# Simplify the expression using Boolean algebra?

• Aug 31st 2010, 06:24 PM
Mattpd
Simplify the expression using Boolean algebra?
(AB`+AC`)(B`+C`)+AB`D

First, I factored the A out:

A(B`+C`)(B`+C`)+AB`D

x.x=x, so:

A(B`+C`)+AB`D

Now I am stuck. Any suggestion on the next step to take?
• Aug 31st 2010, 08:30 PM
Matter_Math
Now expand :

A(B`+C`)+AB`D
AB` + AC` + AB`D
[AB` + AB`D] + AC //Use the identity X + XY = X + Y
• Aug 31st 2010, 10:27 PM
Isomorphism
Quote:

Originally Posted by Mattpd
(AB`+AC`)(B`+C`)+AB`D

First, I factored the A out:

A(B`+C`)(B`+C`)+AB`D

x.x=x, so:

A(B`+C`)+AB`D

Now I am stuck. Any suggestion on the next step to take?

I have to nitpick when the word 'simplify' appears. I think it is not well defined....

The word 'simplify' is especially not well defined for Boolean Algebra. The questions should be of the form prove one expression is another. Certain awful looking expressions in minterm expansion look compact in the maxterm expansions.

A(B'+C') + AB'D = AB'(1 + D) + AC' = AB' + AC' = A(B'+C') = A(BC)' = (A' + BC)'

You see, all the three expressions (namely A(B'+C'), A(BC)', (A' + BC)' ) are equal. So how do we decide which is "simple"?