Why does (A'+ AC) = (A'+C)?
I can understand this via a truth table, but I cannot see why this works in boolean algebra. Conceptually I understand that A doesn't matter, but I can not seem to prove this. Can someone lend some advice?
Asked
Active
Viewed 1,569 times
2
-
Try this using Truth table. – Sep 18 '15 at 18:57
-
I understand the truth table. Would you like me to show you the truth table I have? – Adam Sep 18 '15 at 18:58
-
I am trying to see how to prove this, using the distributive law doesn't work for me. – Adam Sep 18 '15 at 19:00
-
See this answer, I think, this should satisfy you. – Sep 18 '15 at 19:06
2 Answers
4
$$A'+AC=A'(1+C)+AC=A'+(A'+A)C=A'+C$$
-
-
2$1+C=1$, for whatever $C$ be ($0$ or $1$). So, $A'=A'\times 1=A'\times (1+C)$. Now, did you got? – Sep 18 '15 at 19:10
-
-
I never told $A=(A'+A)$, it is $A'(1+C)+AC=A'+A'C+AC=A'+(A+A')C$. Now, $A+A'=1$. So, your result. – Sep 18 '15 at 19:18
-
