my mind.) I'll go back and correct them.
March 12th, 2014, 05:55 AM   #9
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Re: Simplify set expression

Quote:
Originally Posted by soroban
Hello, Zyous!

Quote:
 $\text{Simplify: }\:\big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B'\,\cap\,C)'$

$\begin{array}{ccccc}& \big[(B\,\cup\,C)\,\cap\,(A\,-\,B)\big]\,\cap\,(B'\,\cap\,C)' && \text{Given} \\ \\ \\
= & \big[(B\,\cup\,C) \,\cap\,(A\,\cap\,B'
\big]\,\cap\,(B&#39;\,\cap\,C)&#39; && \text{Def. subtr&#39;n} \\ \\ \\
= & (B\,\cup\,C)\,\cap\,(A\,\cap\,B&#39\,\cap\,(B\,\ cup\,C&#39 && \text{DeMorgan} \\ \\ \\
= & (A\,\cap\,B&#39\,\cap\,(B\,\cup\,C)\,\cap\,(B\,\ cap\,C&#39 && \text{Comm., Assoc.} \\ \\ \\
= & (A\,\cap\,B&#39\,\cap\,\big[(B\,\cup\,(C\,\cap\,C&#39\big] && \text{Distr.}\\ \\ \\
= & (A\,\cap\,B&#39\,\cap\,(B\,\cup\,\emptyset) && P\,\cap\,P&#39; \,=\,\emptyset \\ \\ \\
= & (A\,\cap\,B&#39\,\cap\,B && P\,\cup\,\emptyset \,=\,P \\ \\ \\
= & A\,\cap\,(B&#39;\,\cap\,B) && \text{Assoc.} \\ \\ \\
= & A\,\cap\,\emptyset && P&#39;\,\cap\,P \,=\,\emptyset \\ \\ \\
= & \emptyset && P\,\cap\,\emptyset\,=\,\emptyset \end{array}" />
Well! you have shared very useful solution. I'm suffering in this question while solving.

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