# couple proofs; simplifying expressions help!

Discussion in 'Homework Help' started by PolishPower, Oct 13, 2006.

1. ### PolishPower Thread Starter New Member

Oct 13, 2006
1
0
hey guys, i got a couple questions.....

1) (A¹ + (B¹ * C)¹ * D)¹ = A (B¹ C + D¹)
when i try to simplify the left side, i get.... A * B¹ * C + D¹
so they look equal, but the right side has brackets..does that mean they are not equal?? dont know if its true or not....

2) I have no clue really how to simplify the following 2:

a) AB¹ + ABD + ABD¹ + A¹C¹D¹ + A¹BC¹
b) BD + BCD¹ + AB¹C¹D¹

any help would be greatly appreciated!! thanks!!

2. ### Mr-Synapse New Member

Nov 1, 2006
5
0
Polishpower!,
what you should be attemting to do is to gather similar terms, e.g. in 1.
Note that you show powers (?). I'm not certian just what you are intending to
illustrate, but any number to the first power is itself. Any number to the zero power is 1.
1) (A¹ + (B¹ * C)¹ * D)¹ = A (B¹ C + D¹)
simplifies to:
A+(B*C*D)

2.)
a) AB¹ + ABD + ABD¹ + A¹C¹D¹ + A¹BC¹
simplifies to:
A[B*(1+2D)+C(D+1)]

b) BD + BCD¹ + AB¹C¹D¹
simplifies to:
BD(1+A2C)

I hope that helps! You must apply the rules of Association & Commutation from Algebra.

I hope this is clearer to you.

Cheers

3. ### Mr-Synapse New Member

Nov 1, 2006
5
0
Polishpower the first answer could also be:
A*(1+BCD)

Cheers

4. ### Papabravo Expert

Feb 24, 2006
9,898
1,722
From the Department of Ambiguity and Redundancy Department. Is it possible that the superscript 1 is not an exponent but a crude notation for the boolean NOT function. The conventional ASCII notation for which is a single quote.