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I understand the basics of boolean algebra and how to simplify them. What I am confused about is how do we know what to call a value after simplifying? Imagine we create a boolean algebra expression in canonical form. When simplifying we come across this: $$A[B(C+C')]$$. What do I do when I simplify that? On the inner most parentheses $C+C' = 1$. Does it then become the same variable name only as a one, ie. true? After that do we have $A[BC]$ which is equivalent to $ABC = 1$?

If I understand this correctly when you simplify the result just becomes the variable name but it will be zero or one as the result?

Another thing that bugs me is how do I know how to factor? If I had $ABC'+AB'C$, would I factor out the A because the A is always one in each term while the other variables are different? What if I couldn't find a common variable in the smallest number of terms? Would it be unfactorable and is that even possible in Boolean algebra?

Thank you for your time, I'm a complete novice to Boolean algebra but I am quite good at decimal algebra.

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2 Answers 2

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$$A[B(C+C')] = A[B(1)] = AB$$

(Because $B1 = B$.)


$$ABC'+AB'C = A (BC'+ B'C) \tag{distributive property}$$

If there were no common factors, e.g. $$A'BC' + AB'C$$ then the expression itself is simplified.

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  • $\begingroup$ The value of the simplified expressions depend entirely upon the truth values (0 or 1) of A, B in the first case, and of A, B, and C in the second case. e.g., AB = 1 if and only if A = 1 and B = 1. Else, AB = 0. $\endgroup$
    – amWhy
    Commented Feb 18, 2013 at 22:50
  • $\begingroup$ :-) Thanks, Babak! $\endgroup$
    – amWhy
    Commented Feb 20, 2013 at 2:34
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If you have $A[B(C+C')]$ you are correct that $C+C'=1$. Now we have $A[B1]$. But $B1=B$ so the final result is $AB$. You can confirm this with a truth table easily.

In your second, yes, you can only distribute out a common term. So $ABC'+AB'C=A(BC'+B'C)$. It is quite possible to have a combination of terms that cannot be factored. $A'BC'+AB'C$ would be one such.

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