# Homework Help: Simplify boolean algebra

1. May 3, 2010

### ming2194

1. The problem statement, all variables and given/known data
[PLAIN]http://img340.imageshack.us/img340/7690/123rk.gif [Broken]

2. Relevant equations
The answer should be A'C' + A'D'

3. 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: May 4, 2017
2. May 4, 2010

### Staff: Mentor

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.

Last edited by a moderator: May 4, 2017
3. May 4, 2010

### 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?

4. May 4, 2010

### Staff: Mentor

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.

5. May 4, 2010

### Staff: Mentor

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

6. May 6, 2010

### Staff: Mentor

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.