How to Simplify Boolean Equations with Continuous Bars?

[cact]

I would like to thank everyone for the help on the last Boolean Equation but know I seem to be stuck on simplifing the last two.
any help in simplifing these would be excellent:
Here they are;

1. (A+B)(A+B)= X THE FIRST (A+B) IN THE EQUATION HAS A CONTINIOUS BAR ABOVE BOTH LETTERS




2. (AB)(A+B)= X THE (AB) IN THE EQUATION HAS A CONTINOUS BAR ABOVE BOTH LETTERS



THANKS AGAIN

 

[AKG]

1. So

\overline{(A + B)}(A + B) = X\mbox{?}

If this is the case, this should be really, really, really easy. Have you even tried? For a "hint", let C = A + B, so \overline{C}C = X. It should be pretty obvious what \overline{C}C is. I have to assume you know how to deal with truth tables. You could always just solve these using truth tables, so I really wonder if you've tried anything at all.

2. You really aren't trying. What do you know about Boolean Algebra? Surely you know things like A(B + C) = AB + AC. The basic properties like distributivity should have been taught to you or should be in your book or whatever source you're learning from. If these are the problems you got stuck on, then I'm assuming there are problems you know how to do. If you know how to do any problems, you should know how to do these as they don't really get any more basic. I really suggest you try again before asking for help.

 

[cact]

K, I have tried and I'm pretty sure i understand the last one however I feel that it won't hurt to ask. Basic ones I understand however when it comes to the bars above the letter and truth tables I'm still unsure about since our teacher sucks at explaining it. Which is why I'm trying to find a tutor in the area, but in the meantime I'm trying to understand it.
for the third one this is what I got so if you could help it would be great, exspecially for the 2nd question...


3. (AB)(A+B)=X
AbarAB+ABBar=X
(0)B+A(0)=X
AB=X

Does that seem right?

 

[K.J.Healey]

which in that has bars, any?
AB(A+B)
ABA + ABB
AB + AB
AB

 

[cact]

yes, the first set of brackets (AB) has a bar above it.

 

[K.J.Healey]

X= (AB)_ * ( A+B )
we know (AB)_ = A_ + B_
X= (A_ + B_ ) ( A+B)
X= AA_ +AB_ +BA_ +BB_
X= 0+AB_ + BA_ + 0
X= AB_ + BA_

the logic table for that is an "exclusive or"
X = A XOR B

 

[Hurkyl]

There are a couple standard ways to write \bar{A} in ascii: ~A or !A are direct translations, depending on context, and 1+A is an equivalent expression. (When + is XOR)

 

[cact]

okay after reading through this textbook again I'm really confused;
would the rules for solving AB(A+B)=X and (AB)(A+B)=X not be the same?