# Boolean simplification of a larger term?

• Comp Sci

## Homework Statement:

Not a specific hwk problem, but I keep getting stuck at a certain point in my Boolean simplifications. I'm hoping someone can help me figure out what steps/rules I need to apply to further simplify my Boolean expressions.

## Relevant Equations:

The basic rules/laws/theorems of Boolean Algebra
I'm suspecting consensus theorem in particular.
I was asked to use De Morgans law to find the complement of a particular equation, I applied the law correctly and simplified my solution down to

A'B'CD'+CA+CB'+D'A+D'B'

I ran the problem through a boolean simplifier to check my work. (http://tma.main.jp/logic/logic.php?lang=en&type=eq&eq=~((A+++~(~BC~D))+(~AB+++~CD)+)) and it gave me

CA+CB'+D'A+D'B'

I can't figure out how they got rid of the A'B'CD' term (is that called a term or something else?)

I have run into this situation multiple times and I just can't figure out how to get rid of the larger term (A'B'CD' term in this case). I'm sure it's simple but sometimes you just need a push in the right direction. I suspect it takes multiple steps of manipulation to pull off and that I'm just not seeing how to do that.

I've attached my work so far

Related Engineering and Comp Sci Homework Help News on Phys.org
pbuk
Gold Member
I suspect it takes multiple steps of manipulation to pull off and that I'm just not seeing how to do that.
Why do you think that? The term A'B'CD' has simply disappeared from the sentence, what does this tell you?

garthenar
That it's redundant. But I need to know what laws cause it to be redundant. (In my class we're required to list the laws we use with each step.)

pbuk
Gold Member
First of all, write your terms down consistently: A'B'CD'+CA+CB'+D'A+D'B' = A'B'CD'+ AC+B'C+AD'+B'D'. Now look at A'B'CD' alongside B'C; what do you notice?

garthenar
I notice that I forgot about the absorption law.

Thank you. I knew it would be something obvious.

pbuk
Joshy
Gold Member
You could agree that ##X+X## is still ##X##?

To simplify your problem (a counter intuitive approach) add another ##\bar A \bar B C \bar D## to the equation so that you have

$$(\bar A \bar B C \bar D) + AC + \bar B C + (\bar A \bar B C \bar D) + A \bar D + \bar B \bar D$$

Pull the ##C## on the left side and the ##\bar D## on the right side for

$$C(\bar A \bar B \bar D + A + \bar B) + \bar D (\bar A \bar B C + A + \bar B)$$

Simplify the terms within the parenthesis, which I think you'll be able to follow.

Yikes! You already got it. Great job.

garthenar