I'm currently learning about boolean expressions. Currently, we're looking at how to simplify them. For this question I need to show the intermediate steps using the laws of boolean algebra.
However, I'm not sure where to start with this one (overline is negation):
$A\overline{B} + \overline{B}C + AB\overline{C}$
needs to be simplified to
$\overline{B}C + A\overline{C}$
Can anyone help me out? Thank you!
On
There is a mistake in your question. The two expressions cannot possibly be the same, as the first is clearly degree 3 and the second is degree 2.
They are nearly the same, and only vary for $A=1, B=0, C=1$ (you may substitute $1=$True, $0=$False if you prefer that) In this case $$ AB' + B'C + ABC' = 1 + 1 + 0 = 0 $$ $$ B'C + AC' = 1 + 0 = 1 $$ I believe your question should be to show that $AB' + B'C + ABC' + AB'C$ simplifies to $B'C + AC'$. The easiest way is to expand both expressions to show they possess the same clauses.
On
Under the assumption that $+$ stands for disjunction (OR), the given formula is---as noted---indeed an identity. One quick and painless way to see it is to apply the consensus theorem (which, incidentally, is the same as resolution in the propositional case) twice and absorption.
The consensus theorem says that $xy + \overline{y}z = xy + \overline{y}z + xz$. Absorption says that $x + xy = x$.
In our case the consensus of $A\overline{B}$ and $AB\overline{C}$ is $A\overline{C}$, which absorbs $AB\overline{C}$. Then $A\overline{B}$ is the consensus of $A\overline{C}$ and $\overline{B}C$.
I will use $'$ for negation, it's just more convenient.
\begin{split}AB' + B'C + ABC' & = AB'(C'+C) + B'C + ABC'\\ & = AB'C' + AB'C + B'C + ABC' \\ & =AB'C' + ABC' + AB'C + B'C \quad (\text{rearranging terms}) \\ & = AC'(B'+B) + B'C(A+1) \\ & = AC' +B'C\end{split}