Simplifying Boolean Algebra: How to Simplify Complex Boolean Expressions

In summary, Jaehyun is trying to simplify a rule that states (A+B)' = A'B'. However, he is confused because he keeps arriving at a false equation. He recommends going a bit slower and doing it right.
  • #1
Jaehyun
5
0
Homework Statement
(A OR C) AND NOT(C AND A AND B OR C AND A AND NOT B)
or
(A + C) (CAB + CAB')'

Relevant Equations
(A+B)' = A'B'
A(B+C) = (AB) + (AC)
(AB)' = A' + B'

The attempt at a solution
I'm not sure how I'm suppose to expand (CAB + CAB')' for simplifying. I keep arriving at false which shouldn't be the case.

(A + C) (CA)' (B + B')' (I'm not sure if this is what your suppose to do)

or

(A + C) (C'A'B'C'A'B) (Not sure if i used (A+B)' = A'B' correctly)

Thanks - Jay
 
Last edited:
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  • #2
Hello Jaehyun, :welcome:

Pity you deleted part of the template: relevant equations are needed to do what you want. List a few and you'll see which you need
 
  • #3
BvU said:
Hello Jaehyun, :welcome:

Pity you deleted part of the template: relevant equations are needed to do what you want. List a few and you'll see which you need

I'm confused on how this rule: (A+B)' = A'B' is used to help with (CAB + CAB')'.
 
  • #4
I was searching for (xyp + xyq) = xy (p+q)
 
  • #5
BvU said:
I was searching for (xyp + xyq) = xy (p+q)

I don't think this rule would work as it leads to a false (CA)' (B + B')' = (CA)' (1)' = 0 unless this does not work for NOT.
 
  • #6
No. Do one step at a time.
 
  • #7
By the way, (XY)' ≠ X'Y' !
 
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Likes Jaehyun
  • #8
(A + C) (CAB + CAB')'
(A + C) (CA)' (B + B')'
(A + C) (C' + A') (B + B')'
(AC' + AA' + CC' + CA') (B + B')'
(AC' + CA') (B + B')'

So then (B + B')' = 0 meaning there is no simplified expression?
 
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  • #9
Jaehyun said:
I'm confused on how this rule: (A+B)' = A'B' is used to help with (CAB + CAB')'.
My "I was searching for (xyp + xyq) = xy (p+q)" : was meant to lure you into (CAB + CAB') = CA(B+B') = CA

(A + C) (CAB + CAB')' does NOT lead to (A + C) (CA)' (B + B')' !
 
  • #10
BvU said:
My "I was searching for (xyp + xyq) = xy (p+q)" : was meant to lure you into (CAB + CAB') = CA(B+B') = CA

(A + C) (CAB + CAB')' does NOT lead to (A + C) (CA)' (B + B')' !

But (CAB + CAB') = CA(B+B') = CA is missing the NOT portion (CAB + CAB')'.

Edit: I was rushing the question so much and I finally realized what I was doing wrong thanks BvU for putting up with me it's 2am where I live and I'm clearly not in the right mind at the moment. Solved.
 
Last edited:
  • #11
Yes. The NOT portion comes afterwards. (CA)' is easier to do than what you had before.

Great. Advice: go a bit slower :smile: The speed will come with experience; in the beginning taking small steps and doing it right are more important.
 

1. What is Boolean algebra?

Boolean algebra is a branch of mathematics that deals with logic and boolean values (true and false). It is used to simplify and manipulate logical expressions, making it easier to analyze and solve complex problems.

2. Why is simplifying Boolean algebra important?

Simplifying Boolean algebra is important because it allows us to reduce complex logical expressions into simpler forms, making it easier to understand and work with them. It also helps in reducing the number of logic gates required to implement a logical expression in a circuit, resulting in cost and time savings.

3. What are the basic laws of Boolean algebra?

The basic laws of Boolean algebra include the commutative, associative, distributive, identity, and complement laws. These laws govern how logical expressions can be simplified and manipulated to obtain equivalent expressions.

4. How do I simplify Boolean algebra expressions?

To simplify a Boolean algebra expression, you can use the basic laws of Boolean algebra to manipulate the expression. You can also use truth tables or Karnaugh maps to visualize and simplify the expression. Additionally, there are various online tools and software available that can help with simplifying Boolean algebra expressions.

5. What are some real-world applications of Boolean algebra?

Boolean algebra has many applications in computer science and engineering, such as in digital logic design, computer programming, and circuit analysis. It is also used in fields such as telecommunications, control systems, and artificial intelligence, where logical operations are necessary for decision making and problem solving.

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