# Need help with Boolean Expressions

Discussion in 'Homework Help' started by Ryuk, Apr 14, 2013.

1. ### Ryuk Thread Starter New Member

Oct 9, 2012
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0
How do I simplify this expression?

ABCD + A'B'C'D' + A'BC'D + AB'CD'

The first two terms simply to 1, right?

Could the second two terms be simplified to AC xor BD?

So...1 + ACxorBD is a simplified version of this expression?

Another one: AC' + BD' + B'D + A'C

Can this be simplified?

Last edited: Apr 14, 2013
2. ### WBahn Moderator

Mar 31, 2012
18,065
4,905
Why would they simplify to 1? And, if they did, then wouldn't that make any other terms go away since (1+anything) is (1)?

Write out a truth table for both and see if they are the same.

Again, if one of the terms is a 1, then don't all of the other terms just not matter?

One problem at a time. Let's get you to a correct solution on the first one.

screen1988 likes this.
3. ### Ryuk Thread Starter New Member

Oct 9, 2012
18
0
Nevermind, I don't know what I was thinking.

They are not. I tried this expression in a K map too. It wasn't possible to simplify.

4. ### WBahn Moderator

Mar 31, 2012
18,065
4,905
If by "simplify" you are restricted to an SOP form, then it cannot be simplified. If you are allowed to use XORs and XNORs, then it can be "simplified".

One thing can only be said to be "simplier" than another if the metrics by which "simple" are measured are clearly understood by all involved. Sadly, this is something largely lost on people that write Boolean logic optimization problems.

5. ### Ryuk Thread Starter New Member

Oct 9, 2012
18
0
Well, I believe the second expression can be written as AxorC + BxorD. I don't know about the first one though...

6. ### WBahn Moderator

Mar 31, 2012
18,065
4,905
Post your K-map and let's chip away at it some.

Apr 17, 2013
4
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ABCD + A'B'C'D' + A'BC'D + AB'CD' = X+Y (assumed)

lets consider the situations as follows;

B'D' is the common part of 2nd & 4th terms which they are A'B'C'D' + AB'CD' = X
that helps us to change the way aha aha i like it; B'D'(A'C' + AC) = X

and the other common part is BD for 1th & 3th terms at balance as ABCD + A'BC'D = Y
So; we see more easy-mind view to catch the picture as, BD(AC + A'C') = Y

finally we see the light of the tunnel; (AC + A'C')(BD + B'D') = X +Y at rest knowledge is yours..

Last edited: Apr 17, 2013