# Complex Boolean Expression

Discussion in 'Math' started by ibcoding, Apr 13, 2013.

1. ### ibcoding Thread Starter New Member

Apr 13, 2013
7
0
Can someone step me through the minimization of this expression:

((ab)'(b'c)' + a'bpc')

2. ### ibcoding Thread Starter New Member

Apr 13, 2013
7
0
The reason I'm asking is I'm getting only a true value at the fifth row of a truth table. i.e. 0101

3. ### ibcoding Thread Starter New Member

Apr 13, 2013
7
0
Does it break down to a' + p'

4. ### justtrying Active Member

Mar 9, 2011
330
804
did you apply the following rule:

(ab)' = a' + b'

?

5. ### WBahn Moderator

Mar 31, 2012
23,089
6,941
Don't you think it would help if you told us how {a,b,c,p} map to the bits in 0101? We are not mind readers.

6. ### WBahn Moderator

Mar 31, 2012
23,089
6,941
As near as I can see, you should have two two-factor terms.

You need to show us your work so that we can see where you are going wrong.

7. ### WBahn Moderator

Mar 31, 2012
23,089
6,941
Do a sanity check. If only one row of the table is true, that means that there is exactly one combination of inputs that can make the result true. Looking at the original expression, that term has to be (a'bpc'), meaning that you have

a=0
b=1
p=1
c=0

So now see if you can get the other term, (ab)'(b'c)', to be true from some combination of inputs other than the one above. Well, in order to make (xy)' a 1, we need to make (xy) a 0 which only takes making one or the other a 0. So we can trivially set a=0 and b=0 and not it doesn't matter what values b or p take on. Hence we have just identified FOUR rows for which the result is a 1. In fact, I believe you will find there are eight.