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(a + b)(b' + c')(a + b' + c)

where b' = b not and c' = c not.

I tried distributive because I'm not very good at applying the properties when multiplication is applied but I can with addition.

(a + b)(b' + c') = ab' + ac' + b'b + bc' = ab' + ac' + bc' I think it can be simplified further but I'm not sure how. It would just result in two different terms in the distributive property no matter how you look at it, and combining it with the other equation via distributive would just create a mess.

Am I missing something?

Sal
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3 Answers3

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Yes your guess that it can be simplified further is right, but not so easy to see.

$ab'+ac'+bc' = ab'+bc'$ because the middle term is covered in all cases ($b$ or $b'$) by one of the other two.

That said, it's still a slight chore to multiply $(ab'+bc')(a+b'+c)$. I don't see an easy way out of this. In general, beyond a few variables it does get very difficult to minimize boolean expressions.

user21820
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  • I hear ya. It's annoying that POS are a lot harder to do since you have to multiply them out than SOP, but thanks a bunch! – Sal Feb 05 '16 at 06:10
  • @Sal: If you're interested, the general problem is NP-hard. Also, if my answer works for you then you can accept it by clicking the tick. =) – user21820 Feb 05 '16 at 06:12
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Recall that each operation distributes over the other, so that $(x+y)(x+z)=x+yz$. This suggests that expanding $(a+b)(a+b'+c)$ might be a good approach, since part of the expansion will be the product $(a+b)(a+b')$. And in fact we find that

$$\begin{align*} (a+b)(a+b'+c)&=(a+b)(a+b')+(a+b)c\\ &=a+bb'+(a+b)c\\ &=a+0+ac+bc\\ &=a(1+c)+bc\\ &=a+bc\;. \end{align*}$$

Multiplying this by $b'+c'$ now yields

$$\begin{align*} (a+bc)(b'+c')&=ab'+ac'+bcb'+bcc'\\ &=ab'+ac'\;, \end{align*}$$

which can be checked with a truth table.

Brian M. Scott
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    Would the downvoter care to indicate what s/he thinks is wrong here? – Brian M. Scott Feb 06 '16 at 16:05
  • I don't think there's anything wrong here. In fact, your answer should be accepted. I noticed that a lot of my correct answers on Math SE always get downvoted by people who don't really know mathematics, while blatantly incorrect answers get upvoted. Kind of annoying sometimes. – user21820 Feb 15 '16 at 02:59
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The expression

$$(a + b)(b' + c')(a + b' + c) $$

can be expediently simplified by application of the consensus theorem (i.e., by application of $(x+y)(y'+z) = (x+y)(y'+z)(x+z)$). The consensus of the last two terms is $(a + b')$, which absorbs the third term:

$$ (a + b)(b' + c')(a + b') \enspace.$$

The consensus of the first and third terms is $a$, which absorbs the first and third terms:

$$ a (b' + c') \enspace. $$

Fabio Somenzi
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