Simplifying a Boolean Function: Is This the Correct Solution?

In summary, the original function F(A,B,C,D) = BC + (A + C'D') needs to be simplified. De Morgan's Law is used to simplify the second step, resulting in F(A,B,C,D) = BC + (A + (CD)'). However, the function should actually be F'(A,B,C,D) = BC + (A + (CD)'). Taking the complement of the function is not necessary and results in an incorrect solution.
  • #1
kachilous
15
0
I have the following function to be reduced/simplified.

F(A,B,C,D) = BC + (A + C'D') where ' denotes the complement

Here's my solution:

= BC + (A + C'D')'

= BC + (A + (C+D)

= BC + (A + C + D)

= BC + C + A + D

= C(B + 1) + A + D

= C*1 + A + D

= C + A + D

Is this correct?
 
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  • #2
Is the 1st line wrong or is the original equation wrong? There's an extra complement in there at the end.

Is the 2nd line complete? There's at the very least a bracket missing.

De Morgans Law: A'B' = A' + B'
 
  • #3
on the second step, I did De Morgans
 
  • #4
We got that, but the first line of your work doesn't follow from the problem statement.
Zryn was asking about the last complement below. It shouldn't be there.
BC + (A + C'D') is not equal to BC + (A + C'D')' <--- this one
 
  • #5
My apologies. The function should actually be F'(A,B,C,D) = BC + (A + (CD)')
 
  • #7
because I took the complement of the function F. So I'm reducing F'
 
  • #8
You aren't supposed to take the complement of the function - just write its formula in the simplest form. Am I missing something here?
 
  • #9
Perhaps you could write the original question verbatim and we can start again with your reasoning and results?
 

Related to Simplifying a Boolean Function: Is This the Correct Solution?

1. What is Boolean Algebra Reduction?

Boolean Algebra Reduction is a process used to simplify Boolean expressions. It involves using basic Boolean algebra rules to reduce a complex expression into a simpler form that is easier to analyze and understand.

2. What are the benefits of using Boolean Algebra Reduction?

Boolean Algebra Reduction can help to make Boolean expressions more compact, which can save time and effort when analyzing and designing digital circuits. It also helps to identify redundant terms and eliminate errors in expressions.

3. What are the steps involved in Boolean Algebra Reduction?

The steps involved in Boolean Algebra Reduction include identifying and applying basic Boolean algebra rules, such as the commutative, associative, distributive, and De Morgan's laws. The goal is to manipulate the expression into a simplified form using these rules.

4. Can Boolean Algebra Reduction be applied to any Boolean expression?

Yes, Boolean Algebra Reduction can be applied to any Boolean expression. However, the complexity of the expression may vary, and some expressions may require more steps to simplify than others.

5. How can Boolean Algebra Reduction be used in real-world applications?

Boolean Algebra Reduction is commonly used in digital logic design, as well as in computer programming and database query optimization. It can also be applied in other fields such as circuit analysis, electrical engineering, and computer science.

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