Simplify this Boolean Algebra. Can use Karnaugh Maps. ('A = Not A)

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB)

which makes just 'C'D.

Is this right?

2 posts
• Page **1** of **1**

Simplify this Boolean Algebra. Can use Karnaugh Maps. ('A = Not A)

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB)

which makes just 'C'D.

Is this right?

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB)

which makes just 'C'D.

Is this right?

- stumpy
**Posts:**1**Joined:**Tue Dec 03, 2013 7:42 pm

stumpy wrote:Simplify this Boolean Algebra.

'A'B'C'D+'AB'C'D+A'B'C'D+AB'C'D

I have 'C'D('A'B+'AB+A'B+AB) which makes just 'C'D.

Is this right?

The factorization is valid, and the summands in the parentheses cancel out pairwise. The canceled pairs equal "1", and the sum (in Boolean algebra) of 1 + 1 is itself equal to 1. So you have 'C'D(1), which equals 'C'D.

Your answer looks correct to me.

- anonmeans
**Posts:**52**Joined:**Sat Jan 24, 2009 7:18 pm

2 posts
• Page **1** of **1**