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.

Answer 1

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:

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