prove A+BC =(A+B)(A+C)

>>  prove A+BC =(A+B)(A+C)

It's not - not for general A, B, C and D anyway.

Umm, it doesn't - not for all values of A, B, and C.

if A = 7, B=3 and C=12

A+BC = 7+(3*12) = 43

(A+B)(A+C) = (7+3)*(7+12) = (10)*(84) = 840.

prove A+BC =(A+B)(A+C)    As has been mentioned - You can not.
However it is true IF
A + B + C = 1

????????????????

>> ????????????????

Are you sure you asked the question correctly ?

There are only certain values of A, B and C that allow A+BC = (A+B)(A+C). It's not true for all values.

hmm.. this question asked me for interview!!

Are you sure they weren't asking

Prove A(B+C) = (A+B)(A+C)  ?

You should have asked your interviewer whether in his/her system distributive rule applies from + towards *? It is obviously a trick question.

>> Prove A(B+C) = (A+B)(A+C)  ?

That's not true either ;)

I used to know all this stuff... ;-)

it asked me for Adobe ..interview :-)

If + means or, and concatenation means and
then
(A|B)&(A|C)
=
(A&B)|(A&A)|(B&A)|(B&C)
=
(A&A)|(A&B)|(B&A)|(B&C)
=
A | (A&B) | (B&C)
=
A | (B&C)

Is this arithmetic or logic???

   A or (B and C)  =  (A or B) and (A or C)  ==>  is true

A+BC =(A+B)(A+C)
--multiply right side out

A+BC = A^2 + BA + AC + BC
--subtract BC from each side

A = A^2 + ba + AC
--divide each side by A

1 = A + B + C

So, the original statement is true only where A+B+C = 1.

(That means A=1, B=1, C=1 is not true.  2 not equal to 4)

Perhaps the question was not phrased correctly?

(A|B)&(A|C)
=
(A&C)|(A&A)|(B&A)|(B&C)
=
(A&A)|(A&C)|(A&B)|(B&C)
=
A&(A|B|C) | (B&C)
=
A | (B&C)

A&(A|B|C) | (B&C)
=
A | (B&C)


There's a huge gap between your last two steps.
Everything is fine until the last line.

Effectively, you are stating that:

A = A*(A+B+C)

Which is true ONLY when (A+B+C) = 1.

So...we're back to the same spot.  The statement is not proven.

is there any slight change in question ..will make a good answer?

If (A|B|C)=0 then A=0,
A = 0 = A*(A+B+C)
If X = 0
A&(A|0) = A&A = A
If X=1
A&(A|1) = A&1 = A
so it works in either case

If we interpret + to mean OR and concatenation to mean AND

http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/#redundancelaw

hmm.. i don't have the base of these theorems in this context ..so i believe first i have to start with that :-(

A = A*(A+B+C) is true when (A+B+C) = 1 or when A=0
If we are dealing with booleans, at least one of those conditions will hold.

>> is there any slight change in question ..will make a good answer?

Did you see my post ?  (http:#21488351)

+ -> |
* -> & 
seems like a more plausible transformation than
+ -> *
* -> +

>> seems like a more plausible transformation

Yes, if this is about logic rather than arithmetic.

My post was just to give an interpretation in the arithmetic domain.

jaisonshereen, if this is a logic problem, then my answer is not good.

The question is pertinent to Binary Arithmentic

->A*A = A
->A(1+B) = A

(A+B)(A+C) = A*A + A*(B+C) + B*C
=A*(1+B+C) + B*C     [1+B+C =1]
=A+BC