Simplify Boolean Algebra: A'C' + A'D' Solution Example

In summary, you algebraically solve for A by multiplying each term of the equation by its respective Y value.
  • #1
ming2194
32
0

Homework Statement


[PLAIN]http://img340.imageshack.us/img340/7690/123rk.gif [Broken]

Homework Equations


The answer should be A'C' + A'D'

The Attempt at a Solution


Shown in above in the question.

I want to know how can i get the correct answer. As you see above, in my last step, I wonder whether B'+BD and B'+BC are equal to 1 ? If so, then I can get the answer. But if dont, where is my problem?

Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
ming2194 said:

Homework Statement


[PLAIN]http://img340.imageshack.us/img340/7690/123rk.gif [Broken]
You have a mistake in the second line, in the second term. How did you get
[tex]\bar{A}\bar{B}\bar{D}(\bar{C} + C)[/tex]?

I would try factoring [tex]\bar{A}\bar{D}[/tex] from the 3rd, 4th, and 6th terms of the line above there.
ming2194 said:

Homework Equations


The answer should be A'C' + A'D'


The Attempt at a Solution


Shown in above in the question.

I want to know how can i get the correct answer. As you see above, in my last step, I wonder whether B'+BD and B'+BC are equal to 1 ? If so, then I can get the answer. But if dont, where is my problem?

Thanks.
 
Last edited by a moderator:
  • #3
I am trying to extract the A'B'D' from the 1st and 3nd term, what's mistake?

and I tried your method and found A'C'(B'D'+B'D+BD)+A'D'(B'C'+BC'+BC). Seemingly not help enough. What can I do?
 
  • #4
ming2194 said:
I am trying to extract the A'B'D' from the 1st and 3nd term, what's mistake?
The mistake is that you have already used A'B'D'. You can't use it twice. In the first line, the 1st and 2nd terms on the right are A'B'C'D' + A'B'C'D, which factor into A'B'C'(D' + D). The remaining terms in the first line are A'B'CD' + A'BC'D' + A'BC'D + A'BCD'. The first of these terms (which is the 3rd term in the first line) has a factor of A'B'D', but none of the other three terms has this factor.
ming2194 said:
and I tried your method and found A'C'(B'D'+B'D+BD)+A'D'(B'C'+BC'+BC). Seemingly not help enough. What can I do?
 
  • #5
Also, I get A'C' + A'D'C for the final answer. Are you sure you have posted this problem exactly as given?
 
  • #6
After working on this some more, and with a tip from another forum member who sent me a PM, I have arrived at A'C' + A'D', the answer you posted earlier.
P(A, B, C, D) = A'B'C'D' + A'B'C'D + A'B'CD' + A'BC'D' + A'BC'D + A'BCD'
= A'B'C'(D' + D) ; 1 and 2 - the numbers refer to the position in the equation above
+ A'D'(B'C + BC) ; 3 and 6
+ A'C'(BD' + BD) ; 4 and 5
= A'B'C' + A'D'C(B' + B) + A'C'B(D' + D)

After working with the expression above some more, you can arrive at the following expression after a few steps:
A'(C' + CD')

An identity can be used to work with the expression in parentheses.

1 = 1 + Y, so X(1) = X(1 + Y),
so X = X + XY

This means that X + X'Y = X + XY + X'Y = X + Y(X + X') = X + Y(1) = X + Y.
 

1. What is boolean algebra?

Boolean algebra is a mathematical system used to simplify and manipulate logical expressions involving variables that can only have two values: true or false. It is named after mathematician George Boole and is often used in digital logic design and computer programming.

2. Why is simplifying boolean algebra important?

Simplifying boolean algebra is important for several reasons. It can help reduce complex logical expressions to simpler forms, making them easier to understand and work with. It can also improve the efficiency of digital circuits by reducing the number of logic gates needed. Additionally, simplifying boolean algebra can help identify redundant or contradictory statements in a logical expression.

3. What are the basic laws of boolean algebra?

The basic laws of boolean algebra include the commutative, associative, and distributive laws. The commutative law states that the order of operands does not affect the result of a logical operation. The associative law states that the grouping of operands does not affect the result. The distributive law states that a logical operation can be distributed over another logical operation.

4. How do I simplify boolean algebra expressions?

To simplify boolean algebra expressions, you can use the basic laws, such as commutative, associative, and distributive laws, along with other rules such as De Morgan's laws and the identity laws. You can also use Karnaugh maps and truth tables to help with simplification. It is important to follow a systematic approach and use a combination of these techniques to simplify the expression as much as possible.

5. Are there any tools that can help with simplifying boolean algebra?

Yes, there are several tools available that can help with simplifying boolean algebra, such as online boolean algebra simplifiers, software programs like Mathematica and Matlab, and truth table generators. These tools can quickly and accurately simplify complex boolean expressions, saving time and effort in manual simplification.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
11
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
14
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
Back
Top