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

[ming2194]

Homework Statement

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

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.

 

[Mark44]

Homework Statement

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

You have a mistake in the second line, in the second term. How did you get
\bar{A}\bar{B}\bar{D}(\bar{C} + C)?

I would try factoring \bar{A}\bar{D} from the 3rd, 4th, and 6th terms of the line above there.

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.

 

[ming2194]

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?

 

[Mark44]

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.

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?

 

[Mark44]

Also, I get A'C' + A'D'C for the final answer. Are you sure you have posted this problem exactly as given?

 

[Mark44]

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.