# Can You prove this ??

Discussion in 'Homework Help' started by just-gal, May 25, 2009.

1. ### just-gal Thread Starter New Member

May 25, 2009
5
0
hi there ,

(A+B) (A`+C) = AC+A`B

and

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

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Take a moment to review the material on Boolean Algebra contained in the AAC ebook. I think you will find it a good supplement to any information in your course textbook.

hgmjr

3. ### steveb Senior Member

Jul 3, 2008
2,433
469
Both problems are the same. So when you solve the first one, the second one is done.

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Try laying out a truth table for each, this will constitute a proof.

5. ### just-gal Thread Starter New Member

May 25, 2009
5
0
I know how to prove it using truth table but I need to prove it mathematically

can you help me in this ?

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
Do you know how to multiply out the left hand side of the first equation?

If you multiply it out, you will have 4 terms. Two of these terms will match the right hand side. Can you figure out why the extra two terms are redundant and not necessary?

7. ### Ratch New Member

Mar 20, 2007
1,068
3
just-gal,

Yes, I can.
Code ( (Unknown Language)):
1.
2. (A+B)(A'+C) = AC+A'B                     ; equation to be proved
3. (A+B)(A'+C) = AA'+AC+A'B+BC = AC+A'B+BC  ; AA' = 0
4. A(B+B')C+A'B)(C+C')+(A+A')BC             ; (X+X') = 1
5. ABC+AB'C+A'BC+A'BC'+ABC+A'BC             ; expand expression
6. ABC+AB'C+A'BC+A'BC'                      ; eliminate redundant terms
7. AC(B+B')+A'B(C+C') = AC +A'B             ; factor out (X+X')
8.
Therefore the Boolean equation has been proven correct.

AB+A'C is the dual of the left side of the original equation (A+B) (A'+C). and (A+C)(A'+B) is the dual of the right side of the original equation AC+A'B. Therefore, AB+A'C = (A+C)(A'+B) because the two expressions are duals of an Boolean equation that was proven correct.

Ratch

Last edited: May 26, 2009
8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Does this list of standard Boolean results help?

0.0 = 0
0+0 = 0

1.1 = 1
1+1 = 1

1.0 = 0
1+0 = 1

0'= 1
1'= 0

A+0 = A
A+1 = 1
A.0 = 0

A.1 = A
A+A =A
A.A = A
A+A' = 1
A.A' = 0
(A')' = A

A+A.B = A
A.(A+B) = A
A.(A'+B) = A.B
A+A'.B = A+B

A+B.C = (A+B).(A+C)

(A.B)'= A'+B'

A'.B' = (A+B)'

Last edited: May 26, 2009
9. ### just-gal Thread Starter New Member

May 25, 2009
5
0
That's really helped me soo much , thank you soo much

10. ### mango matto Member

Jun 2, 2009
13
0
holy crap! I hope that one day I will understand a single word of this thread.

11. ### Ratch New Member

Mar 20, 2007
1,068
3
mango matto,

There are no advanced concepts covered in this thread. Everything is basic logic principles covered in any good textbook or course in digital logic. This knowledge is readily available from several sources. All you have to do is read and understand. If one source does not explain it to your satisfaction, try another.

Ratch