# Boolean Algebra simplification - odd number terms

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.

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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.

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