# Boolean simplification of a larger term?

• Comp Sci
• garthenar
In summary, the conversation discusses using De Morgan's law to find the complement of a given equation. The speaker correctly applies the law and simplifies the solution, but is unable to figure out how the term A'B'CD' disappears. The expert points out that it is due to the absorption law and suggests a counterintuitive approach to simplify the problem. The final solution is then provided.
garthenar
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

garthenar said:
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.)

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
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

## What is Boolean simplification?

Boolean simplification is the process of reducing a larger, more complex logical expression into a simpler form using Boolean algebra rules and laws.

## Why is Boolean simplification important?

Boolean simplification allows us to analyze and manipulate logical expressions more easily, making it an essential tool in digital logic design, computer programming, and other scientific fields.

## What are the basic rules of Boolean simplification?

The basic rules of Boolean simplification include the commutative, associative, and distributive laws, as well as De Morgan's laws and the identities for AND, OR, and NOT operations.

## How do I simplify a Boolean expression?

To simplify a Boolean expression, you can use truth tables, logic laws, or Karnaugh maps. First, identify redundant terms and eliminate them using the basic rules. Then, combine terms with the same variables using the appropriate laws.

## What are some applications of Boolean simplification?

Boolean simplification is used in various applications such as digital circuit design, computer programming, database querying, and information retrieval. It is also utilized in mathematical proofs and logical reasoning.