Understanding the simplification of a boolean expression

Question

$ABC+AB'C+A'BC+A'BC'= AB'C+A'BC+A'BC$

I was reading a book about digital systems and it said that i could supress $ABC$ because it belongs to $BC$ but i can't just look to the expression and have a clear understanding of how this works.

I can look for this expression $A+AB'$ and imagine that $AB'$ can be a extension of $A$ but in the case above exist more than just $2$ variables to see the explanation given by the book clear.

Answer 1

$ABC +BC = ABC +(A + A') BC$ (Since, $A+A'=1$)

$= (ABC+ ABC)+ A'BC$

$= ABC+A'BC$ (Since $A+A=A$)

$= BC$

Here, $A'$ is the complement of $A$.

This is basically the absorption law of Boolean Algebra. $A$ consists of two cases- ($A$ and $B$), as well as ($A$ and $B'$), i.e. $A = AB + AB'$. Since $AB$ is already included in $A$, it's presence or absence makes no difference to the final value of the expression. In this sense, you can "absorb" $AB$ in $A$.

The above can extended to greater numbers of variables, for instance -

$A= ABC+ABC'+AB'C+AB'C'$

so any of these $4$ terms can be absorbed into $A$.

Answer 2

I think what you mean is that you can "factor" (it is the right word, but in the context of boolean algebra, however it works the same as the usual factorization) $ABC+\neg ABC = (A+\neg A)BC = BC$ Since $A+\neg A=1$