# Boolean Algebra

#### Davemolly

Joined Apr 11, 2009
6
Hello,
I have been tring to simplify a Boolean problem and have gotten a bit stuck. So far I have:
F = '(a'.'(c'+a)+b)
F = a''+''(c'.a).b'
F = a+(c'.a).b'
Next I think it should be:
a+(c'.1).b'
I'm lost after that.
Any help would be greatly appreciated.

Cheers,

Dave.

#### Ratch

Joined Mar 20, 2007
1,068
Davemolly,

I have been tring to simplify a Boolean problem and have gotten a bit stuck. So far I have:
F = '(a'.'(c'+a)+b)
F = a''+''(c'.a).b'
F = a+(c'.a).b'
First you have to write the Boolean expression so that we can read it. What does '( mean? What does "+" mean? What does a period mean? What does " mean? Search for other questions on Boolean algebra in this forum, and note how Boolean expressions should be written.

Ratch

#### hgmjr

Joined Jan 28, 2005
9,029
For a suggested boolean expression notation method take a look at the material at this link.

hgmjr

#### Davemolly

Joined Apr 11, 2009
6
Ok , I have added an attachment that is in a better format. Hope it makes more sense.

Cheers, Dave

I've just checked and it isn't looking quite right. The question marks should be B and all of the ??/B terms should have a - NOT function.
The A should have an additional NOT and the (C+A) should also have a NOT. Apparently I suck at using computers!

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#### Wendy

Joined Mar 24, 2008
21,840
It's not exactly what you asked for, but it applies...

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#### Davemolly

Joined Apr 11, 2009
6
It's not exactly what you asked for, but it applies...

OK. I understand all of what you have said Bill, but I don't see how that helps with the simplification of the original equation? Am I being particularly dense?

#### Davemolly

Joined Apr 11, 2009
6
In case my last attempt wasn't understandable, I thought I'd try this.

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#### Ratch

Joined Mar 20, 2007
1,068
Davemolly,

In case my last attempt wasn't understandable, I thought I'd try this.
I would help you in a heartbeat if you only would write the Boolean expression without the overbars and periods. Something like this: a' + (b+c)' +abc . Then you would not have to send a scrawly attachment for someone to decipher.

Ratch

#### Davemolly

Joined Apr 11, 2009
6
Davemolly,

I would help you in a heartbeat if you only would write the Boolean expression without the overbars and periods. Something like this: a' + (b+c)' +abc . Then you would not have to send a scrawly attachment for someone to decipher.

Ratch
Right.

f = (a'(c'+a)'+b)'

That looks nothing like the way I have been shown to write Boolean Algebra. I hope it is correct?

#### studiot

Joined Nov 9, 2007
5,003
Ratch, I think you are being a tad hard on Dave.

The primed notation for negation is perfectly valid, if old.

Dave, you do need to be consistent however. The prime should appear after the negated object, not sometimes after and sometime before. I see you have achieved this in your last post. Also decide to consistently use or not use the dot.

Anyway try this.

F = {(a'.(c'+a)) +b}'

= {a'.(c'+a)}' . b'.....……....[De Morgan (R+S)' = R'.S']

= {a'' + (c'+a)'} . b'.....…....[De M (RS)' = R' + S' ]

= { a'' + c''.a' } . b'.............[De M (R+S)' = R'.S']

= {a + c.a'} . b'…………..[R'' = R]

= {a + c} . b'…………….[R+R'.S = R+S]

= a.b' + c.b'

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#### Dave

Joined Nov 17, 2003
6,960
= {a'.(c'+a)}' . b'.....……....[De Moivre (R+S)' = R'.S']
studiot,

I think you are confusing the name De Moivre with De Morgan. The principles as outlined and used in context are correct, however.

Sorry to be pedantic!

Dave

#### studiot

Joined Nov 9, 2007
5,003
I did indeed mean De Morgan, but my mind ( well what I laughingly call my mind) was obviously elsewhere.

DaveMolly you might like to look closely at my application of De Morgan.

I have split the theorem into two lesser statements, called De Morgan's Laws.

(R + S )' = R'.S'

&

(R.S)' = R' + S'

It is often convenient to use DM in this lesser form. It allows you to make one change or step at a time.

Of course you can still shorten things by using the full blown theorem, but you need to be very careful at keeping track of the negations and + to . transformations.

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#### Dave

Joined Nov 17, 2003
6,960
I did indeed mean De Morgan, but my mind ( well what I laughingly call my mind) was obviously elsewhere.
You're supposed to tell me to stop being so picky!

DaveMolly you might like to look closely at my application of De Morgan.

I have split the theorem into two lesser statements, called De Morgan's Laws.

(R + S )' = R'.S'

&

(R.S)' = R' + S'

It is often convenient to use DM in this lesser form. It allows you to make one change or step at a time.

Of course you can still shorten things by using the full blown theorem, but you need to be very careful at keeping track of the negations and + to . transformations.
These two equations are the most useful forms of De Morgan's theorem for use in simplification of Boolean Equations. Something to note, which will help immensely when using these two forms of DM, is that you can double-NOT an expression without changing that expression.

For example, take the following:

A.B

Double-NOT the expression (liberal use of brackets to separate out the two NOTs):

((A.B)')'

You can now use the second DM expression studiot stated above:

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

Therefore:

(A' + B')'

Which by another DM cycle will actually get you back to the original A.B.

Ok, so this is a trivial example that actually renders a result that requires more gates, however it illustrates the point. You will find double-NOTing very useful in more complex expressions where you may not have a negation present in the current expression.

Dave

#### Ratch

Joined Mar 20, 2007
1,068
studiot,

Ratch, I think you are being a tad hard on Dave.

The primed notation for negation is perfectly valid, if old.
The primed notation is valid if it is after the right parenthesis. The dot notation for Boolian multiplication is redundant, and clutters the expression.

I ran Dave's problem and your solution through a program I wrote, and knew immediately that you made a mistake. More about that later.

Here is the way I would solve it without my program.

f = (a'(c'+a)'+b)'
= (a'(a'c)+b)'
= (a'c+b)'
= (a+c')b'
= ab'+b'c'

It has been my experience that is it simpler and faster to work from the inside out rather than the outside in. Dave's problem has a set of minterms of 0,4-5 , as does my solution. Your solution has a set of minterms of 1,4-5 , so I knew it was wrong.

I will be posting my program shortly after I finish it up. It spits out the minterms of a Boolean expression so that it can be K-mapped and simplified. I am sure it will help a lot of folks. Stay tuned.

Ratch

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#### studiot

Joined Nov 9, 2007
5,003
Well spotted Ratch.

I looked again at the original and see that I have missed a prime when copying, so my working is correct, but applied to the wrong expression.

I did agree with you that Davemolly did not correctly and consistently apply his prime or dot notation.

#### Davemolly

Joined Apr 11, 2009
6
Thank you very much Studiot. It's nice to see that someone will actually help. It is very much appreciated. I only stopped using the dots because that was one of Ratch's gripes.

#### Ratch

Joined Mar 20, 2007
1,068
Davemolly,

I only stopped using the dots because that was one of Ratch's gripes.
Thank you, I appreciate not looking through clutter. Just ask yourself, what value added do the dots give following equivalent expression? (a+b).(c+d) <===> (a+b)(c+d)

Do folks write fractional amounts like this, 4+1/2 instead of 4½ ? What would you prefer to see?

Ratch