# Simplification of a boolean algebraic expression

#### Blue_turnip

Joined Mar 5, 2007
8
Hello members of the All About Circuits forum!

I am quite confused with boolean algebra and I have to simplify this:

SF + PB + DM + DB = 1

Would the simplification process be:

SF + PB + DM + DB = 1
SF + PB + D(M + B) = 1
SF + PB + B(M + D) = 1
SF + B(P +M + D) = 1

is this correct? If not, could you explain where I've gone wrong? thanks,

Ollie

#### Papabravo

Joined Feb 24, 2006
19,263
Step 3 is highly questionable. I don't think the rules of algebra allow you to do that. Are you sure that a simplification is even possible? What you have is a boolean equation, not a boolean expression. Usually we simplify expressions, not equations.

#### Blue_turnip

Joined Mar 5, 2007
8
I am sure simplification is possible.

If you're fussy over it being an equation and not an expression then lets take that '1' out:

SF + PB + DM + DB
SF + PB + D(M + B)
SF + PB + B(M + D)
SF + B(P + (M + D))
SF + B(P + M + D)

There, its an expression now. I also added another step to make it simpler.

While I thank you for your help, the fact that it was highly questionable was the reason I made the post. I need a definitive answer. Yes or no.

appreciated,

Ollie

#### Dave

Joined Nov 17, 2003
6,969
SF + PB + D(M + B)
SF + PB + B(M + D)
I think you are confusing this step with the Associative Property of Addition which states that:

D + (M + B) = (D + M) + B

The correct evaluation of you above fragment is dictated by the Distributive Property which states:

D(M + B) = DM + DB

Dave

#### Blue_turnip

Joined Mar 5, 2007
8
So does that mean i can't simplify it further than:

SF + PB + D(M + B)

?

Thanks,

Ollie

#### Papabravo

Joined Feb 24, 2006
19,263
I am sure simplification is possible.

If you're fussy over it being an equation and not an expression then lets take that '1' out:

SF + PB + DM + DB
SF + PB + D(M + B)
SF + PB + B(M + D)
SF + B(P + (M + D))
SF + B(P + M + D)

There, its an expression now. I also added another step to make it simpler.

While I thank you for your help, the fact that it was highly questionable was the reason I made the post. I need a definitive answer. Yes or no.

appreciated,

Ollie
Any simplification that you arrived at is meaningless when you perfrom a questionable step such as step three in your original post. Since your tone is tending toward rudeness I'll come right out and say that step three was wrong and I see no further simplification. That doesn't mean ther isn't one it's just that you won't be able to find them until you correctly apply the rules. Making this stuff up as you go along won't get you a correct result. I was trying to be polite, but I see that did absolutely no good whatsoever.

#### Dave

Joined Nov 17, 2003
6,969
Ok guys lets calm it down and get back on track. It doesn't matter whether its an equation, expression, or just some dumb algebra.

So does that mean i can't simplify it further than:

SF + PB + D(M + B)

?

Thanks,

Ollie
Not that I can see. You could go for the alternative simplification of:

SF + PB + DM + DB
SF + B(P + D) + DM

Its as long as it is round.

Dave

#### Blue_turnip

Joined Mar 5, 2007
8
Thanks alot for your help guys.

Sorry if I came across as rude, Papabravo. I have no parents and was raised on the street - I never went to school. I have recently been accepted into a 2 month electronics course so that I may look good when I apply to work at my local bulk electronics parts store. While I try my best, my language skills aren't top notch, and when i focus really hard on hitting the right buttons on the keyboard, I compromise my ability to express my thoughts perfectly. I did my best to be polite by mentioning my appreciation of your help twice and I'm sorry if i failed. I asked for a definitive yes no answer because I am very much a logical yes no person. Thats why I am seeking a career in this sort of field.

Thanks again,

Oliver