# Simplify boolean expression.

• Feb 27th 2011, 10:30 PM
Phrass
Simplify boolean expression.
i'm having difficulty solving these few problems

1. solve (BC ' + A ' D) ( A B ' + C D ')

2. Simplify the Boolean expression
A C ' + B ' D + A ' C D + A B C D

i got to the first part or what i think is the first part.. but seem to be stuck from there on.
• Feb 27th 2011, 11:09 PM
emakarov
What do you mean by solving problem 1?

Quote:

Originally Posted by Phrass
i got to the first part or what i think is the first part.. but seem to be stuck from there on.

What do you mean by the "first part"? Are you talking about problem 1 or 2?

Quote:

2. Simplify the Boolean expression
A C ' + B ' D + A ' C D + A B C D
This can be done, for example, using Karnaugh maps. I got AC' + CD + A'B'D. It requires some time to explain the solution, so please tell us what ways of simplifying Boolean expressions you have studied and what you have tried.
• Feb 28th 2011, 03:03 AM
Phrass
RE: Problem #1
(BC ' + A ' D) ( A B ' + C D ')
therefore: BC'+A'D AB'+CD'
BC'+D+B'............(Thinking)

Learn Simplification in these ways.. however i'm losing it where the A' is concerned
(A  B)  C = A  (B  C)
(A + B) + C = A + (B + C)
A  B = B  A
A + B = B + A
A  (B + C) = A  B + A  C
A + (B  C) = (A + B)  (A + C)
• Feb 28th 2011, 05:50 AM
emakarov
Quote:

RE: Problem #1
(BC ' + A ' D) ( A B ' + C D ')
therefore: BC'+A'D AB'+CD'
BC'+D+B'............
I still don't understand what you mean by solving a Boolean formula. Do you mean simplifying? Presenting it as a sum of products or of minterms? For example, in regular algebra, what does it mean to solve (x + y)z? This is not even an equation. If it were (x + y)z = 0, then you could try to find roots, even though there are many of them.

Quote:

Learn Simplification in these ways.. however i'm losing it where the A' is concerned
(A  B)  C = A  (B  C)
(A + B) + C = A + (B + C)
A  B = B  A
A + B = B + A
A  (B + C) = A  B + A  C
A + (B  C) = (A + B)  (A + C)
First, my browser does not correctly render characters you are using for multiplication, or conjunction. Use * or /\ or just omit it, as they do for multiplication.

Below I am using these identities.

A = A * 1 = A(B + B') = AB + AB' (1)
A + AB = A * 1 + AB = A(1 + B) = A * 1 = A (2)

I also enclose the expression that is changed from one line to the next in square brackets. So,

AC' + [B'D] + A'CD + ABCD = (by (1))
AC' + [AB'D] + A'B'D + A'CD + ABCD = (by (1))
[AC'] + AB'CD + [AB'C'D] + A'B'D + A'CD + ABCD = (by (2))
AC' + [AB'CD] + A'B'D + [A'CD] + [ABCD] = (by distributivity)
AC' + A'B'D + ([AB'] + A' + [AB])CD = (by (1))
AC' + A'B'D + ([A + A'])CD =
AC' + A'B'D + CD

In finding this sequence of transformations, I was guided by the Karnaugh map. I have not thought about how to find it otherwise, though probably there are ways.
• Mar 1st 2011, 02:00 AM
Phrass
Quote:

Originally Posted by emakarov
I still don't understand what you mean by solving a Boolean formula. Do you mean simplifying?

Yes simplifying..

Im following the steps in the one here to understand, so distributive rule was used to breakdown the fourth line
• Mar 1st 2011, 02:27 AM
emakarov
Quote:

(BC' + A'D) (AB' + CD')
Proceed as in regular algebra. You get BC'AB' + BC'CD' + A'DAB' + A'DCD'. Each of the four terms contains complementary literals. For example, BC'AB' contains B and B', BC'CD' contains C and C'. For any X, XX' = 0, and for any formula F, F0 = 0, so the whole formula equals 0.
• Mar 1st 2011, 02:58 AM
Phrass
SINCE :BC'AB' + BC'CD' + A'DAB' + A'DCD'
therefore
B(C'AB')+C(BC'CD)+A(A'DAB')+D(A'DCD')
C'A+BD+DAB'+AC
CA+1+A+AC
(A+1)/A
• Mar 1st 2011, 03:41 AM
emakarov
Quote:

B(C'AB')+C(BC'CD)+A(A'DAB')+D(A'DCD')
C'A+BD+DAB'+AC
CA+1+A+AC
(A+1)/A
I don't understand the relationship between different lines in the quote above. I also don't know what / means in the context of Boolean formulas. I am not sure whether you agree with my statement that the formula equals 0 or if you are trying to prove that it is equal to something different.

Looking at the first two lines, you may confuse product, which is conjunction ("and"), with sum, which is disjunction ("or"). For example, BC'AB' = AC'(BB'). Here BB' is "B and not-B", i.e., this is always false (0). Therefore, AC'(BB') = AC'0 = 0. If we had A + C' + (B + B'), this is equal to A + C' + 1 = A + C' because B + B' = 1 ("B or not-B" is always true).
• Mar 1st 2011, 05:57 PM
Phrass
oh by the / i meant "or" sorry about that...
i think i am confused.. im going through the other one step by step i need to get this