$A$$\bar{B}$ + $\bar{B}$$C$ + $\bar{A}$$C$ is the equation I am stuck on. It says it can be simplified down to $A$$\bar{B}$ + $\bar{A}$$C$. I can't find any rules that can match this. The steps(and the name of the steps eg. $1$+$A$ = $1$) to solve this would be appreciated.
Asked
Active
Viewed 51 times
1
-
Do you know Karnaugh maps? – ArsenBerk Aug 20 '18 at 17:51
-
I do. I used it to confirm that this is not fully solved. I would really preferred it if I could solve it with algebra. – John Smith Aug 20 '18 at 18:25
2 Answers
2
The rule that matches this is the Consensus Theorem, from where the following algebraic manipulations can be adapted $$\begin{aligned}A\overline{B}+\overline{B}C+\overline{A}C&=A\overline{B} + \overline{A}C+\overline{B}C\\&=A\overline{B} + \overline{A}C+(A+\overline{A})\overline{B}C\\&=A\overline{B}(1+C) + \overline{A}C(1+\overline{B})\\&=A\overline{B} + \overline{A}C \end{aligned}$$
Alijah Ahmed
- 11,609
0
Using the following equivalences:
Absorption
$P + PQ = P$
Adjacency
$PQ + P\overline{Q}=P$
you can do this:
$$A\overline{B}+\overline{B}C+\overline{A}C \overset{Adjacency}{=} $$
$$A\overline{B}+A\overline{B}C+\overline{A}\overline{B}C+\overline{A}C \overset{Absorption \ x \ 2}{=}$$
$$A\overline{B}+\overline{A}C$$
Bram28
- 100,612
- 6
- 70
- 118