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Multiplying out negated terms in boolean algebra?

Answer 1

+1 A:

You think the truth table is different?

Try evaluating it yourself.

Anon. 2009-12-16 19:10:42

Doh! I'm not sure how I convinced myself that they were different. So basically it should never make any difference then if the negation applies to the whole bracketed expression or individual component(s)?

Martin Smith 2009-12-16 19:25:01

Oh no, big diff. That's what DeMorgan is all about. If you move the 'not' into the parentheses, you have to invert the conjunction inside the parentheses.

Carl Smotricz 2009-12-16 19:27:20

Yes I had thought that I had got myself into trouble with doing something like that earlier. I'll try and revisit that example with the help of Wolfram Alpha!

Martin Smith 2009-12-16 19:38:17

Answer 2

+1 A:

You need DeMorgan's Law

Chris Dodd 2009-12-16 19:16:35

Answer 3

+3 A:

Your first expression isn't an xor, try making this substitution:

 Z = A+B

klochner 2009-12-16 19:16:58

I don't know what good the substitution would do, but I believe you're right about the first expression not being an XOR.

Carl Smotricz 2009-12-16 19:24:23

what do you get when you make the substitution in the first expression?

klochner 2009-12-16 19:26:27

(It just makes the truth value more obvious)

klochner 2009-12-16 19:32:56

Ah, I see... you're demonstrating that (1) is always false. Yep, that could be part of the problem.

Carl Smotricz 2009-12-16 19:33:37

Well spotted. That is actually a typo in the book I'm using I think.He originally states it as (A + B) . ¬(A.B) but then proceeds to start with (A + B) . ¬(A + B) in the multiplying out example.

Martin Smith 2009-12-16 19:33:57

OK. Using De Morgan (A+B). ¬(A.B) = (A+B).(¬A+¬B) and I can follow the multiplication out from there. Maybe the author quietly used De Morgan without specifying it and just took it as read the equivalence would be apparent. The book doesn't actually use the ¬ symbol but rather a continuous bar over the whole second A+B expression that I couldn't figure out how to reproduce here. Maybe I misinterpreted that as equivalent to ¬(A+B) in my attempted ASCII translation.

Martin Smith 2009-12-16 20:01:31

sounds like a typo to me. \bar{(A+B)} == ¬(A+B)

klochner 2009-12-16 20:06:49

Thanks. It all makes more sense now!

Martin Smith 2009-12-16 20:23:04

Answer 4

+1 A:

You can throw this kinda thing at Wolfram Alpha. Here's what I did:

http://www.wolframalpha.com/input/?i=truth+table+%28a+or+b%29+and+%28not+a+or+not+b%29

Please click on the link to view the results! Does that truth table look like what you thought it should, or not?

Carl Smotricz 2009-12-16 19:22:17

Thanks very useful :-)

Martin Smith 2009-12-16 19:36:39

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Multiplying out negated terms in boolean algebra?

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