I'm new to boolean algebra and having problems simplifying expressions with odd number terms,
Expressions such as:
1.
A'B'C'D + A'B'CD + AB'C'D + AB'CD + ABC'D
2.
A'BC + AB'C' + A'B'C' + AB'C + ABC
Here is my logic for both expressions:
A'B'C'D + A'B'CD + AB'C'D + AB'CD + ABC'D
A'B'D'(C+C') + AB'D(C+C') + ABC'D
A'B'D + AB'D + ABC'D
B'D(A'+A) + ABC'D
B'D + ABC'D
I never touch the last term, and I don't know what rule I'm missing. Same thing happens on the second expression:
A'BC + AB'C' + A'B'C' + AB'C + ABC
A'BC + B'C'(A+A') + AC(B'+B)
A'BC + B'C + AC
Again, one term untouched..
For 1. the result should be AC'D + B'D
For 2. the result should be B'C + BC + AC
I could maybe use Karnaugh maps but I also would like to understand the algebra logic.
Best Answer
It’s not too hard to check the results after you know them:
$$\begin{align*}B'D+ABC'D&=(B'+ABC')D\\ &=(B'+AB'C'+ABC')D\\ &=\Big(B'+A(B'+B)C')\Big)D\\ &=(B'+AC')D\\ &=B'D+AC'D \end{align*}$$
and
$$\begin{align*}A'BC + B'C + AC&=B'C+(A'B+A)C\\ &=B'C+(A'B+A+AB)C\\ &=B'C+\Big(A+(A'+A)B\Big)C\\ &=B'C+(A+B)C\\ &=B'C+AC+BC\;. \end{align*}$$
That second one can be further simplified to $AC+C=C$, since $B'C+BC=(B'+B)C=C$.
In both calculations I used the absorption law: $B'=B'+AB'C'$ in the first, and $A=A+AB$ in the second.
When you have no more than three or four proposition letters, you may find Venn diagrams helpful.