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so this might sound stupid but I have a problem when it comes to simplifying terms. lets take this simple term as example (it was taken from a truth-table where these are the 0 results, that means DNF right?): $$\neg a * b * \neg c * \neg d + \neg a * b * c * \neg d + a * b * \neg c * \neg d + a * b * c * \neg d$$

My first step would be looking for multiple appearances of a variable, I took b because it is in every block. I would do it like this:

  1. $b*(\neg a + \neg c + \neg d + \neg a + c + \neg d + a +\neg c + \neg d +a + c +\neg d)$, distributive law?
  2. sort: $b*(a+a+\neg a+\neg a+c+c+\neg c+\neg c+\neg d+\neg d+\neg d+\neg d)$, commutative law?
  3. $b*(a+\neg a+c+\neg c+\neg d)$, because $a+a = a$
  4. $b*(1+1+\neg d)$, because $a+ \neg a = 1 $
  5. $b*(1+\neg d)$
  6. $b*(1)$, because $a+1 = 1$
  7. $b$

so my result would be $b$ but it seems wrong. What am I doing wrong? it feels like I am always using the wrong rule!

I appreciate your help, I have looked through other asked questions but they couldn't answer my question! I know that I have knowledge gaps in terms of simple mathematical term-laws

Frenzy Li
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Zesa Rex
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2 Answers2

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You're changing multiplication into addition! It's like doing $$ abcd+xbyd=b(a+c+d+x+y+d) $$ whereas it should be $$ b(acd+yxd) $$

Starting from $$ \neg a * b * \neg c * \neg d + \neg a * b * c * \neg d + a * b * \neg c * \neg d + a * b * c * \neg d $$ collect $b*\neg d$ and go on: \begin{align} b*\neg d*(\neg a*\neg c+\neg a*c+a*\neg c+a*c) &=b*\neg d*\bigl(\neg a*(\neg c+c)+a*(\neg c+c)\bigr)\\ &=b*\neg d*(\neg a+a) \\ &=b*\neg d \end{align}

In slow motion: consider the expression in parentheses $$ \neg a*\neg c+\neg a*c+a*\neg c+a*c $$ You can collect $\neg a$ from the first two terms and $a$ from the other two, getting $$ \neg a*(\neg c+c)+a*(\neg c+c) $$ Now recall that $\neg c+c=1$, so we get $$ \neg a*1+a*1=\neg a+a=1 $$

egreg
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  • I could need an answer telling me what I am doing wrong, not the correct answer. the first problem I found is that I replaced every ∗ with a + in my first try. I will try to work on from that, even though a little correction where my logical failures are would be nice – Zesa Rex Apr 26 '17 at 11:10
  • @ZesaRex Sorry, but I thought you'd understand the mistake from the solution. Anyway, I added some words. Once you get the feeling, the simplification is straightforward, as I showed. – egreg Apr 26 '17 at 11:13
  • thanks, I think the only problem left is that I have to understand what you did in this step: b∗¬d∗(¬a∗(¬c+c)+a∗(¬c+c)), I am sorry, this is probably very, very basic stuff but I just need to understand the logic behind it – Zesa Rex Apr 26 '17 at 11:16
  • @ZesaRex Added a few details – egreg Apr 26 '17 at 11:26
  • ahh thank you, it makes sense now, so we try to factorize as much as possible first? I think that will solve most of my problems, i will try to simplify some terms now – Zesa Rex Apr 26 '17 at 12:05
  • @ZesaRex If you can collect terms, do it, of course; but it's not generally possible and other strategies such as “partial collecting” become useful. – egreg Apr 26 '17 at 12:14
  • yes, I think if you do it often enough you will encounter almost every possibility and after that you certainly know how it works. that's what I will do for now – Zesa Rex Apr 26 '17 at 12:19
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Your mistake is in the very first step.

Taking out the $b$ from:

$$\neg a * b * \neg c * \neg d + \neg a * b * c * \neg d + a * b * \neg c * \neg d + a * b * c * \neg d$$

you should get:

$$b*(\neg a * \neg c * \neg d + \neg a * c * \neg d + a * \neg c * \neg d + a * c * \neg d)$$

Bram28
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