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# Boolean Algebra

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I’m having problems grasping the concepts of Boolean simplification. I got stuck on the first rule A + AB = A The tutorial I was reading shows and example like this: A + AB // Factor A out of both terms A(1 + B) // Applying identity A + 1 = 1 A(1) // Applying identity 1A = A I’m having problems with the factoring. I do not see how A is factored out of both terms. How is A being factored? I’m just really stumped on this right now. Someone please help me because I really need it

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Okay, I think I get this. One of the Boolean identities is A + A = A. So then A would factor out of A + AB because of the ‘A + A’B. So then A + AB would first factor down into something like AB not A( 1 + B ).

So I’m having problems seeing how A + A = A is used to factor A + AB into A( 1 + B ).

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It''s just like "regular" algebra.

Both sides are divisible by A:

(A+AB) =  A------   ---  A       A

for the left side:
A/A = 1.

AB / A = B

for the right side:

A/A = 1

which means:

(A)(1 + B) = (A)1

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Yeah I''m fairly sure that isn''t it . There is a distributive property of boolean logic.. isn''t that enough proof how?

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A + AB = A1 + AB (because A = A1)
= A(1+B) (Factorization)
= A1 (because 1+B=1)
= A

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Okay, so if both sides are divided by A. Why are you dividing both A and B by A? I’m trying to see what algorithm is used to break down the formula. Why would you choice to divide both sides by A instead of subtracting both sides from A? Is it because division has a higher precedence then addition? I though that there was now such thing as division or subtraction in Boolean algebra.

They way I see it right now is, if I divide both sides by A then I the quotient will be 1 + B; with out any As. So were is, or how are you getting the extra A? I’m sorry, but you are helping me out actually, even though it may not seam like it. Thanks though

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regular algebra:

x + xy
x(1 + y)

boolean algebra:

A + AB
A(1 + B)

the factoring works because AND is distributive over OR.

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You''re not dividing "A and B" by A.

You''re dividing both parts of the equation:

A is one part.

AB is another part.

That''s a principle of all algebra; not just the boolean variety. You have to divide all factors on both sides of the equation. A basic principle of any highschool algebra class.

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I think I got it! I don’t know why I didn’t look a things
this way before, but I am now.

A + AB = A

I had to remember "Boolean, Think binary damn you!!", so
there can only be two values: 0 + 1, TRUE + FALSE, A + B. In Boolean
Algebra, there is an identity postulate the specifies: any number multiplied by
zero is zero; the AND operator. I think this is the second multiplication
identity in Boolean Algebra.

So then I got to thinking that if A and B are
the only two values, then either A or B must equal zero, but not
both. Then this must mean that AB, in fact, is zero i.e., 1 · 0 = 0.

Now the first Boolean Algebra Addition identity states
that: Any value added with zero results in the original value i.e., 1 + 0 = 0,
and 0 + 0 = 0. So A + 0 = A

Here is how I broke it down...

A + AB =

A + (1 · 0) =

A + 0 =

A

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Oops, I was posting while you guys were But I didn’t know that AND and OR are distributive with respect to each other. The tutorial that I was reading didn’t say any thing about that. But, another document I found did. The way I figured it out makes more sense to me. But thanks for all the help that you guys have given me on this little hick-up.

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quote:
Original post by sakky
both. Then this must mean that AB , in fact, is zero i.e., 1 · 0 = 0 .

Not exactly, since A is or''d with A AND B, it reduces to simply A. Draw out a truth diagram and look at it.

  let X = A and B  let V = A or  X  A   B   X  V  0   0   0  0  1   0   0  1  0   1   0  0  1   1   1  1

Can you see that V = A or A and B is equal to A?

I assume + means or here, I am unfamiliar with using the + symbol in boolean algebra.

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I''m closing the thread, thought its too late. Please review the Forum FAQ. This problem seems academic, e.g., could easily be a schoolwork/homework problem. Lets try and either follow the forum guidelines or stay on the topic of game development.

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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