Convert a'+b(a+b')(b+c') to sum of products and product of sums using boolean algebra.

Question

Sum of products

F = a'+b(a+b')(b+c')
  = a'+ ab + bb' (b+c')
  = a'+ab

So that's about as far as I've gotten. I'm trying to find a step by step guide on how to do these sorts of problems but still unclear as to how. Any sort of directions would be appreciated.

What did you do with the term (b+c')? I can't quite get if you want to multiply it or add it. — najayaz, Sep 30 '15 at 08:31
I thought bb' evaluates to 0 and 0 and anything can be disregarded? — borderlineNovice, Sep 30 '15 at 14:37
Yeah that's okay but what I was talking about was that you didn't multiply the term $ab$ with $(b+c')$ — najayaz, Sep 30 '15 at 15:27

score 3 · Accepted Answer · answered Sep 30 '15 at 19:41

You can find a good summary of the laws of boolean algebra at http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/. Note that their notation uses $\overline{A}$ instead of $A'$ for the NOT operator.

Then $F$ can be reduced to a much simpler form:

$$\begin{align} F &= a'+b(a+b')(b+c') \\ &= a'+(ba+bb')(b+c') &(\text{distributive law}) \\ &= a'+(ab+0)(b+c') &(\text{commutative law, redundance law (T9)}) \\ &= a'+ab(b+c') &(\text{redundance law (T7)}) \\ &= a'+abb+abc' &(\text{distributive law}) \\ &= a'+ab+abc' &(\text{redundance law (T6)}) \\ &= a'+ab(1+c') &(\text{distributive law}) \\ &= a'+ab &(\text{redundance law (T8)}) \\ &= a'+b &(\text{redundance law (T10)}) \\ \end{align}$$

Convert a'+b(a+b')(b+c') to sum of products and product of sums using boolean algebra.

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